CBSE · NCERT · Class 12 Physics · Chapter 2

NCERT Solutions: Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance

11 textbook Q&A11 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Electrostatic Potential and Capacitance, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 11
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1Exercises11 questions
Q.2.1Two charges 5 × 10–8 C and –3 × 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.v
Solution

Place $+5\times10^{-8}\,\text{C}$ at $x=0$ and $-3\times10^{-8}\,\text{C}$ at $x=0.16\,\text{m}$. For zero potential, $5/r_1=3/r_2$. Between the charges, $r_1=x$ and $r_2=0.16-x$, so $5(0.16-x)=3x$, giving $x=0.10\,\text{m}$. To the right of the negative charge, $r_1=x$ and $r_2=x-0.16$, so $5(x-0.16)=3x$, giving $x=0.40\,\text{m}$. No solution lies to the left of the positive charge.

Answer:

The potential is zero at a point between the charges, $10\,\text{cm}$ from the $+5\times10^{-8}\,\text{C}$ charge, and at a point outside on the side of the negative charge, $40\,\text{cm}$ from the positive charge.

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Q.2.2A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.v
Solution

For a regular hexagon, the distance from the centre to each vertex equals the side, $r=0.10\,\text{m}$. Potential is a scalar, so $V=6(kq/r)=6(9\times10^9)(5\times10^{-6})/0.10=2.7\times10^6\,\text{V}$.

Answer:

$2.7\times10^6\,\text{V}$.

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Q.2.3Two charges 2 μC and –2 μC 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?v
Solution

For any point on the perpendicular bisector plane of AB, the distances from $+2\,\mu\text{C}$ and $-2\,\mu\text{C}$ are equal, so their potentials cancel. Hence the plane is an equipotential surface. Since electric field is normal to an equipotential surface, its direction is along AB, from the positive charge at A to the negative charge at B.

Answer:

(a) The plane perpendicular to AB through its midpoint is an equipotential surface of zero potential. (b) The electric field is perpendicular to this plane and directed from the positive charge towards the negative charge.

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Q.2.4A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?v
Solution

Inside a conductor in electrostatic equilibrium, $E=0$. Outside a charged spherical conductor, the field is as if all charge were at the centre. Just outside, $r=0.12\,\text{m}$, so $E=kq/r^2=9\times10^9(1.6\times10^{-7})/(0.12)^2=1.0\times10^5\,\text{N C}^{-1}$. At $r=0.18\,\text{m}$, $E=9\times10^9(1.6\times10^{-7})/(0.18)^2=4.4\times10^4\,\text{N C}^{-1}$.

Answer:

(a) $0$; (b) $1.0\times10^5\,\text{N C}^{-1}$, radially outward; (c) $4.4\times10^4\,\text{N C}^{-1}$, radially outward.

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Q.2.5A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?v
Solution

For a parallel-plate capacitor, $C\propto K/d$. Initially $K\approx1$ and $C=8\,\text{pF}$. Reducing $d$ to $d/2$ doubles the capacitance, and inserting dielectric $K=6$ multiplies it by 6. Thus $C'=12C=96\,\text{pF}$.

Answer:

$96\,\text{pF}$.

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Q.2.6Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?v
Solution

For three equal capacitors $C$ in series, $C_{\text{eq}}=C/3=9/3=3\,\text{pF}$. The charge on each capacitor in series is the same, and since the capacitances are equal, the voltage divides equally. Therefore each capacitor has $120/3=40\,\text{V}$ across it.

Answer:

(a) $3\,\text{pF}$; (b) $40\,\text{V}$ across each capacitor.

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Q.2.7Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.v
Solution

In parallel, capacitances add: $C_{\text{eq}}=2+3+4=9\,\text{pF}$. Each capacitor has the same potential difference, $100\,\text{V}$. Thus $Q_1=2\,\text{pF}\times100\,\text{V}=200\,\text{pC}$, $Q_2=300\,\text{pC}$ and $Q_3=400\,\text{pC}$.

Answer:

(a) $9\,\text{pF}$; (b) charges are $200\,\text{pC}$, $300\,\text{pC}$ and $400\,\text{pC}$ respectively.

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Q.2.8In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?v
Solution

For air, $C=\epsilon_0A/d=(8.854\times10^{-12})(6\times10^{-3})/(3\times10^{-3})=1.77\times10^{-11}\,\text{F}=17.7\,\text{pF}$. With $V=100\,\text{V}$, $Q=CV=1.77\times10^{-11}\times100=1.77\times10^{-9}\,\text{C}$.

Answer:

$17.7\,\text{pF}$; charge on each plate has magnitude $1.77\times10^{-9}\,\text{C}$.

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Q.2.9Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.v
Solution

The mica fills the full separation, so the capacitance becomes $C'=KC=6C=106\,\text{pF}$. (a) If the battery remains connected, $V$ stays $100\,\text{V}$, so $Q'=C'V=6CV=1.06\times10^{-8}\,\text{C}$; extra charge flows from the battery. (b) If the battery is disconnected before insertion, $Q$ remains $1.77\times10^{-9}\,\text{C}$, while $C$ becomes $6C$. Hence $V'=Q/C'=V/6=16.7\,\text{V}$ and the field and stored energy decrease.

Answer:

(a) With the supply connected, the voltage remains $100\,\text{V}$ and the capacitance and charge become 6 times larger. (b) With the supply disconnected, charge remains fixed, capacitance becomes 6 times larger, and voltage falls to one-sixth of its original value.

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Q.2.10A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?v
Solution

The stored energy is $U=\frac12CV^2$. With $C=12\times10^{-12}\,\text{F}$ and $V=50\,\text{V}$, $U=\frac12(12\times10^{-12})(50)^2=1.5\times10^{-8}\,\text{J}$.

Answer:

$1.5\times10^{-8}\,\text{J}$.

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

Initial energy is $U_i=\frac12CV^2=\frac12(600\times10^{-12})(200)^2=1.2\times10^{-5}\,\text{J}$. After connection to an identical uncharged capacitor, charge is shared equally and the final voltage is $100\,\text{V}$. The equivalent capacitance is $1200\,\text{pF}$, so $U_f=\frac12(1200\times10^{-12})(100)^2=6.0\times10^{-6}\,\text{J}$. Energy lost $=U_i-U_f=6.0\times10^{-6}\,\text{J}$.

Answer:

$6.0\times10^{-6}\,\text{J}$.

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