Tag: Field Lines

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-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 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

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

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

Tag: Field Lines

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-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)

1️⃣ Electric Dipole (Revision)

2️⃣ Electric Dipole Moment (𝒑)

Definition:

ELECTRIC FIELD DUE TO AN ELECTRIC DIPOLE (DERIVATIONS)

🔵 Case I: Electric Field on the Axial Line

🔹 Geometry:

🔹 Step 1: Electric field due to +q at point P

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

🔹 Step 3: Net electric field

🔹 Step 4: Simplification

✅ Final Result (Axial Line):

🔹 Direction:

🟢 Case II: Electric Field on the Equatorial Line

🔹 Geometry:

🔹 Step 1: Field due to +q

🔹 Step 2: Field due to –q

🔹 Step 3: Resolution of Fields