Class 12 Physics - Electrostatic Potential & Capacitance MCQs | Set - 01
Electrostatics 50 Important MCQs
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📗 Top 50 MCQs
1 Mark EachSolution:
Correct Option: (B)
Electric Potential \( V = \frac{\text{Work}}{\text{Charge}} = \frac{[M L^2 T^{-2}]}{[A T]} = [M L^2 T^{-3} A^{-1}] \).
Solution:
Correct Option: (B)
Electric field is the negative gradient of electric potential. So, \( E = -\frac{dV}{dx} \).
Solution:
Correct Option: (B)
Work done \( W = q \times \Delta V = 2 \text{ C} \times 5 \text{ V} = 10 \text{ J} \).
Solution:
Correct Option: (B)
Any point on the equatorial line is equidistant from both \( +q \) and \( -q \). Thus, potentials due to both charges cancel out, resulting in zero potential.
Solution:
Correct Option: (D)
Inside a hollow conducting sphere, the electric field is zero (\( E = 0 \)). Hence, the potential is constant and equals the potential on its surface.
Solution:
Correct Option: (C)
Charge resides only on the outer surface of a conductor. Since both have the same outer radius, their capacitance is same (\( C = 4 \pi \epsilon_0 R \)). Since \( Q = CV \), both will have equal charge.
Solution:
Correct Option: (C)
Potential difference between any two points on an equipotential surface is zero (\( \Delta V = 0 \)). Therefore, work done \( W = q \Delta V = 0 \).
Solution:
Correct Option: (B)
For a point charge, potential \( V = \frac{kQ}{r} \). Points with equal \( r \) form concentric spheres around the charge.
Solution:
Correct Option: (A)
\( E_x = -\frac{\partial V}{\partial x} = -8x \). At \( x = 1 \), \( E_x = -8 \text{ V/m} \). The field is \( 8 \text{ V/m} \) along the negative x-axis.
Solution:
Correct Option: (B)
Potential energy \( U = -pE \cos\theta = -\vec{p} \cdot \vec{E} \).
Solution:
Correct Option: (B)
Battery is disconnected, so charge \( Q \) is constant. Distance \( d \) increases, so capacitance \( C = \frac{\epsilon_0 A}{d} \) decreases. Since \( V = \frac{Q}{C} \), voltage \( V \) increases.
Solution:
Correct Option: (B)
Capacitance with dielectric is \( C' = \frac{K \epsilon_0 A}{d} = K C \).
Solution:
Correct Option: (C)
The energy stored is given by the formula \( U = \frac{1}{2} C V^2 \).
Solution:
Correct Option: (B)
In series, \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \), which simplifies to \( C_s = \frac{C_1 C_2}{C_1 + C_2} \).
Solution:
Correct Option: (C)
Two \( 4 \, \mu\text{F} \) in series give \( 2 \, \mu\text{F} \). Connecting the third \( 4 \, \mu\text{F} \) in parallel with them gives \( 2 + 4 = 6 \, \mu\text{F} \).
Solution: Correct Option: (C)
Farad (F) is the SI unit of capacitance. \( 1 \text{ F} = 1 \text{ Coulomb/Volt} \).
Solution: Correct Option: (D)
For conductors (metals), the dielectric constant \( K \) is infinity because the induced electric field completely cancels the external electric field inside the metal.
Solution: Correct Option: (B)
\( C = \frac{\epsilon_0 A}{d} \). If \( d \) becomes \( d/2 \), \( C \) becomes \( 2C \) (Doubles).
Solution: Correct Option: (B)
Work done \( W = pE(\cos\theta_1 - \cos\theta_2) \). Stable equilibrium \( \theta_1 = 0^\circ \), unstable \( \theta_2 = 180^\circ \). \( W = pE(1 - (-1)) = 2pE \).
Solution: Correct Option: (D)
Capacitors are used in fan motors for starting, in camera flashes for sudden energy release, and in radios for frequency tuning.
Solution: Correct Option: (B)
Energy \( = qV = e \times 1 \text{ V} = 1.6 \times 10^{-19} \text{ C} \times 1 \text{ V} = 1.6 \times 10^{-19} \text{ J} \).
Solution: Correct Option: (C)
\( V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} \). So, \( V \propto 1/r \).
Solution: Correct Option: (A)
Energy density (energy per unit volume) in an electric field is \( u = \frac{1}{2} \epsilon_0 E^2 \).
Solution: Correct Option: (B)
In parallel, voltage \( V \) is same for both. Since \( Q = CV \), \( \frac{Q_1}{Q_2} = \frac{C_1 V}{C_2 V} = \frac{C_1}{C_2} \).
Solution: Correct Option: (A)
For an isolated spherical conductor, \( C = 4 \pi \epsilon_0 R \).
Solution: Correct Option: (B)
Volume of big drop = \( n \times \) Volume of small drop. So \( R = n^{1/3} r \). Since \( C \propto R \), \( C_{\text{big}} = n^{1/3} C_{\text{small}} \).
Solution: Correct Option: (B)
Polarization creates bound charges that produce an induced electric field which always opposes the applied external field, reducing the net field inside.
Solution: Correct Option: (A)
In polar molecules, the center of positive and negative charges do not coincide, leading to a permanent dipole moment (e.g., \( \text{H}_2\text{O} \)).
Solution: Correct Option: (A)
The standard derivation yields \( C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}} \).
Solution: Correct Option: (D)
Since the capacitor is isolated (battery disconnected), there is no path for the charge to flow. Hence, charge \( Q \) remains conserved (unchanged).
Solution: Correct Option: (B)
All charge given to a conductor resides on its outer surface. Hence, charge inside is zero.
Solution: Correct Option: (D)
Earth is a huge conductor and taking or giving charge does not significantly affect its state, so its potential is conventionally taken as zero.
Solution: Correct Option: (A)
Since the battery is connected, voltage \( V \) is constant. Capacitance increases (\( C' = KC \)). Energy \( U = \frac{1}{2} C V^2 \) will increase.
Solution: Correct Option: (A)
The electric field is along the x-axis, and equipotential surfaces are always perpendicular to the electric field lines, i.e., parallel to the yz-plane.
Solution: Correct Option: (B)
Formula for potential energy is \( U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} \).
Solution: Correct Option: (A)
\( p = q \times 2a = 3 \times 10^{-6} \text{ C} \times 2 \times 10^{-3} \text{ m} = 6 \times 10^{-9} \text{ C-m} \).
Solution: Correct Option: (B)
Capacitive reactance \( X_c = \frac{1}{\omega C} = \frac{1}{2\pi f C} \). For DC, \( f = 0 \), so \( X_c = \infty \), effectively blocking DC.
Solution: Correct Option: (C)
Electric field lines cannot exist inside a conductor, and they must always be perpendicular (normal) to the surface of a conductor.

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