Exercises — Electrostatic Potential and Capacitance

1. 14 marksNCERT Exercises, Chapter 2

   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.

   Hint. Potential is a scalar, so simply add the two contributions with their signs and set the sum to zero. Check both between the charges and beyond them.

   Because potential is a scalar, the two contributions add algebraically with no direction to worry about.

   Let the $5\times 10^{-8}$ C charge sit at the origin and the $-3\times 10^{-8}$ C charge at $x=16$ cm. Setting $V=0$:

   $\frac{k(5\times 10^{-8})}{r_1}+\frac{k(-3\times 10^{-8})}{r_2}=0\Rightarrow \frac{5}{r_1}=\frac{3}{r_2}$

   Between the charges, at distance $x$ from the first, $r_1=x$ and $r_2=16-x$:

   $5(16-x)=3x$, so $80=8x$ and $x=10$ cm.

   Beyond the negative charge, at distance $x>16$ cm, $r_2=x-16$:

   $5(x-16)=3x$, so $2x=80$ and $x=40$ cm.

   There are therefore two such points. Note there is no point on the far side of the positive charge, since there its larger magnitude and shorter distance both dominate.

   ✦ Two points: 10 cm from the 5 x 10^-8 C charge (between the charges), and 40 cm from it on the far side of the negative charge

2. 23 marksNCERT Exercises, Chapter 2

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

   Hint. For a regular hexagon, the distance from the centre to each vertex equals the side length.

   The key geometric fact is that a regular hexagon is made of six equilateral triangles, so the distance from the centre to each vertex equals the side, here $0.10$ m.

   All six charges are identical and equidistant from the centre, and potential is a scalar, so the contributions simply add:

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

   $V=6\times 4.5\times 10^5=2.7\times 10^6\text{V}$

   Because potential is scalar, no vector resolution is needed — contrast this with the field at the centre, which is zero by symmetry.

   ✦ V = 2.7 x 10^6 V

3. 33 marksNCERT Exercises, Chapter 2

   Two charges $2\mu$C and $-2\mu$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?

   Hint. An equipotential surface is one on which the potential is the same everywhere; look for the locus where the two contributions cancel.

   (a) The two charges are equal in magnitude and opposite in sign, so at any point equidistant from both, their potentials cancel exactly:

   $V=\frac{kq}{r}+\frac{k(-q)}{r}=0$

   The locus of points equidistant from A and B is the plane perpendicular to AB passing through its midpoint — the perpendicular bisector plane. Every point on it has $V=0$, so it is an equipotential surface.

   (b) The electric field is always perpendicular to an equipotential surface. If it had a component along the surface, work would be done moving a charge between two points at the same potential, which is impossible.

   So the field is normal to this plane at every point on it, directed along AB from the positive charge A towards the negative charge B.

   ✦ (a) The plane perpendicular to AB through its midpoint, on which V = 0 (b) Normal to that plane at every point, directed from A to B

4. 43 marksNCERT Exercises, Chapter 2

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

   Hint. Apply Gauss's law in each region, noting how much charge is enclosed.

   Gauss's law settles all three parts once the enclosed charge is identified in each region.

   (a) Inside the conductor a Gaussian surface encloses no charge, since all the charge resides on the outer surface. Hence $E=0$.

   This is true throughout the interior of any charged conductor in electrostatic equilibrium.

   (b) Just outside, the whole charge is enclosed and the sphere behaves as a point charge at its centre, with $r=R=0.12$ m:

   $E=\frac{kq}{R^2}=\frac{9\times 10^9\times 1.6\times 10^{-7}}{(0.12{)}^2}=\frac{1440}{0.0144}\approx 1.0\times 10^5\text{N/C}$

   (c) At $r=0.18$ m the same point-charge formula applies:

   $E=\frac{9\times 10^9\times 1.6\times 10^{-7}}{(0.18{)}^2}=\frac{1440}{0.0324}\approx 4.4\times 10^4\text{N/C}$

   ✦ (a) E = 0 (b) E = 1.0 x 10^5 N/C (c) E = 4.4 x 10^4 N/C

5. 53 marksNCERT Exercises, Chapter 2

   A parallel plate capacitor with air between the plates has a capacitance of 8 pF ($1$ pF $=10^{-12}$ 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?

   Hint. Track how each change scales C separately, rather than recomputing from the area.

   Reason by scaling rather than recomputing, since the plate area is never given.

   For a parallel plate capacitor $C=\frac{{\epsilon }_{0}A}{d}$, and with a dielectric $C=\frac{K{\epsilon }_{0}A}{d}$.

   Halving $d$ doubles the capacitance, because $C$ is inversely proportional to the separation.

   Introducing a dielectric of constant $K=6$ multiplies the capacitance by 6.

   The two effects combine multiplicatively:

   $C^{\prime }=8\times 2\times 6=96\text{pF}$

   The area cancels out entirely, which is why the question can be answered without it.

   ✦ C' = 96 pF

6. 63 marksNCERT Exercises, Chapter 2

   Three 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?

   Hint. In series the reciprocals add, and the charge is the same on every capacitor.

   (a) In series the reciprocals of the capacitances add:

   $\frac{1}{C_s}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}\Rightarrow C_s=3\text{pF}$

   Note the series combination is smaller than any individual capacitor, which is always the case.

   (b) In series every capacitor carries the same charge, and since the three capacitances are equal, the supply voltage divides equally:

   $V=\frac{120}{3}=40\text{V across each}$

   As a check, the three potential differences sum to $120$ V as they must.

   ✦ (a) 3 pF (b) 40 V across each capacitor

7. 73 marksNCERT Exercises, Chapter 2

   Three 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.

   Hint. In parallel the capacitances add directly, and every capacitor has the same potential difference.

   (a) In parallel the capacitances add directly:

   $C_p=2+3+4=9\text{pF}$

   The parallel combination is larger than any individual capacitor, the opposite of the series case.

   (b) In parallel every capacitor has the same potential difference, here the full $100$ V. Using $Q=CV$ for each:

   $Q_1=2\times 10^{-12}\times 100=2\times 10^{-10}\text{C}$ $Q_2=3\times 10^{-12}\times 100=3\times 10^{-10}\text{C}$ $Q_3=4\times 10^{-12}\times 100=4\times 10^{-10}\text{C}$

   The charges are in the ratio of the capacitances, and their total is $9\times 10^{-10}$ C, consistent with $C_{p}V$.

   The charges differ because each capacitor stores according to its own capacitance, since the voltage across them is common.

   ✦ (a) 9 pF (b) 2 x 10^-10 C, 3 x 10^-10 C and 4 x 10^-10 C respectively

8. 83 marksNCERT Exercises, Chapter 2

   In a parallel plate capacitor with air between the plates, each plate has an area of $6\times 10^{-3}$ m${}^2$ 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?

   Hint. Convert the plate separation from millimetres to metres before substituting.

   Convert the separation first: $3$ mm $=3\times 10^{-3}$ m. Leaving it in millimetres is the usual slip here.

   For a parallel plate capacitor with air between the plates:

   $C=\frac{{\epsilon }_{0}A}{d}=\frac{8.854\times 10^{-12}\times 6\times 10^{-3}}{3\times 10^{-3}}$

   $C=8.854\times 10^{-12}\times 2=1.77\times 10^{-11}\text{F}\approx 17.7\text{pF}$

   The charge on each plate then follows from $Q=CV$:

   $Q=1.77\times 10^{-11}\times 100=1.77\times 10^{-9}\text{C}$

   The two plates carry equal and opposite charges of this magnitude.

   The capacitance is small because the plate separation is large relative to the area, which is why practical capacitors use very thin gaps.

   ✦ C = 1.77 x 10^-11 F (about 17.7 pF); Q = 1.77 x 10^-9 C on each plate

9. 94 marksNCERT Exercises, Chapter 2

   Explain 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.

   Hint. Decide in each case which quantity is held fixed — the voltage or the charge — and let the rest follow.

   In both cases the dielectric multiplies the capacitance by $K=6$, giving $C^{\prime }=6\times 17.7=106$ pF. What differs is which quantity stays fixed.

   (a) Supply still connected. The battery holds the potential difference at $V=100$ V.

   Since $Q=CV$ and $C$ has increased sixfold while $V$ is unchanged, the charge increases sixfold:

   $Q^{\prime }=6\times 1.77\times 10^{-9}\approx 1.06\times 10^{-8}\text{C}$

   The extra charge is supplied by the battery.

   (b) Supply disconnected. Now the charge is trapped on the plates and cannot change, so $Q$ stays at $1.77\times 10^{-9}$ C.

   Since $V=Q/C$ and $C$ has increased sixfold, the voltage falls to one sixth:

   $V^{\prime }=\frac{100}{6}\approx 16.7\text{V}$

   Deciding which quantity is held constant is the whole content of this question.

   ✦ (a) V stays 100 V, C becomes about 106 pF and Q rises sixfold to 1.06 x 10^-8 C (b) Q stays 1.77 x 10^-9 C, C becomes about 106 pF and V falls to about 16.7 V

10. 102 marksNCERT Exercises, Chapter 2

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

    Hint. Use the energy stored in a charged capacitor, taking care to square the voltage.

    The energy stored in a charged capacitor is:

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

    Substituting $C=12\times 10^{-12}$ F and $V=50$ V:

    $U=\frac{1}{2}\times 12\times 10^{-12}\times (50{)}^2=\frac{1}{2}\times 12\times 10^{-12}\times 2500$

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

    Remember to square the voltage — halving the supply voltage would quarter the stored energy, not halve it.

    The voltage is squared because energy accumulates as charge is added against a rising potential difference, so the two contributions compound.

    ✦ U = 1.5 x 10^-8 J

11. 115 marksNCERT Exercises, Chapter 2

    A 600 pF capacitor is charged by a 200 V 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?

    Hint. Charge is conserved when the capacitors are joined, but energy is not. Find the common final voltage first.

    Once disconnected, the charge is conserved but the energy is not — that asymmetry is the whole point of the question.

    Initial state. $Q=CV=600\times 10^{-12}\times 200=1.2\times 10^{-7}$ C, and

    $U_i=\frac{1}{2}C{V}^2=\frac{1}{2}\times 600\times 10^{-12}\times (200{)}^2=1.2\times 10^{-5}\text{J}$

    After connection. The charge redistributes over both capacitors, giving a total capacitance of $1200$ pF while the charge stays at $1.2\times 10^{-7}$ C. The common voltage is:

    $V_f=\frac{Q}{C_{\text{total}}}=\frac{1.2\times 10^{-7}}{1200\times 10^{-12}}=100\text{V}$

    $U_f=\frac{1}{2}\times 1200\times 10^{-12}\times (100{)}^2=6\times 10^{-6}\text{J}$

    Energy lost $=1.2\times 10^{-5}-6\times 10^{-6}=6\times 10^{-6}$ J, exactly half the original energy.

    The missing energy is dissipated as heat in the connecting wires and as electromagnetic radiation during the transient current. Energy is not destroyed, merely converted out of electrostatic form.

    Exactly half is lost because the two capacitors are equal, which halves the voltage while conserving the charge.

    ✦ Energy lost = 6 x 10^-6 J, which is half the initial energy, dissipated as heat in the connecting wires

Solutions written by the tuition.in editorial team and checked against leph102.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.