Class 12 Electrostatic Potential and Capacitance Mock Test (CBSE)

Topics in Electrostatic Potential and Capacitance

• Electrostatic potential
• Potential due to a point charge
• Potential due to an electric dipole
• Potential due to a system of charges
• Equipotential surfaces
• Potential energy of a system of charges
• Potential energy in an external field
• Electrostatics of conductors
• Dielectrics and polarisation
• Capacitors and capacitance
• The parallel plate capacitor
• Effect of dielectric on capacitance
• Combination of capacitors
• Energy stored in a capacitor

Electrostatic Potential and Capacitance practice questions with answers

1. Q1Multiple choice·1 mark
   Four equal point charges, each of magnitude $q$, are placed at the four corners of a square of side $a$. The total electric potential at the centre of the square is:

   1. (a)$\frac{2q}{\pi\varepsilon_0 a}$
   2. (b)$\frac{4\sqrt{2}q}{\pi\varepsilon_0 a}$
   3. (c)$\frac{4q}{\pi\varepsilon_0 a}$
   4. (d)$\frac{\sqrt{2}q}{\pi\varepsilon_0 a}$
   Show answer and step marks

   1. Correct option: (d) $\frac{\sqrt{2}q}{\pi\varepsilon_0 a}$. The distance of each corner from the centre of the square of side $a$ is $r = \frac{a}{\sqrt{2}}$. The total potential is $V = 4 \times \frac{1}{4\pi\varepsilon_0} \frac{q}{(a/\sqrt{2})} = \frac{\sqrt{2}q}{\pi\varepsilon_0 a}$.+1

   Answer: (d)
2. Q2Multiple choice·1 mark
   The electric potential at a distance $r = 1\text{ m}$ from the centre of a short electric dipole of dipole moment $p = 1 \times 10^{-7}\text{ C m}$ along a line making an angle of $60^\circ$ with the dipole axis is: (Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$)

   1. (a)$225\text{ V}$
   2. (b)$900\text{ V}$
   3. (c)$450\text{ V}$
   4. (d)$1800\text{ V}$
   Show answer and step marks

   1. Correct option: (c) $450\text{ V}$. Using the formula $V = \frac{1}{4\pi\varepsilon_0} \frac{p \cos\theta}{r^2}$, substituting $p = 1 \times 10^{-7}\text{ C m}$, $r = 1\text{ m}$, $\theta = 60^\circ$, and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$ gives $V = 9 \times 10^9 \times \frac{10^{-7} \times 0.5}{1^2} = 450\text{ V}$.+1

   Answer: (c)
3. Q3Short answer·2 marks
   Why is the work done in moving a test charge from one point to another on an equipotential surface zero? Hence, show that the electric field is always directed normal to the equipotential surface at any point.
   Show answer and step marks

   1. On an equipotential surface $V$ is the same at every point, so $\Delta V = 0$ between any two points; since $W = q\Delta V$, the work done is $W = 0$.+1
   2. For any small displacement $d\vec{l}$ along the surface, $dV = -\vec{E} \cdot d\vec{l} = -E\, dl \cos\theta = 0$; since $E \ne 0$ and $dl \ne 0$, $\cos\theta = 0$, so $\theta = 90^\circ$ and $\vec{E}$ is normal to the surface at every point.+1

   Answer: $V$ is constant on the surface, so $W = q\Delta V = 0$; for any displacement along it $dV = -\vec{E} \cdot d\vec{l} = 0$, so $\vec{E}$ has no component along the surface and is normal to it.

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