Class 12th Physics Chapter-2 Solutions

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Question 2.1
Two charges $+5\times {10}^{-8}\text{C}$ and $-3\times {10}^{-8}\text{C}$ 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.

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Solution:

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

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

$$V=\frac{1}{4\pi {\epsilon }_0}(\frac{q_1}{r_1}+\frac{q_2}{r_2})$$

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

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

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

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

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

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

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

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

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

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Question 2.2

A regular hexagon of side 10 cm has a charge $5\mu \text{C}$ at each of its vertices. Calculate the electric potential at the centre of the hexagon.

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Given

• Charge at each vertex:

  $$q=5\mu \text{C}=5\times {10}^{-6}\text{C}$$

• Side of hexagon:

  $$a=10\text{cm}=0.10\text{m}$$

• Number of vertices (charges): $6$

For a regular hexagon, the distance from the centre to each vertex is equal to the side length:

$$r=a=0.10\text{m}$$

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

Potential due to one charge at the centre

$$V_1=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$$

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

$$V=6V_1=6\times \frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$$

Substitute values:

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

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

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

$$\boxed{{\displaystyle V=2.7\times {10}^6\text{ V}}}$$

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Question 2.3
Two charges $+2\mu \text{C}$ and $-2\mu \text{C}$ 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?

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(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 $+2\mu \text{C}$ and $-2\mu \text{C}$ 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.

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Question 2.4
A spherical conductor of radius 12 cm has a charge $1.6\times {10}^{-7}\text{C}$ 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.

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Given

• Radius of sphere:

  $$R=12\text{cm}=0.12\text{m}$$

• Charge on sphere:

  $$Q=1.6\times {10}^{-7}\text{C}$$

• Coulomb constant:

  $$k=\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9{\text{N m}}^2{\text{/C}}^2$$

(a) Electric field inside the sphere

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

$$\boxed{{\displaystyle E=0}}$$

(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.

$$E=\frac{kQ}{R^2}$$

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

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

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

(c) Electric field at 18 cm from the centre

Distance:

$$r=18\text{cm}=0.18\text{m}$$

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

$$E=\frac{kQ}{r^2}$$

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

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

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

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Question 2.5
A parallel plate capacitor with air between the plates has a capacitance of $8\text{pF}$. 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 $K=6$?

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Given

• Initial capacitance (air):

  $$C_0=8\text{pF}$$

• New separation:

  $$d^{\mathrm{\prime }}=\frac{d}{2}$$

• Dielectric constant:

  $$K=6$$

Key formula

For a parallel plate capacitor:

$$C=\frac{K{\epsilon }_{0}A}{d}$$

Capacitance is:

• inversely proportional to plate separation $d$

• directly proportional to dielectric constant $K$

Effect of changes

1. Halving the distance $(d\to d\mathrm{/}2)$

   $$C\to 2C$$

2. Filling dielectric of constant $K=6$

   $$C\to 6C$$

New capacitance

$$C^{\mathrm{\prime }}=2\times 6\times C_0$$

$$C^{\mathrm{\prime }}=12\times 8\text{pF}$$

$$\boxed{{\displaystyle C^{\mathrm{\prime }}=96\text{pF}}}$$

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Question 2.6
Three capacitors, each of capacitance $9\text{pF}$, 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 $120\text{V}$ supply.

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Given

• Each capacitor:

  $$C=9\text{pF}$$

• Number of capacitors: $3$

• Supply voltage:

  $$V=120\text{V}$$

(a) Total capacitance in series

For capacitors in series:

$$\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$

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

$$\boxed{{\displaystyle C_{\text{eq}}=3\text{pF}}}$$

(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.

$$V_{\text{each}}=\frac{V_{\text{total}}}{3}=\frac{120}{3}$$

$$\boxed{{\displaystyle V_{\text{each}}=40\text{V}}}$$

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

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Given

• $C_1=2\text{pF}$

• $C_2=3\text{pF}$

• $C_3=4\text{pF}$

• Applied voltage:

  $$V=100\text{V}$$

(a) Total capacitance (parallel combination)

For capacitors in parallel, capacitances add directly:

$$C_{\text{eq}}=C_1+C_2+C_3$$

$$C_{\text{eq}}=2+3+4=\boxed{{\displaystyle 9\text{pF}}}$$

(b) Charge on each capacitor

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

Using $Q=CV$:

Charge on $2\text{pF}$ capacitor

$$Q_1=C_{1}V=2\times {10}^{-12}\times 100=2\times {10}^{-10}\text{C}$$

Charge on $3\text{pF}$ capacitor

$$Q_2=3\times {10}^{-12}\times 100=3\times {10}^{-10}\text{C}$$

Charge on $4\text{pF}$ capacitor

$$Q_3=4\times {10}^{-12}\times 100=4\times {10}^{-10}\text{C}$$

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Question 2.8
In a parallel plate capacitor with air between the plates:

• Area of each plate $A=6\times {10}^{-3}{\text{m}}^2$

• Separation $d=3\text{mm}=3\times {10}^{-3}\text{m}$

(a) Calculate the capacitance.
(b) If connected to a $100\text{V}$ supply, find the charge on each plate.

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Given

• ${\epsilon }_0=8.85\times {10}^{-12}{\text{F m}}^{-1}$

• Dielectric: air ($K\approx 1$)

(a) Capacitance of a parallel plate capacitor

$$C=\frac{{\epsilon }_{0}A}{d}$$

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

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

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

or$$\boxed{{\displaystyle C=17.7\text{pF}}}$$

(b) Charge on each plate

$$Q=CV$$

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

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

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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.)

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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

  $$V=100\text{V}$$

• New capacitance:

  $$C^{\mathrm{\prime }}=KC=6C$$

• Since $Q=CV$,

  $$Q^{\mathrm{\prime }}=C^{\mathrm{\prime }}V=6CV=6Q$$

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:

  $$Q^{\mathrm{\prime }}=Q$$

• Capacitance still increases:

  $$C^{\mathrm{\prime }}=6C$$

• Since $V=Q\mathrm{/}C$,

  $$V^{\mathrm{\prime }}=\frac{Q}{C^{\mathrm{\prime }}}=\frac{V}{6}$$

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)

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Question 2.10
A $12\text{pF}$ capacitor is connected to a $50\text{V}$ battery. How much electrostatic energy is stored in the capacitor?

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Given

• Capacitance:

  $$C=12\text{pF}=12\times {10}^{-12}\text{F}$$

• Voltage:

  $$V=50\text{V}$$

Formula for energy stored in a capacitor

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

Calculation

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

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

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

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

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Given

• $C_1=600\text{pF}=600\times {10}^{-12}\text{F}$

• $C_2=600\text{pF}$ (uncharged)

• Initial voltage:

  $$V=200\text{V}$$

Step 1: Initial energy stored

Energy in a charged capacitor:

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

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

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

$$\boxed{{\displaystyle U_i=1.2\times {10}^{-5}\text{J}}}$$

Step 2: Situation after connection

• The two capacitors are identical and connected together.

• Total capacitance:

$$C_{\text{total}}=600+600=1200\text{pF}$$

• Total charge is conserved.

• Final voltage becomes half:

$$V_f=\frac{200}{2}=100\text{V}$$

Step 3: Final energy stored

$$U_f=\frac{1}{2}{C}_{\text{total}}{V}_f^2$$

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

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

$$\boxed{{\displaystyle U_f=6.0\times {10}^{-6}\text{J}}}$$

Step 4: Energy lost

$$\text{Energy lost}=U_i-U_f$$

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

$$\boxed{{\displaystyle =6.0\times {10}^{-6}\text{J}}}$$

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