Tag: Physics for Class 12 Students

Class 12th Physics Chapter-1 Notes Potential due to a point charge
Derivation: Electrostatic Potential Due to a Point Charge

Step 1: Physical situation

Consider a point charge $Q$ placed at the origin O.
We want to find the electrostatic potential $V$ at a point P, which is at a distance $r$ from the charge.

By definition, electrostatic potential at a point is the work done per unit positive test charge in bringing it from infinity to that point, without acceleration.

Electric force on a unit test charge

For a point charge $Q$ , electric field at distance $r^{'}$ is

$E = \frac{1}{4 π ε_{0}} \frac{Q}{r^{' 2}}$

Since the test charge is unit positive charge, the force is

$F = E = \frac{1}{4 π ε_{0}} \frac{Q}{r^{' 2}}$

Step 2: Small work done (why the minus sign appears)

Work done by external force for a small displacement $d r^{'}$ :

$d W = {\vec{F}}_{e x t} \cdot d \vec{r}$

The electric force pushes the charge away from Q, but we move it towards Q, so

$d W = - F d r^{'}$

$d W = - \frac{1}{4 π ε_{0}} \frac{Q}{r^{' 2}} d r^{'}$

This minus sign is very important conceptually.

Step 3: Setting up the definite integral

We bring the charge:
- from infinity → where potential is zero
- to distance $r$
So,

$W = \int_{\infty}^{r} d W$

$W = - \frac{Q}{4 π ε_{0}} \int_{\infty}^{r} \frac{d r^{'}}{r^{' 2}}$

Step 4: Actual integration (expanded)

Recall basic calculus:

$\int r^{' - 2} d r^{'} = - r^{' - 1}$

So,

$W = - \frac{Q}{4 π ε_{0}} {[- \frac{1}{r^{'}}]}_{\infty}^{r}$

Minus × minus becomes plus:

$W = \frac{Q}{4 π ε_{0}} [\frac{1}{r} - \frac{1}{\infty}]$

Since $\frac{1}{\infty} = 0$ ,

$W = \frac{1}{4 π ε_{0}} \frac{Q}{r}$

This is the total work done in bringing one unit charge from infinity to distance $r$

2. Connecting Work Done and Potential (Key Concept)

Definition of Potential

$V = \frac{W}{q}$

Here:
- $W$ = work done
- $q$ = test charge
Since we used a unit charge ( $q = 1$ ),

$V = W$

So,

$V (r) = \frac{1}{4 π ε_{0}} \frac{Q}{r}$

👉 Same mathematical expression, but different physical meaning.

3. Why Work and Potential Have Different Units

This is where students usually mix things up.

(a) Unit of Work Done

$W = Force \times distance$

$\Rightarrow unit of W = newton \times metre$

$[W] = joule (J)$

(b) Unit of Potential

From definition,

$V = \frac{W}{q}$

$\Rightarrow [V] = \frac{joule}{coulomb}$

$[V] = volt (V)$
January 24, 2026
Class 12th Physics Chapter-2 Notes on Electrostatic Potential
ELECTROSTATIC POTENTIAL

(Study Material & Notes )

1. Why do we need Electrostatic Potential?
- Electrostatic force is a conservative force.
- For conservative forces, work done depends only on initial and final positions, not on the path.
- Hence, we can define:
  - Electrostatic Potential Energy
  - Electrostatic Potential
This is similar to gravitational potential energy and gravitational potential.

2. Electrostatic Potential Energy

Definition

Electrostatic potential energy of a charge is the work done by an external force in bringing the charge from infinity to a given point in an electric field without acceleration.

Important Points
- Test charge must be very small (so it does not disturb the field).
- External force must be equal and opposite to electric force.
- Work done is stored as potential energy.
Mathematical Expression

$Δ U = - W_{electric}$
- Only change in potential energy is physically meaningful.
- Absolute value depends on the chosen reference point.
3. Reference Point for Potential Energy
- Potential energy is defined up to an additive constant.
- Convenient choice:
  $U = 0 at infinity$
This choice is universally used in electrostatics.

4. Electrostatic Potential

Definition (Most Important)

Electrostatic potential at a point is the work done by an external force in bringing a unit positive charge from infinity to that point without acceleration.

Symbol
- Electrostatic potential → V
Mathematical Definition

$V = \frac{U}{q}$

or,

$V_{P} - V_{R} = \frac{W_{R P}}{q}$

5. Physical Meaning of Electrostatic Potential
- It represents the potential energy per unit charge.
- It tells us how much work is required to bring a charge to a point.
- Higher potential → more work needed for a positive charge.
6. Unit of Electrostatic Potential

SI Unit: Volt (V)

$1 Volt = 1 Joule per Coulomb$

$1 V = 1 \frac{J}{C}$

7. Potential Difference

Definition

Potential difference between two points is the work done per unit charge in moving a test charge from one point to another.

Expression

$V_{P} - V_{R} = \frac{W_{R P}}{q}$

Important Notes
- Only potential difference is measurable.
- Absolute potential has no physical meaning unless a reference is chosen.
8. Potential at Infinity
- By convention:
  $V (\infty) = 0$
- Then,
$V = Work done in bringing unit charge from infinity$

9. Electrostatic Potential Due to a Point Charge

Consider a point charge Q at the origin.

Expression

$V (r) = \frac{1}{4 π ε_{0}} \frac{Q}{r}$

where:
- $r$ = distance from the charge
- $ε_{0}$ = permittivity of free space
Nature of Potential
- If $Q > 0$ → $V > 0$
- If $Q < 0$ → $V < 0$
Key Difference from Electric Field

Quantity Depends on

Electric Field $\frac{1}{r^{2}}$

Potential $\frac{1}{r}$

10. Potential Due to a System of Charges

Principle Used

👉 Superposition Principle

Expression

For charges $q_{1}, q_{2}, q_{3}, . . .$

$V = \frac{1}{4 π ε_{0}} (\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}} + \frac{q_{3}}{r_{3}} + \dots)$
- Potential is a scalar, so algebraic sum is taken.
- Easier to calculate than electric field.
11. Potential Due to a Dipole (Short Note – Pre-Equipotential)

For a dipole with moment p:

$V = \frac{1}{4 π ε_{0}} \frac{p \cos θ}{r^{2}}$

Special Cases
- Axial line: Potential is maximum
- Equatorial plane: Potential = 0
12. Relation Between Potential and Work

$W = q Δ V$
- Positive charge moves naturally from higher to lower potential.
- Negative charge moves from lower to higher potential.
13. Important Exam Points (Very High Yield)
- Electrostatic potential is a scalar quantity
- Defined only for conservative fields
- Depends on position, not path
- Potential can be positive, negative, or zero
- Easier to calculate than electric field
This material is directly aligned with NCERT Class 12 Physics, suitable for CBSE Board, NEET, and JEE (conceptual) preparation.
January 22, 2026
Class 12th Physics Chapter-1 Notes on Gauss’s Law
Gauss’s Law

1. Charge Density (Starting Point)

In real situations, charge is spread over a body, not concentrated at a single point. To describe this distribution, we define charge density.

(a) Linear Charge Density (λ)
- Charge distributed along a line (e.g., wire)
- Definition:
$λ = \frac{d q}{d l}$
- Unit: C m⁻¹
(b) Surface Charge Density (σ)
- Charge spread over a surface (e.g., spherical shell)
- Definition:
$σ = \frac{d q}{d A}$
- Unit: C m⁻²
(c) Volume Charge Density (ρ)
- Charge distributed throughout a volume
- Definition:
$ρ = \frac{d q}{d V}$
- Unit: C m⁻³
From volume charge density,

$d q = ρ d V$

This idea of distributed charge is essential for Gauss’s Law.

2. Electric Flux (Key Idea Behind Gauss’s Law)

Electric flux measures how much electric field passes through a surface.

For a small area element dA:

$d Φ = \vec{E} \cdot d \vec{A}$

For a complete surface:

$Φ = \oint \vec{E} \cdot d \vec{A}$
- If field lines pass outward, flux is positive
- If field lines pass inward, flux is negative
3. Statement of Gauss’s Law

The total electric flux through any closed surface is equal to
$\frac{1}{ε_{0}}$
times the total charge enclosed by the surface.

Mathematically:

$\oint \vec{E} \cdot d \vec{A} = \frac{Q_{enclosed}}{ε_{0}}$

where

$ε_{0}$ = permittivity of free space

$Q_{enclosed}$ = total charge inside the closed surface

This result follows directly from the inverse-square nature of Coulomb’s law and the concept of electric field and flux (see NCERT discussion in Chapter Electric Charges and Fields

4. Gauss’s Law in Terms of Charge Density

If charge is distributed continuously:

$Q_{enclosed} = \int ρ d V$

So Gauss’s Law becomes:

$\oint \vec{E} \cdot d \vec{A} = \frac{1}{ε_{0}} \int ρ d V$

This form is very important for theoretical understanding.

5. Why Gauss’s Law Is Powerful

Gauss’s Law is useful only when symmetry is high, such as:
- Spherical symmetry
- Cylindrical symmetry
- Planar symmetry
In such cases, E is constant over the surface, making calculations easy.

6. Applications of Gauss’s Law

(A) Electric Field Due to an Infinitely Long Straight Charged Wire

Charge density: λ
Gaussian surface: cylindrical

By symmetry:
- E is radial
- Same magnitude everywhere on curved surface
Flux:

$Φ = E (2 π r l)$

Charge enclosed:

$Q = λ l$

Using Gauss’s Law:

$E (2 π r l) = \frac{λ l}{ε_{0}}$

$E = \frac{λ}{2 π ε_{0} r}$

(B) Electric Field Due to an Infinite Plane Sheet of Charge

Surface charge density: σ
Gaussian surface: pillbox

Flux:

$Φ = 2 E A$

Charge enclosed:

$Q = σ A$

Applying Gauss’s Law:

$2 E A = \frac{σ A}{ε_{0}}$

$E = \frac{σ}{2 ε_{0}}$

Important result:
Electric field is independent of distance.

(C) Electric Field Due to a Uniformly Charged Spherical Shell

(i) Outside the shell (r > R):

$E = \frac{1}{4 π ε_{0}} \frac{Q}{r^{2}}$

Behaves like a point charge at the centre.

(ii) Inside the shell (r < R):

$E = 0$

7. Key Takeaways (Exam-Friendly)
- Gauss’s Law connects electric field and charge directly
- Works best for highly symmetric charge distributions
- Flux depends only on enclosed charge, not on shape
- Charges outside the surface do not affect net flux
January 20, 2026
Class 12th Physics Chapter-6 Solutions
Electromagnetic Induction – NCERT

Question 6.1

Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f).

Basic Principle Used (Very Important)

Lenz’s Law:

The induced current is always in such a direction that it opposes the change in magnetic flux producing it.

So, for every diagram, we follow three steps:
1. Decide whether magnetic flux is increasing or decreasing
2. Decide how the loop will oppose this change
3. Use right-hand thumb rule to get current direction
Fig. 6.15 (a): North pole of magnet moving towards the coil

Analysis
- Magnetic flux increases through the coil.
- Coil must oppose increase in flux.
- So, the face of the coil towards the magnet becomes North pole.
Direction of induced current

➡️ Anticlockwise (as seen from the magnet side)

Fig. 6.15 (b): North pole of magnet moving away from the coil

Analysis
- Magnetic flux decreases.
- Coil tries to oppose decrease in flux.
- Coil face becomes South pole to attract the retreating magnet.
Direction of induced current

➡️ Clockwise (as seen from the magnet side)

Fig. 6.15 (c): South pole of magnet moving towards the coil

Analysis
- Magnetic flux increases.
- Coil must oppose this increase.
- Coil face becomes South pole (repels approaching south pole).
Direction of induced current

➡️ Clockwise (as seen from the magnet side)

Fig. 6.15 (d): South pole of magnet moving away from the coil

Analysis
- Magnetic flux decreases.
- Coil tries to maintain flux.
- Coil face becomes North pole to attract the receding south pole.
Direction of induced current

➡️ Anticlockwise (as seen from the magnet side)

Fig. 6.15 (e): Coil moving towards a stationary magnet

Analysis
- Relative motion causes increase in magnetic flux.
- Situation is equivalent to magnet moving towards coil.
- Coil must oppose increase in flux.
Direction of induced current

➡️ Same as Fig. 6.15(a)
➡️ Anticlockwise (from magnet side)

Fig. 6.15 (f): Coil moving away from a stationary magnet

Analysis
- Magnetic flux decreases.
- Coil tries to oppose decrease.
- Equivalent to magnet moving away from coil.
Direction of induced current

➡️ Same as Fig. 6.15(b)
➡️ Clockwise (from magnet side)

Question 6.2

Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16:

(a) A wire of irregular shape turning into a circular shape
(b) A circular loop being deformed into a narrow straight wire

(Assume a uniform magnetic field perpendicular to the plane of the loop, as shown in the figure.)

Basic Idea (Must write in exam)

According to Lenz’s law,

The induced current always flows in such a direction that it opposes the change in magnetic flux causing it.

Magnetic flux:

$Φ = B A$

So here, only area (A) is changing.

(a) Irregular wire turning into a circular loop

Analysis
- For a given length of wire, a circle encloses the maximum area.
- When the irregular loop becomes circular:
  - Area increases
  - Hence, magnetic flux increases
Application of Lenz’s law
- The induced current must oppose the increase in flux.
- Therefore, the loop produces a magnetic field opposite to the original field.
Direction of induced current
- If the external magnetic field is into the plane of the paper (×):
  - Induced magnetic field must be out of the plane
  - By right-hand thumb rule → anticlockwise current
Answer (a): Induced current is anticlockwise

(b) Circular loop deformed into a narrow straight wire

Analysis
- Deforming a circular loop into a straight wire:
  - Area decreases
  - Hence, magnetic flux decreases
Application of Lenz’s law
- The induced current must oppose the decrease in flux.
- So, it tries to maintain the original magnetic field direction.
Direction of induced current
- If the original magnetic field is into the plane of the paper:
  - Induced field must also be into the plane
  - By right-hand thumb rule → clockwise current
Answer (b): Induced current is clockwise

Question 6.3

A long solenoid with 15 turns per cm has a small loop of area
2.0 cm² placed inside the solenoid normal to its axis.
If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, find the induced emf in the loop while the current is changing.

Given Data
- Turns per unit length of solenoid:
  $n = 15 turns/cm = 1500 turns/m$
- Area of small loop:
  $A = 2.0 {cm}^{2} = 2.0 \times 10^{- 4} m^{2}$
- Initial current:
  $I_{1} = 2.0 A$
- Final current:
  $I_{2} = 4.0 A$
- Time interval:
  $Δ t = 0.1 s$
- Permeability of free space:
  $μ_{0} = 4 π \times 10^{- 7} {H m}^{- 1}$
Step 1: Magnetic field inside the solenoid

For a long solenoid,

$B = μ_{0} n I$

Since current is changing,

$\frac{d B}{d t} = μ_{0} n \frac{d I}{d t}$

Step 2: Rate of change of current

$\frac{d I}{d t} = \frac{I_{2} - I_{1}}{Δ t} = \frac{4.0 - 2.0}{0.1} = 20 {A s}^{- 1}$

Step 3: Induced emf in the loop

Magnetic flux through the loop:

$Φ = B A$

Induced emf:

$ε = ∣ \frac{d Φ}{d t} ∣ = A \frac{d B}{d t}$

Substitute values: $ε = A μ_{0} n \frac{d I}{d t}$

$ε = (2.0 \times 10^{- 4}) (4 π \times 10^{- 7}) (1500) (20)$

Step 4: Calculation

$ε \approx 7.5 \times 10^{- 6} V$

Final Answer

$ε \approx 7.5 μ V$

Question 6.4

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T, directed normal to the loop. The velocity of the loop is 1 cm s⁻¹ in a direction normal to

(a) the longer side
(b) the shorter side

Find
(i) the emf developed across the cut, and
(ii) for how long the induced voltage lasts in each case.

Given Data
- Magnetic field:
  $B = 0.3 T$
- Velocity of loop:
  $v = 1 {cm s}^{- 1} = 1 \times 10^{- 2} {m s}^{- 1}$
- Length of longer side:
  $l_{1} = 8 cm = 8 \times 10^{- 2} m$
- Length of shorter side:
  $l_{2} = 2 cm = 2 \times 10^{- 2} m$
Basic Formula Used

When a loop moves out of a magnetic field,

$ε = B l v$

(where $l$ is the length of the side cutting magnetic field lines)

(a) Velocity normal to the longer side

Induced emf

Here, the longer side (8 cm) cuts the magnetic field.

$ε = B l_{1} v$

$ε = 0.3 \times (8 \times 10^{- 2}) \times (1 \times 10^{- 2})$

$ε = 2.4 \times 10^{- 4} V$

Time for which emf lasts

The emf exists while the loop is leaving the field region.

Distance to be covered = shorter side = 2 cm

$t = \frac{2 cm}{1 {cm s}^{- 1}} = 2 s$

(b) Velocity normal to the shorter side

Induced emf

Now, the shorter side (2 cm) cuts the magnetic field.

$ε = B l_{2} v$

$ε = 0.3 \times (2 \times 10^{- 2}) \times (1 \times 10^{- 2})$

$ε = 6.0 \times 10^{- 5} V$

Time for which emf lasts

Distance to be covered = longer side = 8 cm

$t = \frac{8 cm}{1 {cm s}^{- 1}} = 8 s$

Question 6.5

A 1.0 m long metallic rod is rotated with an angular frequency of
400 rad s⁻¹ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A uniform magnetic field of 0.5 T, parallel to the axis, exists everywhere. Calculate the emf developed between the centre and the ring.

Given Data
- Length of rod:
  $L = 1.0 m$
- Angular frequency:
  $ω = 400 {rad s}^{- 1}$
- Magnetic field:
  $B = 0.5 T$
Concept Used: Motional emf in a rotating rod

For a rod rotating in a uniform magnetic field, the emf developed between the centre and the rim is:

$ε = \frac{1}{2} B ω L^{2}$

(This result comes from integrating the motional emf $d ε = B v d r$ with $v = ω r$ .)

Calculation

$ε = \frac{1}{2} \times 0.5 \times 400 \times (1.0)^{2}$

$ε = \frac{1}{2} \times 200$

$ε = 100 V$

Question 6.6

A horizontal straight wire of length 10 m, extending from east to west, is falling with a speed of 5.0 m s⁻¹, at right angles to the horizontal component of the Earth’s magnetic field,

$B_{H} = 0.30 \times 10^{- 4} {Wb m}^{- 2}$

Answer the following:

(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?

Given Data
- Length of wire:
  $l = 10 m$
- Speed of falling wire:
  $v = 5.0 {m s}^{- 1}$
- Horizontal magnetic field:
  $B = 0.30 \times 10^{- 4} T$
(a) Instantaneous value of induced emf

For a straight conductor moving perpendicular to a magnetic field,

$ε = B l v$

Substitute values:

$ε = (0.30 \times 10^{- 4}) \times 10 \times 5$

$ε = 1.5 \times 10^{- 3} V$

(b) Direction of the induced emf
- Magnetic field is towards north (horizontal component of Earth’s field).
- Wire is falling vertically downward.
- Use Fleming’s right-hand rule (or Lorentz force rule):
  - First finger → Magnetic field (North)
  - Thumb → Velocity (Downward)
  - Middle finger → Induced emf/current
The induced emf is directed from West to East.

(c) Which end is at higher potential?

Since emf is directed from West to East:
- East end is at higher potential
- West end is at lower potential
Question 6.7

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s.
If an average induced emf of 200 V is produced, estimate the self-inductance of the circuit.

Given Data

Initial current: $I_{1} = 5.0 A$

Final current: $I_{2} = 0.0 A$

Time interval: $Δ t = 0.1 s$

Average induced emf: $ε = 200 V$

Formula Used

For self-induction,

$ε = L ∣ \frac{Δ I}{Δ t} ∣$

Calculation

$\frac{Δ I}{Δ t} = \frac{5.0 - 0.0}{0.1} = 50 {A s}^{- 1}$

$L = \frac{ε}{∣ Δ I / Δ t ∣} = \frac{200}{50}$

$L = 4.0 H$

Question 6.8

A pair of adjacent coils has a mutual inductance of 1.5 H.
If the current in one coil changes from 0 to 20 A in 0.5 s, find the change of flux linkage with the other coil.

Given Data
- Mutual inductance:
  $M = 1.5 H$
- Initial current:
  $I_{1} = 0 A$
- Final current:
  $I_{2} = 20 A$
Formula Used

Change in flux linkage is given by:

$Δ (N Φ) = M Δ I$

Calculation

$Δ I = I_{2} - I_{1} = 20 - 0 = 20 A$ $Δ (N Φ) = 1.5 \times 20$ $Δ (N Φ) = 30 Wb-turn$
January 20, 2026
Class 12th Physics Chapter-1 Notes on Dipole, Dipole Moment, Electric Field due to a dipole
Electric Dipole, Dipole Moment & Electric Field Due to a Dipole

(With FULL DERIVATIONS – Class 12 NCERT, One-Go Complete Notes)

4

1️⃣ Electric Dipole (Revision)

An electric dipole consists of:
- Two equal and opposite charges $+ q$ and $- q$
- Separated by a small distance $2 a$
📌 Condition:

$a ≪ r (distance of observation point from dipole)$

2️⃣ Electric Dipole Moment (𝒑)

Definition:

Electric dipole moment is the product of one charge and separation vector.

$\vec{p} = q \cdot 2 \vec{a}$
- SI unit: C m
- Vector quantity
- Direction: from –q to +q
ELECTRIC FIELD DUE TO AN ELECTRIC DIPOLE (DERIVATIONS)

🔵 Case I: Electric Field on the Axial Line

🔹 Geometry:
- Point $P$ lies on the axis of dipole
- Distance of point from centre $O$ = $r$
- Distance from +q = $r - a$
- Distance from –q = $r + a$
🔹 Step 1: Electric field due to +q at point P

$E_{+} = \frac{1}{4 π ε_{0}} \frac{q}{(r - a)^{2}}$

(Direction: away from +q)

🔹 Step 2: Electric field due to –q at point P

$E_{-} = \frac{1}{4 π ε_{0}} \frac{q}{(r + a)^{2}}$

(Direction: towards –q)

🔹 Step 3: Net electric field

Both fields act along the axis, but in opposite directions.

$E = E_{+} - E_{-}$

$E = \frac{1}{4 π ε_{0}} [\frac{q}{(r - a)^{2}} - \frac{q}{(r + a)^{2}}]$

🔹 Step 4: Simplification

$E = \frac{1}{4 π ε_{0}} \frac{4 q a r}{(r^{2} - a^{2})^{2}}$

For short dipole $(r ≫ a)$ :

$(r^{2} - a^{2})^{2} \approx r^{4}$

✅ Final Result (Axial Line):

$E_{axial} = \frac{1}{4 π ε_{0}} \frac{2 p}{r^{3}}$

🔹 Direction:

👉 Along the dipole moment

🟢 Case II: Electric Field on the Equatorial Line

🔹 Geometry:
- Point $P$ lies on the perpendicular bisector
- Distance of point from centre = $r$
- Distance from each charge:
$\sqrt{r^{2} + a^{2}}$

🔹 Step 1: Field due to +q

$E_{+} = \frac{1}{4 π ε_{0}} \frac{q}{r^{2} + a^{2}}$

🔹 Step 2: Field due to –q

$E_{-} = \frac{1}{4 π ε_{0}} \frac{q}{r^{2} + a^{2}}$

🔹 Step 3: Resolution of Fields
- Horizontal components cancel
- Vertical components add
$E = 2 E_{+} \cos θ$

$\cos θ = \frac{a}{\sqrt{r^{2} + a^{2}}}$

🔹 Step 4: Net Field

$E = \frac{1}{4 π ε_{0}} \frac{2 q a}{(r^{2} + a^{2})^{3 / 2}}$

For short dipole $(r ≫ a)$ :

$(r^{2} + a^{2})^{3 / 2} \approx r^{3}$

✅ Final Result (Equatorial Line):

$E_{equatorial} = \frac{1}{4 π ε_{0}} \frac{p}{r^{3}}$

🔹 Direction:

👉 Opposite to dipole moment

4️⃣ Axial vs Equatorial (Very Important Table 🔥)

Property Axial Line Equatorial Line

Formula $\frac{2 p}{4 π ε_{0} r^{3}}$ $\frac{p}{4 π ε_{0} r^{3}}$

Direction Along $\vec{p}$ Opposite to $\vec{p}$

Relative Strength Stronger Weaker

Distance Dependence $1 / r^{3}$ $1 / r^{3}$

5️⃣ One-Line Exam Conclusions ⭐
- Electric field due to dipole decreases rapidly with distance.
- Axial field is twice the equatorial field.
- Dipole moment controls strength and direction.
6️⃣ Memory Shortcut 🧠
- Axial → 2p
- Equatorial → p
- Both → $1 / r^{3}$
January 18, 2026
Class 12th Physics Chapter-1 Notes on Electric Flux
Electric Flux, Area Vector & Total Electric Flux

(Class 12 Physics – NCERT )

1️⃣ Electric Flux – Meaning and Idea

Electric flux gives a measure of how much electric field passes through a given surface.

Think of electric field lines as invisible threads:
- More lines passing through a surface → greater flux
- Fewer lines → smaller flux
👉 Electric flux does not mean electric field is flowing like water.
It is only a mathematical quantity used to describe the electric field.

2️⃣ Area Vector – Why Area is a Vector

The effect of an electric field on a surface depends not only on:
- Size of the surface
  but also on
- Orientation of the surface
So, area is represented as a vector.

✔ Definition of Area Vector
- Magnitude = area of the surface
- Direction = perpendicular (normal) to the surface
$\vec{A} = A \hat{n}$

where
- $A$ = area
- $\hat{n}$ = unit vector normal to the surface
Important Convention
- For a closed surface, the area vector always points outward
3️⃣ Electric Flux Through a Plane Surface

If:
- Electric field = $\vec{E}$
- Area vector = $\vec{A}$
- Angle between $\vec{E}$ and $\vec{A}$ = $θ$
Then electric flux is defined as:

$Φ = \vec{E} \cdot \vec{A} = E A \cos θ$

This is a dot product, so orientation matters.

4️⃣ Special Cases of Electric Flux (Very Important)

🔹 Case 1: Maximum Flux
- $θ = 0^{\circ}$
- Surface is perpendicular to electric field
$Φ = E A$

➡ Maximum number of field lines pass through the surface.

🔹 Case 2: Zero Flux
- $θ = 90^{\circ}$
- Surface is parallel to electric field
$Φ = 0$

➡ Field lines do not cross the surface.

🔹 Case 3: Inclined Surface
- $0^{\circ} < θ < 90^{\circ}$
$Φ = E A \cos θ$

➡ Only the effective area $A \cos θ$ contributes.

5️⃣ Electric Flux Through a Curved Surface

For a curved or irregular surface, we divide it into many small area elements $d A$ .

Flux through a small element:

$d Φ = \vec{E} \cdot d \vec{A}$

Total electric flux:

$Φ = \int \vec{E} \cdot d \vec{A}$

This is the most general definition of electric flux.

6️⃣ Total Electric Flux Through a Closed Surface

A closed surface completely encloses a volume (sphere, cube, cylinder).

Sign Convention
- Field lines leaving the surface → positive flux
- Field lines entering the surface → negative flux
👉 Total electric flux is the algebraic sum of flux over the entire closed surface.

This concept leads directly to Gauss’s Law.

7️⃣ SI Unit of Electric Flux

$Electric flux = {N m}^{2} C^{- 1}$

8️⃣ Key Exam Points (One-Look Revision)
- Electric flux is a scalar quantity
- Depends on:
  - Electric field
  - Area
  - Orientation of surface
- Independent of:
  - Shape of surface (for uniform field)
- Area vector is always normal to the surface
- For closed surfaces, area vector is outward
9️⃣ One-Line Definitions

Electric Flux:

Electric flux through a surface is the surface integral of the normal component of electric field over that surface.

Area Vector:

Area vector of a surface has magnitude equal to the area and direction normal to the surface.
January 17, 2026
Class 12th Physics Chapter-1 Notes on Electric Fields and Field Lines
ELECTRIC FIELD

1. Why Do We Need the Concept of Electric Field?

Imagine this question:

If two charges attract or repel each other, how does one charge “know” the other is there—especially when there is empty space between them?

Early scientists struggled with this idea. The solution came from Michael Faraday, who introduced the revolutionary concept of a field.

👉 A charge does not act directly on another charge.
👉 It first creates an electric field in the surrounding space.
👉 Any other charge placed in this field experiences a force.

This idea is the foundation of electrostatics.

⚡ 2. What Is an Electric Field? (Definition)

Electric Field (E) at a point is defined as:

The force experienced by a unit positive test charge placed at that point.

Mathematically,

$\vec{E} = \frac{\vec{F}}{q}$

where
- F = force on the test charge
- q = magnitude of the test charge
📌 SI Unit: N/C (newton per coulomb)

🔹 The test charge is taken very small so it does not disturb the original field.

📐 3. Derivation of Electric Field Due to a Point Charge

Consider a point charge Q placed at the origin.

If a small test charge q is placed at distance r, the electrostatic force (by Coulomb’s law) is:

$\vec{F} = \frac{1}{4 π ε_{0}} \frac{Q q}{r^{2}} \hat{r}$

Now, electric field is force per unit charge:

$\vec{E} = \frac{\vec{F}}{q}$

Substituting:

$\vec{E} = \frac{1}{4 π ε_{0}} \frac{Q}{r^{2}} \hat{r}$

🔍 Key Observations
- Electric field depends on source charge Q, not on test charge q
- Direction is:
  - Away from Q if Q is positive
  - Towards Q if Q is negative
- Field decreases as 1/r² with distance
🔗 4. Relation Between Electric Field and Electrostatic Force

The most powerful and exam-important relation is:

$\vec{F} = q \vec{E}$

💡 Interpretation
- Electric field describes the electrical environment
- Force depends on:
  - Strength of field (E)
  - Nature and magnitude of charge (q)
🔹 For a positive charge, force is along E
🔹 For a negative charge, force is opposite to E

This relation helps us separate cause (field) from effect (force).

5. Electric Field Lines – Visualising the Invisible

Electric field is invisible, but field lines help us see it.

Definition

Electric field lines are imaginary curves drawn such that:

The tangent at any point gives the direction of the electric field at that point.

4

6. Properties of Electric Field Lines (Very Important)

(1) Start and End
- Start from positive charges
- End on negative charges
- May start or end at infinity if isolated
(2) Direction
- Direction of field = direction of force on a positive test charge
(3) Density of Lines
- Closer lines → stronger field
- Farther lines → weaker field
This explains why:

$E \propto \frac{1}{r^{2}}$

(4) Never Intersect

Field lines never cross, because:
- At a point, electric field has only one direction
(5) No Closed Loops

Electrostatic field lines do not form closed loops, because:
- Electrostatic field is conservative
⚖️ 7. Electric Field Due to Multiple Charges (Superposition)

If more than one charge is present:

${\vec{E}}_{net} = {\vec{E}}_{1} + {\vec{E}}_{2} + {\vec{E}}_{3} + \dots$

🔹 Electric field follows the principle of superposition
🔹 Add fields vectorially, not algebraically

This makes electric field more powerful than force calculations.

✨ 8. Why Electric Field Is a Brilliant Concept
- Explains action at a distance
- Works even when charges are not present
- Becomes essential in:
  - Electromagnetic waves
  - Time-varying fields
  - Modern physics
👉 That’s why field, not force, is the fundamental idea in physics.

📝 Exam-Ready Summary Box
- $\vec{E} = \vec{F} / q$
- $\vec{E} = \frac{1}{4 π ε_{0}} \frac{Q}{r^{2}} \hat{r}$
- $\vec{F} = q \vec{E}$
- Field lines show direction + strength
- Density ∝ strength of electric field
January 16, 2026
Class 12th Physics Chapter-5 Solutions
Question 5.1

A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 T$ experiences a torque of magnitude equal to
$4.5 \times 10^{- 2} J$ .
What is the magnitude of the magnetic moment of the magnet?

Solution

The torque on a magnetic dipole in a uniform magnetic field is given by:

$τ = m B \sin θ$ Given:
- $τ = 4.5 \times 10^{- 2} J$
- $B = 0.25 T$
- $θ = 30^{\circ}$ , so $\sin 30^{\circ} = \frac{1}{2}$
$m = \frac{τ}{B \sin θ} = \frac{4.5 \times 10^{- 2}}{0.25 \times \frac{1}{2}} = \frac{4.5 \times 10^{- 2}}{0.125} = 0.36$

Answer:

$m = 0.36 {J T}^{- 1} or 0.36 {A m}^{2}$

Question 5.2

A short bar magnet of magnetic moment

$m = 0.32 {J T}^{- 1}$

is placed in a uniform magnetic field

$B = 0.15 T .$

If the bar is free to rotate in the plane of the field:

(a) Which orientation corresponds to stable equilibrium?
(b) Which orientation corresponds to unstable equilibrium?
What is the potential energy of the magnet in each case?

Concept Used

The magnetic potential energy of a magnetic dipole in a uniform magnetic field is:

$U = - \vec{m} \cdot \vec{B} = - m B \cos θ$

where $θ$ is the angle between $\vec{m}$ and $\vec{B}$ .

(a) Stable Equilibrium
- Stable equilibrium occurs when potential energy is minimum
- This happens when:
  
  $θ = 0^{\circ} (\vec{m} ∥ \vec{B})$
Potential Energy:

$U_{stable} = - m B \cos 0^{\circ} = - m B$

$U_{stable} = - (0.32) (0.15) = - 0.048 J$

(b) Unstable Equilibrium
- Unstable equilibrium occurs when potential energy is maximum
- This happens when:
  
  $θ = 180^{\circ} (\vec{m} antiparallel to \vec{B})$
Potential Energy:

$U_{unstable} = - m B \cos 180^{\circ} = + m B$

$U_{unstable} = + (0.32) (0.15) = + 0.048 J$

Question 5.3

A closely wound solenoid of 800 turns and area of cross-section
$2.5 \times 10^{- 4} m^{2}$ carries a current of 3.0 A.

(a) Explain the sense in which the solenoid acts like a bar magnet.
(b) What is its associated magnetic moment?

(a) Solenoid as a Bar Magnet

A current-carrying solenoid produces a magnetic field pattern similar to that of a bar magnet:
- One end of the solenoid behaves like a north pole
- The other end behaves like a south pole
- Magnetic field lines emerge from one end and enter the other, forming closed loops
- The polarity of the solenoid is given by the right-hand thumb rule:
  - Curl fingers in the direction of current → thumb gives the direction of the north pole
Hence, a solenoid acts like a magnetic dipole, just like a bar magnet.

(b) Magnetic Moment of the Solenoid

The magnetic moment of a solenoid is:

$m = N I A$

Given:
- $N = 800$
- $I = 3.0 A$
- $A = 2.5 \times 10^{- 4} m^{2}$
Calculation:

$m = 800 \times 3.0 \times 2.5 \times 10^{- 4}$

$m = 6000 \times 10^{- 4} = 0.6 {A m}^{2}$

Question 5.4

If the solenoid in the previous exercise is free to turn about the vertical direction and a uniform horizontal magnetic field of

$B = 0.25 T$

is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30^{\circ}$ with the direction of the applied field?

Given (from Exercise 5.3)
- Magnetic moment of the solenoid:
$m = 0.6 {J T}^{- 1}$
- Magnetic field:
$B = 0.25 T$
- Angle:
$θ = 30^{\circ}, \sin 30^{\circ} = \frac{1}{2}$

Formula Used

Torque on a magnetic dipole in a uniform magnetic field:

$τ = m B \sin θ$

Calculation

$τ = 0.6 \times 0.25 \times \frac{1}{2}$

$τ = 0.075 N m$

Question 5.5

A bar magnet of magnetic moment

$m = 1.5 {J T}^{- 1}$

lies aligned with the direction of a uniform magnetic field

$B = 0.22 T .$

(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment:
(i) normal to the field direction,
(ii) opposite to the field direction?

(b) What is the torque on the magnet in cases (i) and (ii)?

Concepts Used
- Magnetic potential energy:
$U = - m B \cos θ$
- Work done by external agent:
$W = Δ U$
- Torque on a magnetic dipole:
$τ = m B \sin θ$

Given Initial Condition

Initially, the magnet is aligned with the field:

$θ_{i} = 0^{\circ}$

Initial potential energy:

$U_{i} = - m B = - (1.5) (0.22) = - 0.33 J$

(a) Work Done by External Torque

(i) Magnet turned normal to the field

$θ = 90^{\circ}, \cos 90^{\circ} = 0$

Final potential energy:

$U_{f} = 0$

Work done:

$W = U_{f} - U_{i} = 0 - (- 0.33) = 0.33 J$

(ii) Magnet turned opposite to the field

$θ = 180^{\circ}, \cos 180^{\circ} = - 1$

Final potential energy:

$U_{f} = + m B = + 0.33 J$

Work done:

$W = U_{f} - U_{i} = 0.33 - (- 0.33) = 0.66 J$

(b) Torque on the Magnet

(i) At $90^{\circ}$ :

$τ = m B \sin 90^{\circ} = (1.5) (0.22) = 0.33 N m$

(ii) At $180^{\circ}$ :

$τ = m B \sin 180^{\circ} = 0$

Final Answer (Summary)

(a) Work Done
- Normal to the field (90°):
$W = 0.33 J$
- Opposite to the field (180°):
$W = 0.66 J$

(b) Torque
- At 90°:
$τ = 0.33 N m$
- At 180°:
$τ = 0$

Question 5.6

A closely wound solenoid of 2000 turns and area of cross-section
$1.6 \times 10^{- 4} m^{2}$ , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of
$7.5 \times 10^{- 2} T$ is set up at an angle of $30^{\circ}$ with the axis of the solenoid?

(a) Magnetic moment of the solenoid

For a solenoid:

$m = N I A$

Given:
- $N = 2000$
- $I = 4.0 A$
- $A = 1.6 \times 10^{- 4} m^{2}$
Calculation:

$m = 2000 \times 4.0 \times 1.6 \times 10^{- 4}$

$m = 1.28 {A m}^{2}$

$m = 1.28 {J T}^{- 1}$

(b) Force and torque on the solenoid

Force on the solenoid

In a uniform magnetic field, the net force on a magnetic dipole is zero.

$F = 0$

Torque on the solenoid

Torque on a magnetic dipole:

$τ = m B \sin θ$

Given:
- $m = 1.28 {J T}^{- 1}$
- $B = 7.5 \times 10^{- 2} T$
- $θ = 30^{\circ}, \sin 30^{\circ} = \frac{1}{2}$
Calculation:

$τ = 1.28 \times 7.5 \times 10^{- 2} \times \frac{1}{2}$

$τ = 0.048 N m$

Final Answer (Summary)

(a) Magnetic moment of the solenoid:

$m = 1.28 {J T}^{- 1}$

(b)
- Force on the solenoid:
$F = 0$
- Torque on the solenoid:
$τ = 4.8 \times 10^{- 2} N m$

Question 5.7

A short bar magnet has a magnetic moment

$m = 0.48 {J T}^{- 1}$

Find the direction and magnitude of the magnetic field produced by the magnet at a distance of

$r = 10 cm = 0.10 m$

from the centre of the magnet on:

(a) the axial line,
(b) the equatorial line (normal bisector) of the magnet.

Given
- Magnetic moment: $m = 0.48 {J T}^{- 1}$
- Distance: $r = 0.10 m$
- Permeability of free space:
$\frac{μ_{0}}{4 π} = 10^{- 7} {T m A}^{- 1}$

(a) Magnetic field on the axis of the magnet

Formula:

$B_{axis} = \frac{μ_{0}}{4 π} \frac{2 m}{r^{3}}$

Calculation:

$B_{axis} = 10^{- 7} \times \frac{2 \times 0.48}{(0.10)^{3}}$

$B_{axis} = 10^{- 7} \times \frac{0.96}{10^{- 3}}$

$B_{axis} = 9.6 \times 10^{- 5} T$

Direction:

Along the axis of the magnet, in the same direction as the magnetic moment (from south pole to north pole outside the magnet).

(b) Magnetic field on the equatorial line (normal bisector)

Formula:

$B_{equatorial} = \frac{μ_{0}}{4 π} \frac{m}{r^{3}}$

Calculation:

$B_{equatorial} = 10^{- 7} \times \frac{0.48}{(0.10)^{3}}$

$B_{equatorial} = 4.8 \times 10^{- 5} T$

Direction:

Along the equatorial line, opposite to the direction of the magnetic moment.

Final Answer (Summary)

Position Magnitude of magnetic field Direction

Axial line $9.6 \times 10^{- 5} T$ Along magnetic moment

Equatorial line $4.8 \times 10^{- 5} T$ Opposite to magnetic moment
January 15, 2026
Class 12th Physics Chapter-1 Notes on Coulomb’s law
COULOMB’S LAW (WITH VECTOR FORM)

Class XII Physics – NCERT Based Study Material

1️⃣ What is Coulomb’s Law?

Coulomb’s law gives the quantitative expression for the electrostatic force between two stationary point charges placed in vacuum (or air).

When the size of charged bodies is much smaller than the distance between them, they are treated as point charges.

2️⃣ Statement of Coulomb’s Law

The magnitude of the electrostatic force between two point charges is
- directly proportional to the product of the magnitudes of the charges, and
- inversely proportional to the square of the distance between them.
- The force acts along the line joining the two charges.
3️⃣ Scalar Form of Coulomb’s Law

Let:
- Charges = $q_{1}, q_{2}$
- Distance between them = $r$
$F = k \frac{∣ q_{1} q_{2} ∣}{r^{2}}$

In SI units,

$F = \frac{1}{4 π ε_{0}} \frac{∣ q_{1} q_{2} ∣}{r^{2}}$

Where:
- $ε_{0} = 8.854 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$
- $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} C^{- 2}$
4️⃣ Nature of Electrostatic Force

Charges Nature of Force

Like charges (+ + or − −) Repulsive

Unlike charges (+ −) Attractive

✔️ Force always acts along the line joining the charges

5️⃣ Why Vector Form is Needed?

According to NCERT:
- Force is a vector quantity
- Direction must be specified
- Newton’s third law must be satisfied
- Required for multiple charge systems
6️⃣ Position Vectors and Relative Vectors

Let:
- Charge $q_{1}$ at position vector ${\vec{r}}_{1}$
- Charge $q_{2}$ at position vector ${\vec{r}}_{2}$
Relative position vectors:

${\vec{r}}_{21} = {\vec{r}}_{2} - {\vec{r}}_{1}$

${\vec{r}}_{12} = {\vec{r}}_{1} - {\vec{r}}_{2} = - {\vec{r}}_{21}$

Magnitude:

$∣ {\vec{r}}_{21} ∣ = ∣ {\vec{r}}_{12} ∣ = r$

7️⃣ Unit Vector (Very Important Definition)

A unit vector gives direction only:

${\hat{r}}_{21} = \frac{{\vec{r}}_{21}}{∣ {\vec{r}}_{21} ∣}$
- Points from q₁ to q₂
- Dimensionless
8️⃣ Vector Form of Coulomb’s Law (NCERT Equation)

Force on charge $q_{2}$ due to charge $q_{1}$ :

${\vec{F}}_{21} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{21}^{2}} {\hat{r}}_{21}$

9️⃣ Direction Interpretation (Built-in Advantage)
- If $q_{1} q_{2} > 0$ → force along ${\hat{r}}_{21}$ → repulsion
- If $q_{1} q_{2} < 0$ → force opposite to ${\hat{r}}_{21}$ → attraction
📌 No separate equations needed for attraction and repulsion

🔟 Newton’s Third Law from Coulomb’s Law

Force on $q_{1}$ due to $q_{2}$ :

${\vec{F}}_{12} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}^{2}} {\hat{r}}_{12}$

Since:

${\hat{r}}_{12} = - {\hat{r}}_{21}$

${\vec{F}}_{12} = - {\vec{F}}_{21}$

✔️ Coulomb’s law automatically satisfies Newton’s third law

Compact Vector Form (Advanced NCERT Form)

${\vec{F}}_{21} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{∣ {\vec{r}}_{2} - {\vec{r}}_{1} ∣^{3}} ({\vec{r}}_{2} - {\vec{r}}_{1})$

📌 Useful for:
- Superposition
- Electric field derivation
- Higher numerical
Important Assumptions of Coulomb’s Law

✔ Charges are point charges
✔ Charges are at rest
✔ Distance ≫ size of charges
✔ Medium is vacuum / air

Experimental Basis (NCERT)
- Established using torsion balance
- Valid from macroscopic scale to subatomic distances (~10⁻¹⁰ m)
One-Line Exam Definitions (Very Important)
- Coulomb’s Law:
  The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
- Unit Vector:
  A vector of unit magnitude used to specify direction.
- Permittivity of Free Space:
  A constant that characterises the electrical properties of vacuum.
Key Takeaways for Exams

✔ Force ∝ $1 / r^{2}$
✔ Vector nature essential
✔ Direction decided by sign of charges
✔ Basis of electric field and superposition principle
January 14, 2026
Class 12th Physics Chapter-4 Solutions
Question 4.1

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field $B$ at the centre of the coil?

Given
- Number of turns, $N = 100$
- Radius of coil, $R = 8.0 cm = 0.08 m$
- Current, $I = 0.40 A$
- Permeability of free space,
  $μ_{0} = 4 π \times 10^{- 7} {T m A}^{- 1}$
Formula Used

For a circular coil with $N$ turns, the magnetic field at the centre is:

$B = \frac{μ_{0} N I}{2 R}$

(This result follows from the Biot–Savart law for a circular current loop, as discussed in the NCERT text)

Substitution

$B = \frac{(4 π \times 10^{- 7}) \times 100 \times 0.40}{2 \times 0.08}$

$B = \frac{160 π \times 10^{- 7}}{0.16}$

$B = 1000 π \times 10^{- 7} T$

$B = π \times 10^{- 4} T$

Final Answer

$B \approx 3.14 \times 10^{- 4} T$

Question 4.2

A long straight wire carries a current of 35 A. What is the magnitude of the magnetic field $B$ at a point 20 cm from the wire?

Given
- Current in the wire, $I = 35 A$
- Distance from the wire,
  $r = 20 cm = 0.20 m$
- Permeability of free space,
  $μ_{0} = 4 π \times 10^{- 7} {T m A}^{- 1}$
Formula Used

For a long straight current-carrying wire, the magnetic field at a distance $r$ is:

$B = \frac{μ_{0} I}{2 π r}$

(This expression follows from Ampere’s circuital law / Biot–Savart law as given in NCERT Physics, Moving Charges and Magnetism)

Substitution

$B = \frac{4 π \times 10^{- 7} \times 35}{2 π \times 0.20}$

Cancel $π$ :

$B = \frac{4 \times 35 \times 10^{- 7}}{0.40}$

$B = \frac{140 \times 10^{- 7}}{0.40}$

$B = 350 \times 10^{- 7} T$

$B = 3.5 \times 10^{- 5} T$

Question 4.3

A long straight wire in the horizontal plane carries a current of 50 A in the north-to-south direction. Give the magnitude and direction of the magnetic field $B$ at a point 2.5 m east of the wire.

Given
- Current, $I = 50 A$
- Distance from the wire, $r = 2.5 m$
- Direction of current: North → South
- Permeability of free space:
  $μ_{0} = 4 π \times 10^{- 7} {T m A}^{- 1}$
Magnitude of Magnetic Field

For a long straight current-carrying wire,

$B = \frac{μ_{0} I}{2 π r}$

(Substantiated by Ampere’s circuital law / Biot–Savart law as discussed in NCERT Physics, Moving Charges and Magnetism)

Substitution

$B = \frac{4 π \times 10^{- 7} \times 50}{2 π \times 2.5}$ Cancel $π$ :

$B = \frac{200 \times 10^{- 7}}{5}$

$B = 40 \times 10^{- 7} T$

$B = 4.0 \times 10^{- 6} T$

Direction of Magnetic Field

Use the right-hand thumb rule:
- Point the right-hand thumb in the direction of current (north to south).
- The curl of the fingers gives the direction of magnetic field lines.
At a point east of the wire, the curled fingers point vertically upward, i.e., out of the horizontal plane.

Question 4.4

A horizontal overhead power line carries a current of 90 A in the east-to-west direction. What is the magnitude and direction of the magnetic field due to the current at a point 1.5 m below the line?

Given
- Current, $I = 90 A$
- Distance from the wire, $r = 1.5 m$
- Direction of current: East → West
- Permeability of free space:
  $μ_{0} = 4 π \times 10^{- 7} {T m A}^{- 1}$
Magnitude of Magnetic Field

For a long straight current-carrying wire,

$B = \frac{μ_{0} I}{2 π r}$

(as derived using Ampere’s circuital law / Biot–Savart law in NCERT Physics, Moving Charges and Magnetism )

Substitution

$B = \frac{4 π \times 10^{- 7} \times 90}{2 π \times 1.5}$

Cancel $π$ :

$B = \frac{360 \times 10^{- 7}}{3}$

$B = 120 \times 10^{- 7} T$

$B = 1.2 \times 10^{- 5} T$

Direction of Magnetic Field

Apply the right-hand thumb rule:
- Thumb → direction of current (east to west)
- Curled fingers → direction of magnetic field lines around the wire
At a point below the wire, the magnetic field is directed towards the south.

Question 4.5

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of $30^{\circ}$ with the direction of a uniform magnetic field of 0.15 T?

Given
- Current in the wire, $I = 8 A$
- Magnetic field, $B = 0.15 T$
- Angle between current and magnetic field,
  $θ = 30^{\circ}$
Formula Used

The magnetic force on a current-carrying conductor of length $l$ in a uniform magnetic field is:

$F = I l B \sin θ$

Therefore, force per unit length is:

$\frac{F}{l} = I B \sin θ$

(This relation follows from the Lorentz force law as discussed in NCERT Physics, Moving Charges and Magnetism)

Substitution

$\frac{F}{l} = 8 \times 0.15 \times \sin 30^{\circ}$

$\sin 30^{\circ} = \frac{1}{2}$

$\frac{F}{l} = 8 \times 0.15 \times 0.5$

$\frac{F}{l} = 0.6 {N m}^{- 1}$

Question 4.6

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is 0.27 T. What is the magnetic force on the wire?

Given
- Current in the wire, $I = 10 A$
- Length of wire,
  $l = 3.0 cm = 0.03 m$
- Magnetic field inside solenoid,
  $B = 0.27 T$
- Angle between current and magnetic field,
  $θ = 90^{\circ} (wire is perpendicular to solenoid axis)$
Formula Used

Magnetic force on a current-carrying conductor in a uniform magnetic field:

$F = I l B \sin θ$

(This follows from the Lorentz force law for a current-carrying conductor, as discussed in NCERT Physics, Moving Charges and Magnetism )

Substitution

$F = 10 \times 0.03 \times 0.27 \times \sin 90^{\circ}$

$\sin 90^{\circ} = 1$

$F = 10 \times 0.03 \times 0.27$

$F = 0.081 N$

Question 4.7

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Given
- Current in wire A, $I_{A} = 8.0 A$
- Current in wire B, $I_{B} = 5.0 A$
- Separation between wires,
  $d = 4.0 cm = 0.04 m$
- Length of wire A considered,
  $l = 10 cm = 0.10 m$
- Permeability of free space,
  $μ_{0} = 4 π \times 10^{- 7} {T m A}^{- 1}$
Formula Used

Force between two long parallel current-carrying conductors:

$F = \frac{μ_{0} I_{A} I_{B} l}{2 π d}$

(This relation is derived using the magnetic field due to a straight wire and the Lorentz force law, as discussed in NCERT Physics, Moving Charges and Magnetism )

Substitution

$F = \frac{4 π \times 10^{- 7} \times 8.0 \times 5.0 \times 0.10}{2 π \times 0.04}$

Cancel $π$ :

$F = \frac{4 \times 10^{- 7} \times 8 \times 5 \times 0.10}{2 \times 0.04}$

$F = \frac{16 \times 10^{- 7}}{0.08}$

$F = 200 \times 10^{- 7} N$

$F = 2.0 \times 10^{- 5} N$

Direction of Force

Since the currents in the two wires are in the same direction, the wires attract each other.

Force on wire A is directed towards wire B.

Question 4.8

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of the magnetic field $B$ inside the solenoid near its centre.

Given
- Length of solenoid,
  $l = 80 cm = 0.80 m$
- Number of layers = 5
- Turns per layer = 400
$\Rightarrow Total turns N = 5 \times 400 = 2000$
- Current,
  $I = 8.0 A$
- Diameter = 1.8 cm ⇒ Radius = 0.9 cm = 0.009 m
- Permeability of free space,
  $μ_{0} = 4 π \times 10^{- 7} {T m A}^{- 1}$
Since the solenoid is long compared to its diameter, the field near the centre is uniform.

Formula Used

Magnetic field inside a long solenoid:

$B = μ_{0} n I$

where

$n = \frac{N}{l} (number of turns per unit length)$

(This result follows from Ampere’s circuital law as discussed in NCERT Physics, Moving Charges and Magnetism)

Calculation

Turns per unit length:

$n = \frac{2000}{0.80} = 2500 m^{- 1}$

Magnetic field:

$B = (4 π \times 10^{- 7}) \times 2500 \times 8$

$B = 4 π \times 10^{- 7} \times 20000$

$B = 8 π \times 10^{- 3} T$

$B \approx 2.5 \times 10^{- 2} T$

Question 4.9

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of $30^{\circ}$ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Given
- Side of square coil,
  $a = 10 cm = 0.10 m$
- Number of turns,
  $N = 20$
- Current,
  $I = 12 A$
- Magnetic field,
  $B = 0.80 T$
- Angle between normal to the coil and magnetic field,
  $θ = 30^{\circ}$
Formula Used

Torque on a current-carrying coil in a uniform magnetic field:

$τ = N I A B \sin θ$

where

$A = area of the coil$

(This expression follows from the magnetic dipole moment of a current loop, as discussed in NCERT Physics, Moving Charges and Magnetism )

Calculation

Area of square coil:

$A = a^{2} = (0.10)^{2} = 0.01 m^{2}$

Substitute values:

$τ = 20 \times 12 \times 0.01 \times 0.80 \times \sin 30^{\circ}$

$\sin 30^{\circ} = \frac{1}{2}$

$τ = 20 \times 12 \times 0.01 \times 0.80 \times 0.5$

$τ = 0.96 N m$

Question 4.10

Two moving coil meters, $M_{1}$ and $M_{2}$ , have the following particulars:
- Meter $M_{1}$ :
  $R_{1} = 10 Ω, N_{1} = 30, A_{1} = 3.6 \times 10^{- 3} m^{2}, B_{1} = 0.25 T$
- Meter $M_{2}$ :
  $R_{2} = 14 Ω, N_{2} = 42, A_{2} = 1.8 \times 10^{- 3} m^{2}, B_{2} = 0.50 T$
Compare the current sensitivities of the two meters. Which one is more sensitive?

Concept Used: Current Sensitivity of a Moving Coil Meter

For a moving coil galvanometer, current sensitivity is defined as angular deflection per unit current:

$Current sensitivity \propto \frac{N A B}{k}$

where
- $N$ = number of turns
- $A$ = area of the coil
- $B$ = magnetic field
- $k$ = torsional constant of the spring
If the meters are of similar construction, we may assume $k$ is the same for both.
Hence, sensitivity depends on $N A B$ .

(This result follows from the torque on a current loop in a magnetic field, discussed in NCERT Physics, Moving Charges and Magnetism )

Calculation

For meter $M_{1}$ :

$(N A B)_{1} = 30 \times (3.6 \times 10^{- 3}) \times 0.25$

$(N A B)_{1} = 30 \times 0.9 \times 10^{- 3}$

$(N A B)_{1} = 27 \times 10^{- 3}$

For meter $M_{2}$ :

$(N A B)_{2} = 42 \times (1.8 \times 10^{- 3}) \times 0.50$

$(N A B)_{2} = 42 \times 0.9 \times 10^{- 3}$

$(N A B)_{2} = 37.8 \times 10^{- 3}$

Comparison

$\frac{Sensitivity of M_{2}}{Sensitivity of M_{1}} = \frac{37.8}{27} \approx 1.4$

Ratio of voltage sensitivity

$Voltage sensitivity = \frac{current sensitivity}{R}$

$\Rightarrow Voltage sensitivity \propto \frac{N A B}{R}$ $\frac{S_{V 2}}{S_{V 1}} = \frac{(N A B)_{2} / R_{2}}{(N A B)_{1} / R_{1}} = \frac{37.8 / 14}{27 / 10}$

$\frac{S_{V 2}}{S_{V 1}} = \frac{37.8 \times 10}{27 \times 14} = 1$

Final Answer
- Meter $M_{2}$ is more sensitive than meter $M_{1}$ .
- $M_{2}$ is approximately 1.4 times more sensitive than $M_{1}$ .
- Voltage Sensitivity = 1
Question 4.11

In a chamber, a uniform magnetic field of 6.5 G is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} {m s}^{- 1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.

Given:

$B = 6.5 G = 6.5 \times 10^{- 4} T, v = 4.8 \times 10^{6} {m s}^{- 1}$

$e = 1.5 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} kg$

Why is the path circular?

When a charged particle moves perpendicular to a uniform magnetic field, it experiences a magnetic (Lorentz) force:

$\vec{F} = q \vec{v} \times \vec{B}$
- The force is always perpendicular to the velocity of the electron.
- Hence, the force does no work (speed remains constant).
- A force perpendicular to velocity acts as a centripetal force, continuously changing only the direction of motion.
Therefore, the electron moves in a circular path.

Radius of the circular orbit

For circular motion:

$Centripetal force = Magnetic force$ $\frac{m v^{2}}{r} = e v B$

$r = \frac{m v}{e B}$

Substitution

$r = \frac{(9.1 \times 10^{- 31}) (4.8 \times 10^{6})}{(1.5 \times 10^{- 19}) (6.5 \times 10^{- 4})}$

Numerator:

$9.1 \times 4.8 \times 10^{- 25} = 43.68 \times 10^{- 25}$

Denominator:

$1.5 \times 6.5 \times 10^{- 23} = 9.75 \times 10^{- 23}$

$r = \frac{43.68}{9.75} \times 10^{- 2}$

$r \approx 4.48 \times 10^{- 2} m$

Question 4.12

In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Given (from Q. 4.11):

$B = 6.5 G = 6.5 \times 10^{- 4} T, e = 1.5 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} kg$

Frequency of revolution

For a charged particle moving perpendicular to a uniform magnetic field, the cyclotron (revolution) frequency is:

$f = \frac{e B}{2 π m}$

Substitution

$f = \frac{(1.5 \times 10^{- 19}) (6.5 \times 10^{- 4})}{2 π (9.1 \times 10^{- 31})}$

First compute:

$\frac{e B}{m} = \frac{9.75 \times 10^{- 23}}{9.1 \times 10^{- 31}} \approx 1.07 \times 10^{8} s^{- 1}$

$f = \frac{1.07 \times 10^{8}}{2 π} \approx 1.7 \times 10^{7} Hz$

$f \approx 1.7 \times 10^{7} Hz$

Does the frequency depend on the speed of the electron?

No.
From the expression

$f = \frac{e B}{2 π m}$

the frequency depends only on:
- charge of the particle ( $e$ ),
- magnetic field ( $B$ ),
- mass of the particle ( $m$ ).
It is independent of the speed (or energy) of the electron, as long as the speed is non-relativistic.

Question 4.13

(a)

A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of $60^{\circ}$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

Given
- Number of turns, $N = 30$
- Radius, $r = 8.0 cm = 0.08 m$
- Current, $I = 6.0 A$
- Magnetic field, $B = 1.0 T$
- Angle between field and normal to the coil, $θ = 60^{\circ}$
Formula Used

Torque on a current-carrying coil in a uniform magnetic field:

$τ = N I A B \sin θ$

where $A$ is the area of the coil.

Calculation

Area of the circular coil:

$A = π r^{2} = π (0.08)^{2} = π \times 0.0064 \approx 2.01 \times 10^{- 2} m^{2}$

Substitute values:

$τ = 30 \times 6.0 \times (2.01 \times 10^{- 2}) \times 1.0 \times \sin 60^{\circ}$ $\sin 60^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866$ $τ \approx 30 \times 6.0 \times 2.01 \times 10^{- 2} \times 0.866$ $τ \approx 3.1 N m$

Answer (a)

$τ \approx 3.1 N m$

This is the counter torque that must be applied to keep the coil from rotating.

(b)

Would your answer change if the circular coil were replaced by a planar coil of some irregular shape enclosing the same area (all other particulars unchanged)?

Answer

No, the answer would not change.

Explanation

The torque on a planar current-carrying coil depends on:

$τ = N I A B \sin θ$

It depends only on:
- number of turns $N$ ,
- current $I$ ,
- enclosed area $A$ ,
- magnetic field $B$ ,
- angle $θ$ between the field and the normal.
It is independent of the shape of the coil.
Hence, any planar coil (circular or irregular) enclosing the same area will experience the same torque under identical conditions.
January 13, 2026
Class 12th Physics Chapter-1 Notes on Electric Charges
🌟 Electric Charge: The Invisible Power Around Us

Have you ever noticed your hair standing up after combing it on a dry day, or felt a tiny shock while touching a metal door handle? Lightning flashing across the sky during a thunderstorm is another dramatic example. All these everyday experiences happen because of electric charge.

Electric charge is a fundamental property of matter. Although we cannot see it directly, its effects are visible everywhere—from small sparks to powerful lightning bolts.

⚡ What Is Electric Charge?

Every object around us is made up of tiny particles called atoms. Inside each atom are:
- Protons (positively charged)
- Electrons (negatively charged)
- Neutrons (no charge)
Normally, an object has equal numbers of protons and electrons, so it is electrically neutral. But when electrons move from one object to another—often due to rubbing—an imbalance occurs, and the object becomes charged.

✨ Types of Electric Charges

There are two kinds of electric charges:
- Positive charge
- Negative charge
Their interaction follows a simple rule:
- Like charges repel each other
- Unlike charges attract each other
For example:
- Two positively charged objects push each other away.
- A positively charged object attracts a negatively charged one.
🔌 Conductors and Insulators

Not all materials behave the same way when electric charge is involved.

🧲 Conductors

Conductors are materials that allow electric charges (electrons) to move freely through them.

Examples:
- Metals like copper, aluminium, iron
- Human body
- Earth
Because charges can move easily, when a conductor is charged, the charge spreads quickly over its entire surface.

🪵 Insulators

Insulators are materials that do not allow electric charges to move freely.

Examples:
- Plastic
- Glass
- Rubber
- Wood
- Porcelain
When an insulator is charged, the charge stays at the place where it was produced. That is why a plastic comb rubbed on dry hair can attract tiny bits of paper, while a metal spoon usually cannot.

🔍 Why Do Objects Get Charged?

Charging mainly happens due to the transfer of electrons:
- An object losing electrons becomes positively charged
- An object gaining electrons becomes negatively charged
⚠️ Important:
No new charge is created during rubbing—charges are only transferred from one object to another.

📘 Basic Properties of Electric Charge

Electric charge has some important properties that help us understand how electricity works.

1️⃣ Additivity of Charge

Electric charge is a scalar quantity, which means it has magnitude but no direction.

If a system has many charges, the total charge is the algebraic sum of all individual charges.

Example:
If charges are +2 C, −3 C, and +5 C,
Total charge = +2 − 3 + 5 = +4 C

2️⃣ Conservation of Charge

Electric charge is conserved.

This means:
- Charge can neither be created nor destroyed
- It can only be transferred from one object to another
Even when charges move or redistribute within a system, the total charge remains constant.

3️⃣ Quantisation of Charge

Electric charge exists in discrete packets, not in continuous amounts.

The smallest possible unit of charge is the charge of an electron or proton, denoted by e.

$q = n e$

where:

$q$ = total charge

$n$ = integer (positive or negative)

$e = 1.6 \times 10^{- 19} C$

This means every observable charge is an integral multiple of e.

🌈 Why Electric Charge Matters

Electric charge is the foundation of:
- Electricity and electronics
- Lightning and thunderstorms
- Modern technology like mobile phones, computers, and medical devices
Understanding electric charge helps us explore deeper ideas like electric fields, electric current, and electromagnetic waves—all of which shape the modern world.
January 13, 2026
Class 12th Physics Chapter-3 Solutions
Go Back to Class 12th Physics

Question 3.1
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Solution:

The maximum current is drawn when the external resistance is zero (short-circuit condition).

Given:
Emf of the battery, $ε = 12 V$
Internal resistance, $r = 0.4 Ω$

Maximum current:

$I_{max} = \frac{ε}{r}$

$I_{max} = \frac{12}{0.4} = 30 A$

Question 3.2
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor.
If the current in the circuit is 0.5 A,
(i) what is the resistance of the resistor?
(ii) what is the terminal voltage of the battery when the circuit is closed?

Solution:

Given:
Emf of battery, $ε = 10 V$
Internal resistance, $r = 3 Ω$
Current, $I = 0.5 A$

(i) Resistance of the resistor

For a closed circuit:

$ε = I (R + r)$

$10 = 0.5 (R + 3)$

$R + 3 = \frac{10}{0.5} = 20$

$R = 20 - 3 = 17 Ω$

(ii) Terminal voltage of the battery

Terminal voltage $V$ is given by:

$V = ε - I r$

$V = 10 - (0.5 \times 3)$

$V = 10 - 1.5 = 8.5 V$

Question 3.3
At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is
$α = 1.70 \times 10^{- 4} {°C}^{- 1}$ ?

Solution:

Given:
Resistance at $T_{0} = {27.0}^{\circ} C$ , $R_{0} = 100 Ω$
Resistance at temperature $T$ , $R = 117 Ω$
Temperature coefficient, $α = 1.70 \times 10^{- 4} {°C}^{- 1}$

The relation between resistance and temperature is:

$R = R_{0} [1 + α (T - T_{0})]$

Substituting the given values:

$117 = 100 [1 + 1.70 \times 10^{- 4} (T - 27)]$

$1.17 = 1 + 1.70 \times 10^{- 4} (T - 27)$

$0.17 = 1.70 \times 10^{- 4} (T - 27)$

$T - 27 = \frac{0.17}{1.70 \times 10^{- 4}} = 1000$

$T = 1000 + 27 = 1027^{\circ} C$

Question 3.4
A negligibly small current is passed through a wire of length 15 m and uniform cross-section $6.0 \times 10^{- 7} m^{2}$ , and its resistance is measured to be $5.0 Ω.$ What is the resistivity of the material at the temperature of the experiment?

Solution:

Given:
Length of the wire, $l = 15 m$
Cross-sectional area, $A = 6.0 \times 10^{- 7} m^{2}$
Resistance, $R = 5.0 Ω$

The relation between resistance and resistivity is:

$R = ρ \frac{l}{A}$ So,

$ρ = \frac{R A}{l}$

Substituting the values:

$ρ = \frac{5.0 \times 6.0 \times 10^{- 7}}{15}$

$ρ = \frac{30 \times 10^{- 7}}{15} = 2.0 \times 10^{- 7} Ω m$

Question 3.5
A silver wire has a resistance of 2.1 Ω at 27.5 °C and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

Solution:

Given:
Resistance at $T_{1} = {27.5}^{\circ} C$ , $R_{1} = 2.1 Ω$
Resistance at $T_{2} = 100^{\circ} C$ , $R_{2} = 2.7 Ω$

The relation between resistance and temperature is:

$R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$

Substituting the values:

$2.7 = 2.1 [1 + α (100 - 27.5)]$

$2.7 = 2.1 [1 + 72.5 α]$

Divide both sides by 2.1:

$\frac{2.7}{2.1} = 1 + 72.5 α$

$1.2857 = 1 + 72.5 α$

$72.5 α = 0.2857$

$α = \frac{0.2857}{72.5} \approx 3.94 \times 10^{- 3} {°C}^{- 1}$

Question 3.6
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A, which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? The temperature coefficient of resistance of nichrome (average over the range) is
$α = 1.70 \times 10^{- 4} {°C}^{- 1}$

Solution:

Given:
Supply voltage, $V = 230 V$
Initial current, $I_{1} = 3.2 A$
Steady current, $I_{2} = 2.8 A$
Room temperature, $T_{1} = {27.0}^{\circ} C$
Temperature coefficient, $α = 1.70 \times 10^{- 4} {°C}^{- 1}$

Step 1: Find resistance at room temperature

$R_{1} = \frac{V}{I_{1}} = \frac{230}{3.2} = 71.9 Ω$

Step 2: Find resistance at steady state

$R_{2} = \frac{V}{I_{2}} = \frac{230}{2.8} = 82.1 Ω$

Step 3: Use temperature–resistance relation

$R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$

$82.1 = 71.9 [1 + 1.70 \times 10^{- 4} (T_{2} - 27)]$

Divide both sides by 71.9:

$1.142 = 1 + 1.70 \times 10^{- 4} (T_{2} - 27)$

$0.142 = 1.70 \times 10^{- 4} (T_{2} - 27)$

$T_{2} - 27 = \frac{0.142}{1.70 \times 10^{- 4}} \approx 835$

$T_{2} = 27 + 835 = 862^{\circ} C$

Question 3.7
Determine the current in each branch of the network shown in Fig. 3.20.

Solution:

From the figure:
- $A B = 10 Ω$
- $B C = 5 Ω$
- $A D = 5 Ω$
- $D C = 10 Ω$
- $B D = 5 Ω$ (central branch)
- External resistor $= 10 Ω$
- Battery emf $= 10 V$
Let the current supplied by the battery be $I$ .
Assume currents in the branches as follows:
- Current in branch $A B = I_{1}$
- Current in branch $B C = I_{2}$
- Current in central branch $B D = I_{3}$
Step 1: Apply Kirchhoff’s junction rule

At junction B:

$\begin{matrix} I_{1} = I_{2} + I_{3} & (1) \end{matrix}$

Step 2: Apply Kirchhoff’s loop rule

Loop ABDA:

$10 I_{1} + 5 (I_{1} - I_{3}) - 5 (I_{3}) = 0$

$15 I_{1} - 10 I_{3} = 0$ $\begin{matrix} 3 I_{1} = 2 I_{3} & (2) \end{matrix}$

Loop BCDC:

$5 I_{2} + 10 (I_{2} - I_{3}) - 5 I_{3} = 0$

$15 I_{2} - 15 I_{3} = 0$ $\begin{matrix} I_{2} = I_{3} & (3) \end{matrix}$

Step 3: Solve equations

From (3):

$I_{2} = I_{3}$

Substitute into (1): $I_{1} = I_{3} + I_{3} = 2 I_{3}$

Substitute into (2):

$3 (2 I_{3}) = 2 I_{3} \Rightarrow 6 I_{3} = 2 I_{3} \Rightarrow I_{3} = 0$

Hence: $I_{2} = 0, I_{1} = 0$

Step 4: Current from the battery

The Wheatstone bridge is balanced, so no current flows through the central branch.

Only the outer circuit draws current:

$R_{outer} = 10 Ω + 10 Ω = 20 Ω$

$I = \frac{V}{R} = \frac{10}{20} = 0.5 A$

This current divides equally in the two side branches.

Final Answer:
- Current through central branch $B D = 0 A$
- Current through left branch $A B – A D = 0.25 A$
- Current through right branch $B C – C D = 0.25 A$
- Total current from the battery = 0.5 A
Question 3.8
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω.
(i) What is the terminal voltage of the battery during charging?
(ii) What is the purpose of having a series resistor in the charging circuit?

Solution:

Given:
Emf of battery, $ε = 8.0 V$
Internal resistance, $r = 0.5 Ω$
Supply voltage, $V = 120 V$
Series resistance, $R = 15.5 Ω$

Step 1: Find the charging current

During charging, the supply emf opposes the battery emf.

Total resistance of the circuit:

$R_{total} = R + r = 15.5 + 0.5 = 16 Ω$

Net driving voltage:

$V_{net} = 120 - 8 = 112 V$

Current in the circuit:

$I = \frac{112}{16} = 7 A$

Step 2: Terminal voltage of the battery during charging

During charging:

$V_{terminal} = ε + I r$ $V_{terminal} = 8.0 + (7 \times 0.5)$ $V_{terminal} = 8.0 + 3.5 = 11.5 V$

Step 3: Purpose of the series resistor

The series resistor is used to:
1. Limit the charging current to a safe value
2. Prevent damage to the battery due to excessive current
3. Control the rate of charging
4. Protect the dc supply and battery from overheating
Answers:
- Terminal voltage of the battery during charging =
  
  $11.5 V$
- Purpose of series resistor:
  To limit and control the charging current and protect the battery.
Question 3.9
The number density of free electrons in a copper conductor estimated in Example 3.1 is
$n = 8.5 \times 10^{28} m^{- 3}$ .
How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is $2.0 \times 10^{- 6} m^{2}$ and it is carrying a current of 3.0 A.

Solution:

Given:
$n = 8.5 \times 10^{28} m^{- 3}$
Length of wire, $l = 3.0 m$
Area, $A = 2.0 \times 10^{- 6} m^{2}$
Current, $I = 3.0 A$
Charge of electron, $e = 1.6 \times 10^{- 19} C$

Step 1: Find the drift velocity of electrons

Current is given by:

$I = n e A v_{d}$

So,

$v_{d} = \frac{I}{n e A}$

Substitute the values:

$v_{d} = \frac{3.0}{(8.5 \times 10^{28}) (1.6 \times 10^{- 19}) (2.0 \times 10^{- 6})}$

$v_{d} = \frac{3.0}{2.72 \times 10^{4}}$

$v_{d} \approx 1.10 \times 10^{- 4} {m s}^{- 1}$

Step 2: Time taken to drift through the wire

$t = \frac{l}{v_{d}} = \frac{3.0}{1.10 \times 10^{- 4}}$

$t \approx 2.73 \times 10^{4} s$

Step 3: Convert time into hours

$t = \frac{2.73 \times 10^{4}}{3600} \approx 7.6 hours$
January 8, 2026
Class 12th Physics Chapter-2 Solutions
Go Back to Class 12th Physics

Question 2.1
Two charges $+ 5 \times 10^{- 8} C$ and $- 3 \times 10^{- 8} C$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Solution:

Let the positive charge $+ 5 \times 10^{- 8} C$ be at point O (origin).
The negative charge $- 3 \times 10^{- 8} C$ is at point A, 16 cm to the right.

Electric potential at a point $P$ on the line is

$V = \frac{1}{4 π ε_{0}} (\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}})$

For $V = 0$ , the bracket must be zero.

Case 1: Point $P$ between the charges $(0 < x < 16 cm)$

Distance from $+ 5 \times 10^{- 8} C$ = $x$
Distance from $- 3 \times 10^{- 8} C$ = $16 - x$

$\frac{5}{x} - \frac{3}{16 - x} = 0$

$5 (16 - x) = 3 x \Rightarrow 80 - 5 x = 3 x \Rightarrow x = 10 cm$

Case 2: Point $P$ beyond the negative charge $(x > 16 cm)$

Distance from $+ 5 \times 10^{- 8} C$ = $x$
Distance from $- 3 \times 10^{- 8} C$ = $x - 16$

$\frac{5}{x} - \frac{3}{x - 16} = 0$

$5 (x - 16) = 3 x \Rightarrow 5 x - 80 = 3 x \Rightarrow x = 40 cm$

Question 2.2

A regular hexagon of side 10 cm has a charge $5 μ C$ at each of its vertices. Calculate the electric potential at the centre of the hexagon.

Given
- Charge at each vertex:
  $q = 5 μ C = 5 \times 10^{- 6} C$
- Side of hexagon:
  $a = 10 cm = 0.10 m$
- Number of vertices (charges): $6$
For a regular hexagon, the distance from the centre to each vertex is equal to the side length:

$r = a = 0.10 m$

Electric potential is a scalar, so potentials due to all charges add algebraically.

Potential due to one charge at the centre

$V_{1} = \frac{1}{4 π ε_{0}} \frac{q}{r}$

Total potential at the centre (due to 6 identical charges)

$V = 6 V_{1} = 6 \times \frac{1}{4 π ε_{0}} \frac{q}{r}$

Substitute values:

$V = 6 \times 9 \times 10^{9} \times \frac{5 \times 10^{- 6}}{0.10}$

$V = 6 \times 9 \times 10^{9} \times 5 \times 10^{- 5}$

$V = 27 \times 10^{5}$

$V = 2.7 \times 10^{6} V$

Question 2.3
Two charges $+ 2 μ C$ and $- 2 μ C$ are placed at points A and B, 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?

(a) Equipotential surface

For equal and opposite charges (an electric dipole), the plane perpendicular to AB and passing through the midpoint of AB is an equipotential surface.

Reason:
At any point on this plane, the distances from the two charges are equal. Hence, the potentials due to $+ 2 μ C$ and $- 2 μ C$ are equal in magnitude and opposite in sign, so the net potential is zero everywhere on this plane.

(b) Direction of the electric field on this surface.

The electric field is perpendicular (normal) to the equipotential surface at every point.

Specifically here:
On the perpendicular bisector plane, the electric field points along the line AB, from the positive charge toward the negative charge.

Question 2.4
A spherical conductor of radius 12 cm has a charge $1.6 \times 10^{- 7} C$ uniformly distributed on its surface. Find the electric field
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the centre.

Given
- Radius of sphere:
  $R = 12 cm = 0.12 m$
- Charge on sphere:
  $Q = 1.6 \times 10^{- 7} C$
- Coulomb constant:
  $k = \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} {/C}^{2}$
(a) Electric field inside the sphere

For a conducting sphere, the electric field everywhere inside is zero.

$E = 0$

(b) Electric field just outside the sphere

Just outside a charged conducting sphere, the field is the same as that due to a point charge $Q$ at the centre.

$E = \frac{k Q}{R^{2}}$

$E = \frac{9 \times 10^{9} \times 1.6 \times 10^{- 7}}{(0.12)^{2}}$

$E = \frac{14.4 \times 10^{2}}{0.0144} = 1.0 \times 10^{5} N/C$

$E = 1.0 \times 10^{5} N/C (radially outward)$

(c) Electric field at 18 cm from the centre

Distance:

$r = 18 cm = 0.18 m$

Outside the sphere, again treat it as a point charge at the centre:

$E = \frac{k Q}{r^{2}}$

$E = \frac{9 \times 10^{9} \times 1.6 \times 10^{- 7}}{(0.18)^{2}}$

$E = \frac{14.4 \times 10^{2}}{0.0324} \approx 4.44 \times 10^{4} N/C$

$E \approx 4.4 \times 10^{4} N/C (radially outward)$

Question 2.5
A parallel plate capacitor with air between the plates has a capacitance of $8 pF$ . What will be the capacitance if
- the distance between the plates is reduced to half, and
- the space between them is filled with a dielectric of constant $K = 6$ ?
Given
- Initial capacitance (air):
  $C_{0} = 8 pF$
- New separation:
  $d^{'} = \frac{d}{2}$
- Dielectric constant:
  $K = 6$
Key formula

For a parallel plate capacitor:

$C = \frac{K ε_{0} A}{d}$

Capacitance is:
- inversely proportional to plate separation $d$
- directly proportional to dielectric constant $K$
Effect of changes
1. Halving the distance $(d \to d / 2)$
  $C \to 2 C$
2. Filling dielectric of constant $K = 6$
  $C \to 6 C$
New capacitance

$C^{'} = 2 \times 6 \times C_{0}$

$C^{'} = 12 \times 8 pF$

$C^{'} = 96 pF$

Question 2.6
Three capacitors, each of capacitance $9 pF$ , are connected in series.
(a) Find the total capacitance of the combination.
(b) Find the potential difference across each capacitor when the combination is connected to a $120 V$ supply.

Given
- Each capacitor:
  $C = 9 pF$
- Number of capacitors: $3$
- Supply voltage:
  $V = 120 V$
(a) Total capacitance in series

For capacitors in series:

$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

$\frac{1}{C_{eq}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}$

$C_{eq} = 3 pF$

(b) Potential difference across each capacitor

In a series combination:
- The same charge flows on each capacitor.
- For identical capacitors, the total voltage divides equally.
$V_{each} = \frac{V_{total}}{3} = \frac{120}{3}$

$V_{each} = 40 V$

Question 2.7
Three capacitors of capacitances $2 pF$ , $3 pF$ , and $4 pF$ are connected in parallel.
(a) Find the total capacitance.
(b) Find the charge on each capacitor when connected to a $100 V$ supply.

Given
- $C_{1} = 2 pF$
- $C_{2} = 3 pF$
- $C_{3} = 4 pF$
- Applied voltage:
  $V = 100 V$
(a) Total capacitance (parallel combination)

For capacitors in parallel, capacitances add directly:

$C_{eq} = C_{1} + C_{2} + C_{3}$

$C_{eq} = 2 + 3 + 4 = 9 pF$

(b) Charge on each capacitor

In a parallel combination, the potential difference across each capacitor is the same and equals the supply voltage.

Using $Q = C V$ :

Charge on $2 pF$ capacitor

$Q_{1} = C_{1} V = 2 \times 10^{- 12} \times 100 = 2 \times 10^{- 10} C$

Charge on $3 pF$ capacitor

$Q_{2} = 3 \times 10^{- 12} \times 100 = 3 \times 10^{- 10} C$

Charge on $4 pF$ capacitor

$Q_{3} = 4 \times 10^{- 12} \times 100 = 4 \times 10^{- 10} C$

Question 2.8
In a parallel plate capacitor with air between the plates:
- Area of each plate $A = 6 \times 10^{- 3} m^{2}$
- Separation $d = 3 mm = 3 \times 10^{- 3} m$
(a) Calculate the capacitance.
(b) If connected to a $100 V$ supply, find the charge on each plate.

Given
- $ε_{0} = 8.85 \times 10^{- 12} {F m}^{- 1}$
- Dielectric: air ( $K \approx 1$ )
(a) Capacitance of a parallel plate capacitor

$C = \frac{ε_{0} A}{d}$

$C = \frac{8.85 \times 10^{- 12} \times 6 \times 10^{- 3}}{3 \times 10^{- 3}}$

$C = 8.85 \times 10^{- 12} \times 2$

$C = 1.77 \times 10^{- 11} F$

or $C = 17.7 pF$

(b) Charge on each plate

$Q = C V$

$Q = 1.77 \times 10^{- 11} \times 100$

$Q = 1.77 \times 10^{- 9} C$

Question 2.9
Explain what happens if, in the capacitor of Exercise 2.8, a 3 mm thick mica sheet (dielectric constant $K = 6$ ) is inserted between the plates:
(a) while the voltage supply remains connected
(b) after the supply is disconnected

(Recall from Ex. 2.8: plate separation = 3 mm, so the dielectric completely fills the space.)

Basic idea (important for exams)
- Inserting a dielectric increases capacitance by a factor $K$ .
- What changes next depends on whether voltage is fixed or charge is fixed.
(a) Supply connected (Voltage constant)
- Voltage remains fixed at
  $V = 100 V$
- New capacitance:
  $C^{'} = K C = 6 C$
- Since $Q = C V$ ,
  $Q^{'} = C^{'} V = 6 C V = 6 Q$
What happens?
- Capacitance increases 6 times
- Charge on each plate increases 6 times
- Extra charge flows from the battery to the plates
- Electric field decreases inside the capacitor
- Energy stored increases (battery supplies extra energy)
(b) Supply disconnected (Charge constant)
- Charge on the plates remains fixed:
  $Q^{'} = Q$
- Capacitance still increases:
  $C^{'} = 6 C$
- Since $V = Q / C$ ,
  $V^{'} = \frac{Q}{C^{'}} = \frac{V}{6}$
What happens?
- Capacitance increases 6 times
- Voltage across plates decreases to one-sixth
- Electric field decreases
- Energy stored decreases (energy is released as heat/mechanical work)
Question 2.10
A $12 pF$ capacitor is connected to a $50 V$ battery. How much electrostatic energy is stored in the capacitor?

Given
- Capacitance:
  $C = 12 pF = 12 \times 10^{- 12} F$
- Voltage:
  $V = 50 V$
Formula for energy stored in a capacitor

$U = \frac{1}{2} C V^{2}$

Calculation

$U = \frac{1}{2} \times 12 \times 10^{- 12} \times (50)^{2}$

$U = 6 \times 10^{- 12} \times 2500$

$U = 1.5 \times 10^{- 8} J$

Question 2.11
A $600 pF$ capacitor is charged by a $200 V$ supply. It is then disconnected from the supply and connected to another uncharged $600 pF$ capacitor. How much electrostatic energy is lost?

Given
- $C_{1} = 600 pF = 600 \times 10^{- 12} F$
- $C_{2} = 600 pF$ (uncharged)
- Initial voltage:
  $V = 200 V$
Step 1: Initial energy stored

Energy in a charged capacitor:

$U_{i} = \frac{1}{2} C V^{2}$

$U_{i} = \frac{1}{2} \times 600 \times 10^{- 12} \times (200)^{2}$

$U_{i} = 300 \times 10^{- 12} \times 40000$

$U_{i} = 1.2 \times 10^{- 5} J$

Step 2: Situation after connection
- The two capacitors are identical and connected together.
- Total capacitance:
$C_{total} = 600 + 600 = 1200 pF$
- Total charge is conserved.
- Final voltage becomes half:
$V_{f} = \frac{200}{2} = 100 V$

Step 3: Final energy stored

$U_{f} = \frac{1}{2} C_{total} V_{f}^{2}$

$U_{f} = \frac{1}{2} \times 1200 \times 10^{- 12} \times (100)^{2}$

$U_{f} = 600 \times 10^{- 12} \times 10000$

$U_{f} = 6.0 \times 10^{- 6} J$

Step 4: Energy lost

$Energy lost = U_{i} - U_{f}$

$= 1.2 \times 10^{- 5} - 6.0 \times 10^{- 6}$

$= 6.0 \times 10^{- 6} J$
December 23, 2025
Class 12th Physics Chapter-1 Solutions
Go Back to Class 12th Physics

Question 1.1

What is the force between two small charged spheres having charges $2 \times 10^{- 7} C$ and $3 \times 10^{- 7} C$ placed 30 cm apart in air?

Solution

Given:

$q_{1} = 2 \times 10^{- 7} C$

$q_{2} = 3 \times 10^{- 7} C$

Distance, $r = 30 cm = 0.30 m$
1. Medium: air (same as vacuum)
Formula (Coulomb’s Law):

$F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$ where $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} C^{- 2}$

Calculation:

$F = 9 \times 10^{9} \times \frac{(2 \times 10^{- 7}) (3 \times 10^{- 7})}{(0.30)^{2}}$

$F = 9 \times 10^{9} \times \frac{6 \times 10^{- 14}}{0.09}$

$F = 9 \times 10^{9} \times 6.67 \times 10^{- 13}$

$F = 6 \times 10^{- 3} N$

Question 1.2

The electrostatic force on a small sphere of charge

$0.4 μ C$ due to another small sphere of charge $- 0.8 μ C$ in air is $0.2 N$

.(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

Solution

Given:

$q_{1} = 0.4 μ C = 0.4 \times 10^{- 6} C$

$q_{2} = - 0.8 μ C = - 0.8 \times 10^{- 6} C$

Force, $F = 0.2 N$

Medium: air (same as vacuum)

(a) Distance between the two spheres

Formula (Coulomb’s Law):

$F = \frac{1}{4 π ε_{0}} \frac{∣ q_{1} q_{2} ∣}{r^{2}}$

$\Rightarrow r^{2} = \frac{1}{4 π ε_{0}} \frac{∣ q_{1} q_{2} ∣}{F}$

$\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} C^{- 2}$

Substitution: $r^{2} = \frac{9 \times 10^{9} \times (0.4 \times 10^{- 6}) (0.8 \times 10^{- 6})}{0.2}$

$r^{2} = \frac{9 \times 10^{9} \times 0.32 \times 10^{- 12}}{0.2}$

$r^{2} = \frac{2.88 \times 10^{- 3}}{0.2}$

$r^{2} = 1.44 \times 10^{- 2}$

$r = \sqrt{1.44 \times 10^{- 2}} = 0.12 m = 12 cm$

Question 1.3

Check that the ratio

$\frac{k e^{2}}{G m_{e} m_{p}}$

is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?

Solution

(a) Checking whether the ratio is dimensionless

Write the units of each quantity:

$k = \frac{1}{4 π ε_{0}}$ ; Units: ${N m}^{2} C^{- 2}$

$e$ (electronic charge) Units: C

$G$ (gravitational constant) Units: ${N m}^{2} {kg}^{- 2}$

$m_{e}, m_{p}$ (masses of electron and proton) – Units: kg

Now, units of the numerator:

$k e^{2} = ({N m}^{2} C^{- 2}) (C^{2}) = {N m}^{2}$

Units of the denominator:

$G m_{e} m_{p} = ({N m}^{2} {kg}^{- 2}) ({kg}^{2}) = {N m}^{2}$

So, the ratio is: $\frac{{N m}^{2}}{{N m}^{2}} = 1$

Hence, the ratio is dimensionless.

(b) Numerical value of the ratio

Using standard physical constants:

$k = 9.0 \times 10^{9} {N m}^{2} C^{- 2}$

$e = 1.6 \times 10^{- 19} C$

$G = 6.67 \times 10^{- 11} {N m}^{2} {kg}^{- 2}$

$m_{e} = 9.11 \times 10^{- 31} kg$

$m_{p} = 1.67 \times 10^{- 27} kg$

Substitute: $\frac{k e^{2}}{G m_{e} m_{p}} = \frac{(9.0 \times 10^{9}) (1.6 \times 10^{- 19})^{2}}{(6.67 \times 10^{- 11}) (9.11 \times 10^{- 31}) (1.67 \times 10^{- 27})}$

$\approx 2.4 \times 10^{39}$

(c) Physical significance of the ratio
- This ratio represents the ratio of electrostatic force to gravitational force between an electron and a proton.
The value is extremely large ( $\sim 10^{39}$ ), which shows that:

$Electrostatic force is enormously stronger than gravitational force$
- Gravity is negligible at the atomic and subatomic scale compared to electric forces.
Question 1.4

(a) Explain the meaning of the statement “electric charge of a body is quantised”.

(b) Why can one ignore quantisation of electric charge when dealing with macroscopic (large-scale) charges?

Answer

(a) Meaning of quantisation of electric charge

The statement “electric charge is quantised” means that:
- Electric charge exists in discrete packets, not in a continuous manner.
- The charge $q$
$q$ on any body is always an integral multiple of a fundamental unit of charge $e$ .

Mathematically,

$q = \pm n e$ where:

$n = 0, 1, 2, 3, \dots, \dots(an integer)$

$e = 1.6 \times 10^{- 19} C$

(magnitude of charge on an electron or proton)

✔ Hence, electric charge is quantised.

(b) Why quantisation of charge can be ignored for macroscopic charges

In macroscopic (large-scale) situations:
- The amount of charge involved is usually of the order of microcoulombs (μC) or coulombs (C).
- Such charges contain a very large number of electrons.
For example:

$1 C = 6.25 \times 10^{18} electrons$

Compared to this huge number:
- The elementary charge
  $e = 1.6 \times 10^{- 19} C$
  is extremely small.
- Any change in charge by one electron is negligible.
✔ Hence, quantisation of electric charge can be ignored for large-scale charges.

Question 1.5

When a glass rod is rubbed with a silk cloth, charges appear on both.
A similar phenomenon is observed with many other pairs of bodies.
Explain how this observation is consistent with the law of conservation of charge.

Answer

The law of conservation of charge states that:

Electric charge can neither be created nor destroyed; it can only be transferred from one body to another.

When a glass rod is rubbed with a silk cloth, electrons are transferred from the glass rod to the silk cloth. As a result, the glass rod becomes positively charged and the silk cloth becomes negatively charged. No new charge is created; the charge lost by the glass rod is equal to the charge gained by the silk cloth.

Thus, the total charge of the system remains conserved, which is consistent with the law of conservation of charge.

Question 1.6

Four point charges

$q_{A} = + 2 μ C, q_{B} = - 5 μ C, q_{C} = + 2 μ C, q_{D} = - 5 μ C$

are located at the corners of a square ABCD of side 10 cm.
What is the force on a charge of

$1 μ C$ placed at the centre of the square?

Solution

Using symmetry:
- The distance from the centre to each corner of the square is the same.
- The charges at opposite corners are equal:
1. $q_{A} = q_{C} = + 2 μ C$
2. $q_{B} = q_{D} = - 5 μ C$
Forces due to opposite corners:

Force on the central charge due to

$q_{A}$ is equal in magnitude and opposite in direction to that due to $q_{C}$

⇒ they cancel.

Force due to $q_{B}$ is equal in magnitude and opposite in direction to that due to $q_{D}$

⇒ they cancel.

Since each pair of opposite forces cancels, the vector sum of all forces is zero.

Question 1.7

(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

(b) Explain why two field lines never cross each other at any point.

Answer

(a) Why an electrostatic field line cannot have sudden breaks

An electrostatic field line represents the direction of the electric field at every point in space.
The electric field exists continuously in space (except at the location of charges).
- If a field line had a sudden break, it would imply that the electric field suddenly disappears at that point.
- This is not possible because an electric field cannot start or stop abruptly in free space.
✔ Therefore, electrostatic field lines must be continuous curves, starting from positive charges and ending on negative charges (or at infinity).

(b) Why two field lines never cross each other
- The direction of the electric field at a point is unique.
- A field line at any point gives the direction of the electric field at that point.
If two field lines crossed:
- There would be two different tangents at the point of intersection.
- This would mean two different directions of the electric field at the same point, which is impossible.
✔ Hence, two electric field lines can never cross each other.

Question 1.8

Two point charges

$q_{A} = + 3 μ C$ and $q_{B} = - 3 μ C$ are located 20 cm apart in vacuum.

(a) What is the electric field at the midpoint $O$ of the line AB joining the two charges?
(b) If a negative test charge of magnitude $1.5 \times 10^{- 9} C$ is placed at this point, what is the force experienced by the test charge?

Solution

Given:

$q_{A} = + 3 μ C = 3 \times 10^{- 6} C$

$q_{B} = - 3 μ C = - 3 \times 10^{- 6} C$

Distance AB = 20 cm ⇒ distance of each charge from midpoint

$r = 10 cm = 0.10 m$

$k = \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} C^{- 2}$

(a) Electric field at midpoint $O$

Electric field due to a point charge:

$E = \frac{k ∣ q ∣}{r^{2}}$ Field due to $q_{A}$ at $O$ :

$E_{A} = \frac{9 \times 10^{9} \times 3 \times 10^{- 6}}{(0.10)^{2}} = 2.7 \times 10^{6} {N C}^{- 1}$ Field due to $q_{B}$ at $O$ :

$E_{B} = 2.7 \times 10^{6} {N C}^{- 1}$

Direction:
- Field due to is away from A (towards B) + $q_{A}$
- $Field due to - q_{B} is towards B .$
Thus, both fields are in the same direction (from A to B).

$E_{net} = E_{A} + E_{B} = 5.4 \times 10^{6} {N C}^{- 1}$

Answer (a):

$E = 5.4 \times 10^{6} {N C}^{- 1} from A to B$

(b) Force on the test charge

Given test charge: $q_{0} = - 1.5 \times 10^{- 9} C$

Force on a charge in an electric field:

$F = q_{0} E$

$F = (- 1.5 \times 10^{- 9}) (5.4 \times 10^{6}) = - 8.1 \times 10^{- 3} N$

The negative sign indicates that the force is opposite to the direction of the electric field.

Answer (b):

$F = 8.1 \times 10^{- 3} N, directed from B to A$

Question 1.9

A system has two charges

$q_{A} = 2.5 \times 10^{- 7} C$ and $q_{B} = - 2.5 \times 10^{- 7} C$ located at points $A (0, 0, - 15 cm)$ and $B (0, 0, + 15 cm)$ , respectively. Find:
(a) Total charge of the system
(b) Electric dipole moment of the system

Solution

(a) Total charge of the system

$Q_{total} = q_{A} + q_{B}$

$Q_{total} = (2.5 \times 10^{- 7}) + (- 2.5 \times 10^{- 7}) = 0$

$Q_{total} = 0$

✔ The system is electrically neutral.

(b) Electric dipole moment

Electric dipole moment: $\vec{p} = q \vec{d}$

where
- $q = 2.5 \times 10^{- 7} C$ (magnitude of either charge)
- $\vec{d}$ = displacement vector from negative to positive charge
Distance between charges:

$d = 15 cm + 15 cm = 30 cm = 0.30 m$

So, $p = q d = (2.5 \times 10^{- 7}) (0.30)$

$p = 7.5 \times 10^{- 8} C m$

Direction:
From negative charge at $(0, 0, + 15 cm)$ to positive charge at

$(0, 0, - 15 cm)$ , i.e. along the negative z-axis.

Question 1.10

An electric dipole with dipole moment $p = 4 \times 10^{- 9} C m$ is aligned at $30^{\circ}$ with a uniform electric field of magnitude E $= 5 \times 10^{4} {N C}^{- 1}$ $.$ Calculate the magnitude of the torque acting on the dipole.

Solution

Formula (Torque on an electric dipole):

$τ = p E \sin θ$

Substitution:

$τ = (4 \times 10^{- 9}) (5 \times 10^{4}) \sin 30^{\circ}$

$\sin 30^{\circ} = \frac{1}{2}$

$τ = (4 \times 5 \times 0.5) \times 10^{- 9 + 4}$

$τ = 10 \times 10^{- 5}$

$τ = 1.0 \times 10^{- 4} N m$

Question 1.11

A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{- 7} C$

(a) Estimate the number of electrons transferred (from which to which?).
(b) Is there a transfer of mass from wool to polythene?

Solution

(a) Number of electrons transferred

Given:
- Charge on polythene, $q = - 3 \times 10^{- 7} C$
- Charge of one electron, $e = 1.6 \times 10^{- 19} C$
Number of electrons transferred:

$n = \frac{∣ q ∣}{e} = \frac{3 \times 10^{- 7}}{1.6 \times 10^{- 19}} = 1.875 \times 10^{12}$

Direction of transfer:
Since the polythene becomes negatively charged, electrons are transferred from wool to polythene.

$n \approx 1.9 \times 10^{12} electrons, from wool to polythene$

(b) Is there a transfer of mass?

Yes.
Electrons have mass ( $m_{e} = 9.11 \times 10^{- 31} kg$ ). When electrons move from wool to polythene, a very small mass is transferred.

However, this mass is extremely small and negligible in practice.

Question 1.12

(a) Two insulated charged copper spheres A and B have their centres 50 cm apart. Charge on each sphere $= 6.5 \times 10^{- 7} C$ . Radii are negligible compared to separation. Find the mutual force of electrostatic repulsion.

(b) What is the force of repulsion if each charge is doubled and the distance is halved?

Solution

Given (a):
- $q_{1} = q_{2} = 6.5 \times 10^{- 7} C$
- $r = 50 cm = 0.50 m$
- $k = \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} C^{- 2}$
Formula (Coulomb’s law):

$F = k \frac{q_{1} q_{2}}{r^{2}}$

Calculation (a): $F = 9 \times 10^{9} \times \frac{(6.5 \times 10^{- 7})^{2}}{(0.50)^{2}}$

$(6.5 \times 10^{- 7})^{2} = 4.225 \times 10^{- 13}$

$F = 9 \times 10^{9} \times \frac{4.225 \times 10^{- 13}}{0.25} = 1.52 \times 10^{- 2} N$

Answer (a):

$F = 1.52 \times 10^{- 2} N (repulsive)$

(b) Effect of doubling charges and halving distance

New charges: $q_{1}^{'} = q_{2}^{'} = 2 q$

New distance: $r^{'} = \frac{r}{2}$

Using proportionality:

$F \propto \frac{q_{1} q_{2}}{r^{2}}$ So,

$\frac{F^{'}}{F} = \frac{(2 q) (2 q)}{(r / 2)^{2}} = \frac{4 q^{2}}{r^{2} / 4} = 16$

$F^{'} = 16 F = 16 \times 1.52 \times 10^{- 2} = 2.43 \times 10^{- 1} N$

Answer (b):

$F^{'} = 2.43 \times 10^{- 1} N (repulsive)$

Question 1.13

Figure 1.30 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Understanding the situation
- In a uniform electrostatic field,
  - Positive charges bend in the direction of the electric field
  - Negative charges bend opposite to the direction of the electric field
- A neutral particle would move straight (no bending).
- The amount of bending (curvature) depends on the charge to mass ratio (q/m):
  - Greater curvature → larger q/m
  - Smaller curvature → smaller q/m
Signs of the charges

From Figure 1.30 :
- Particle 1 bends upward (along the electric field)
  → Positively charged
- Particle 2 bends downward (opposite to the field)
  → Negatively charged
- Particle 3 bends upward (same direction as Particle 1)
  → Positively charged
Signs of charges
- Particle 1: Positive
- Particle 2: Negative
- Particle 3: Positive
Highest charge to mass ratio (q/m)
- The particle whose path shows the maximum curvature has the highest q/m ratio.
- From the figure, Particle 2 bends the most.
Question 1.14

Consider a uniform electric field

$\vec{E} = 3 \times 10^{3} \hat{i} N/C$

(a) What is the flux of this field through a square of side 10 cm whose plane is parallel to the yz-plane?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

Given
- Electric field:
  $E = 3 \times 10^{3} N/C(along +x direction)$
- Side of square:
  $a = 10 cm = 0.10 m$
- Area of square:
  $A = a^{2} = (0.10)^{2} = 0.01 m^{2}$
- Formula for electric flux:
  $Φ = \vec{E} \cdot \vec{A} = E A \cos θ$
  where $θ$ is the angle between electric field and area vector (normal).
(a) Plane parallel to the yz-plane
- Normal to the yz-plane is along the x-axis.
- Electric field is also along the x-axis.
- Therefore,
  $θ = 0^{\circ}, \cos 0^{\circ} = 1$
Calculation

$Φ = E A \cos θ$

$Φ = (3 \times 10^{3}) (0.01) (1)$

$Φ = 30 {N m}^{2} ⁣ / C$

(b) Normal makes 60° with the x-axis
- Angle between $\vec{E}$ and normal:
  $θ = 60^{\circ}, \cos 60^{\circ} = \frac{1}{2}$
Calculation

$Φ = E A \cos 60^{\circ}$

$Φ = (3 \times 10^{3}) (0.01) (\frac{1}{2})$

$Φ = 15 {N m}^{2} ⁣ / C$

Question 1.15

What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Given
- Uniform electric field (from Exercise 1.14):
  $\vec{E} = 3 \times 10^{3} \hat{i} N/C$
- Side of cube:
  $a = 20 cm = 0.20 m$
- Cube faces are parallel to coordinate planes.
- The cube is a closed surface.
Concept Used

According to Gauss’s law,

$Φ_{net} = \oint \vec{E} \cdot d \vec{A} = \frac{q_{enclosed}}{ε_{0}}$
- A uniform electric field does not enclose any charge inside the cube.
- Hence,
  $q_{enclosed} = 0$
Reasoning (Important for exams)
- Electric flux entering the cube through one face is equal to the flux leaving through the opposite face.
- Flux through the remaining four faces is zero because the electric field is parallel to those faces.
- Therefore, all fluxes cancel out.
Question 1.16

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is
$8.0 \times 10^{3} {N m}^{2} / C$ .

(a) What is the net charge inside the box?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

Concept Used: Gauss’s Law

$Φ = \oint \vec{E} \cdot d \vec{A} = \frac{q_{enclosed}}{ε_{0}}$ where
$Φ$ = net electric flux through the closed surface
$q_{enclosed}$ = net charge inside the surface
$ε_{0} = 8.85 \times 10^{- 12} C^{2} / ({N m}^{2})$

(a) Net charge inside the box

Given

$Φ = 8.0 \times 10^{3} {N m}^{2} / C$

Using Gauss’s law:

$q_{enclosed} = ε_{0} Φ$

$q_{enclosed} = (8.85 \times 10^{- 12}) (8.0 \times 10^{3})$

$q_{enclosed} = 7.1 \times 10^{- 8} C$

(b) If the net outward flux were zero

Can we conclude that there are no charges inside the box?

Answer: No

Reason
- Zero net flux means:
  $q_{enclosed} = 0$
- This only tells us that the algebraic sum of charges inside is zero.
- There may still be charges present, such as:
  - Equal positive and negative charges inside the box.
- Their effects cancel, giving zero net flux.
Question 1.17

A point charge $+ 10 mC$ is at a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

Understanding the idea (key hint explained)
- The square has side 10 cm.
- The charge is 5 cm above the centre of the square.
- If we imagine a cube of edge 10 cm, this square can be taken as one face of the cube.
- Then the charge lies exactly at the centre of the cube.
This allows us to use Gauss’s law and symmetry.

Step 1: Total flux through the cube

By Gauss’s law,

$Φ_{cube} = \frac{q}{ε_{0}}$ Given:

$q = + 10 mC = 10 \times 10^{- 3} = 1.0 \times 10^{- 2} C$

$ε_{0} = 8.85 \times 10^{- 12} C^{2} / ({N m}^{2})$

$Φ_{cube} = \frac{1.0 \times 10^{- 2}}{8.85 \times 10^{- 12}} = 1.13 \times 10^{9} {N m}^{2} / C$

Step 2: Flux through one face of the cube
- A cube has 6 identical faces.
- The charge is at the centre, so flux is equally distributed.
$Φ_{one face} = \frac{1}{6} Φ_{cube}$

$Φ_{square} = \frac{1.13 \times 10^{9}}{6}$

$Φ_{square} \approx 1.9 \times 10^{8} {N m}^{2} / C$

Question 1.18

A point charge of $2.0 μ C$ is at the centre of a cubic Gaussian surface of edge 9.0 cm. What is the net electric flux through the surface?

Concept Used: Gauss’s Law

$Φ_{net} = \oint \vec{E} \cdot d \vec{A} = \frac{q_{enclosed}}{ε_{0}}$

Given
- Charge at centre:
  $q = 2.0 μ C = 2.0 \times 10^{- 6} C$
- Permittivity of free space:
  $ε_{0} = 8.85 \times 10^{- 12} C^{2} / ({N m}^{2})$
(Note: The size of the cube is irrelevant for net flux.)

Calculation

$Φ_{net} = \frac{q}{ε_{0}}$

$Φ_{net} = \frac{2.0 \times 10^{- 6}}{8.85 \times 10^{- 12}}$

$Φ_{net} = 2.26 \times 10^{5} {N m}^{2} / C$

Question 1.20

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is
$1.5 \times 10^{3} N/C$ and points radially inward, what is the net charge on the sphere?

Given
- Radius of sphere = 10 cm (not directly needed)
- Distance from centre:
  $r = 20 cm = 0.20 m$
- Electric field:
  $E = 1.5 \times 10^{3} N/C$
- Direction of field: radially inward
Concept Used

Outside a conducting sphere, the electric field is the same as that due to a point charge at the centre.

$E = \frac{1}{4 π ε_{0}} \frac{∣ Q ∣}{r^{2}}$ or $Q = \frac{E r^{2}}{k} where k = 9 \times 10^{9} {N m}^{2} / C^{2}$

Calculation

$Q = \frac{(1.5 \times 10^{3}) (0.20)^{2}}{9 \times 10^{9}}$

$Q = \frac{(1.5 \times 10^{3}) (0.04)}{9 \times 10^{9}}$

$Q = \frac{60}{9 \times 10^{9}}$

$Q = 6.7 \times 10^{- 9} C$

Sign of the charge
- Electric field points radially inward
- Therefore, the charge on the sphere must be negative
Question 1.21

A uniformly charged conducting sphere of diameter 2.4 m has a surface charge density of $80.0 μ {C m}^{- 2}$ .

(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere?

Given
- Diameter of sphere = 2.4 m
  $\Rightarrow r = 1.2 m$
- Surface charge density:
  $σ = 80.0 μ {C/m}^{2} = 80 \times 10^{- 6} {C/m}^{2}$
- Permittivity of free space:
  $ε_{0} = 8.85 \times 10^{- 12} C^{2} / ({N m}^{2})$
(a) Charge on the sphere

Surface area of a sphere:

$A = 4 π r^{2}$

$A = 4 π (1.2)^{2} = 4 π (1.44) = 5.76 π m^{2}$ Total charge: $Q = σ A$

$Q = (80 \times 10^{- 6}) (5.76 π)$

$Q \approx 1.45 \times 10^{- 3} C$

Question 1.22

An infinite line charge produces an electric field of
$9 \times 10^{4} N/C$ at a distance of 2 cm. Calculate the linear charge density.

Given
- Electric field:
  $E = 9 \times 10^{4} N/C$
- Distance from line charge:
  $r = 2 cm = 0.02 m$
- Permittivity of free space:
  $ε_{0} = 8.85 \times 10^{- 12} C^{2} / ({N m}^{2})$
Concept Used: Gauss’s Law for an Infinite Line Charge

For an infinitely long straight charged wire,

$E = \frac{λ}{2 π ε_{0} r}$

where $λ$ = linear charge density. Rearranging,

$λ = E (2 π ε_{0} r)$ Calculation

$λ = (9 \times 10^{4}) (2 π) (8.85 \times 10^{- 12}) (0.02)$

$λ = (9 \times 10^{4}) (1.11 \times 10^{- 12})$

$λ \approx 1.0 \times 10^{- 7} C/m$

Question 1.23

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude
$σ = 17.0 \times 10^{- 22} {C/m}^{2}$ .
Find the electric field $E$ :
(a) in the outer region of the first plate,
(b) in the outer region of the second plate, and
(c) between the plates.

Concept Used
- Electric field due to a single infinite sheet of charge:
$E = \frac{σ}{2 ε_{0}}$
- For two parallel plates with equal and opposite surface charge densities:
  - Outside the plates → fields cancel
  - Between the plates → fields add
Given

$σ = 17.0 \times 10^{- 22} {C/m}^{2}$

$ε_{0} = 8.85 \times 10^{- 12} C^{2} / ({N m}^{2})$

(a) Electric field outside the first plate
- Field due to the positively charged plate is canceled by the field due to the negatively charged plate.
$E = 0$

(b) Electric field outside the second plate
- Same reasoning as part (a).
$E = 0$

(c) Electric field between the plates
- Fields due to both plates are in the same direction and hence add.
$E = \frac{σ}{ε_{0}}$

Calculation $E = \frac{17.0 \times 10^{- 22}}{8.85 \times 10^{- 12}}$

$E \approx 1.92 \times 10^{- 10} N/C$
December 22, 2025

Property	Axial Line	Equatorial Line
Formula	$\frac{2 p}{4 π ε_{0} r^{3}}$	$\frac{p}{4 π ε_{0} r^{3}}$
Direction	Along $\vec{p}$	Opposite to $\vec{p}$
Relative Strength	Stronger	Weaker
Distance Dependence	$1 / r^{3}$	$1 / r^{3}$

Position	Magnitude of magnetic field	Direction
Axial line	$9.6 \times 10^{- 5} T$	Along magnetic moment
Equatorial line	$4.8 \times 10^{- 5} T$	Opposite to magnetic moment

Charges	Nature of Force
Like charges (+ + or − −)	Repulsive
Unlike charges (+ −)	Attractive

Quantity	Depends on
Electric Field	$\frac{1}{r^{2}}$
Potential	$\frac{1}{r}$

Tag: Physics for Class 12 Students

Class 12th Physics Chapter-1 Notes Potential due to a point charge

Derivation: Electrostatic Potential Due to a Point Charge

Electric force on a unit test charge

Step 2: Small work done (why the minus sign appears)

Step 3: Setting up the definite integral

Step 4: Actual integration (expanded)

2. Connecting Work Done and Potential (Key Concept)

Definition of Potential

3. Why Work and Potential Have Different Units

(a) Unit of Work Done

(b) Unit of Potential

Class 12th Physics Chapter-2 Notes on Electrostatic Potential

ELECTROSTATIC POTENTIAL

1. Why do we need Electrostatic Potential?

2. Electrostatic Potential Energy

Definition

Important Points

Mathematical Expression

3. Reference Point for Potential Energy

4. Electrostatic Potential

Definition (Most Important)

Symbol

Mathematical Definition

5. Physical Meaning of Electrostatic Potential

6. Unit of Electrostatic Potential

SI Unit: Volt (V)

7. Potential Difference

Definition

Expression

Important Notes

8. Potential at Infinity

9. Electrostatic Potential Due to a Point Charge

Expression

Nature of Potential

Key Difference from Electric Field

10. Potential Due to a System of Charges

Principle Used

Expression

11. Potential Due to a Dipole (Short Note – Pre-Equipotential)

Special Cases

12. Relation Between Potential and Work

13. Important Exam Points (Very High Yield)

Class 12th Physics Chapter-1 Notes on Gauss’s Law

Gauss’s Law

1. Charge Density (Starting Point)

(a) Linear Charge Density (λ)

(b) Surface Charge Density (σ)

(c) Volume Charge Density (ρ)

2. Electric Flux (Key Idea Behind Gauss’s Law)

3. Statement of Gauss’s Law

4. Gauss’s Law in Terms of Charge Density

5. Why Gauss’s Law Is Powerful

6. Applications of Gauss’s Law

(A) Electric Field Due to an Infinitely Long Straight Charged Wire

(B) Electric Field Due to an Infinite Plane Sheet of Charge

(C) Electric Field Due to a Uniformly Charged Spherical Shell

(i) Outside the shell (r > R):

(ii) Inside the shell (r < R):

7. Key Takeaways (Exam-Friendly)

Class 12th Physics Chapter-6 Solutions

Electromagnetic Induction – NCERT

Question 6.1

Basic Principle Used (Very Important)

Fig. 6.15 (a): North pole of magnet moving towards the coil

Analysis

Direction of induced current

Fig. 6.15 (b): North pole of magnet moving away from the coil

Analysis

Direction of induced current

Fig. 6.15 (c): South pole of magnet moving towards the coil

Analysis

Direction of induced current

Fig. 6.15 (d): South pole of magnet moving away from the coil

Analysis

Direction of induced current

Fig. 6.15 (e): Coil moving towards a stationary magnet

Analysis

Direction of induced current

Fig. 6.15 (f): Coil moving away from a stationary magnet

$B_{H} = 0.30 \times 10^{- 4} {Wb m}^{- 2}$

$ε = B l v$