Ch-2: Electrostatic Potential and Capacitance25 Questions
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Explanation: Electric potential is defined as work done per unit charge (V = W/q), which has the SI unit Joule/Coulomb, commonly known as Volt (V).
2. Electric potential is a:
A. Scalar quantity
B. Vector quantity
C. Tensor quantity
D. Dimensionless quantity
Answer: A. Scalar quantity
Explanation: Electric potential is defined as work done per unit charge. Since both work and charge are scalars, electric potential is a scalar quantity.
3. Work done in moving a test charge over an equipotential surface is:
A. Maximum
B. Minimum
C. Zero
D. Infinite
Answer: C. Zero
Explanation: On an equipotential surface, potential difference (ΔV) between any two points is zero. Since W = q × ΔV, the work done is zero.
4. The angle between an equipotential surface and the electric field lines at any point is always:
A. 0°
B. 45°
C. 90°
D. 180°
Answer: C. 90°
Explanation: Electric field lines are always perpendicular (at 90°) to equipotential surfaces because no work is done moving a charge along the surface.
5. Electric potential due to a point charge 'q' at a distance 'r' varies as:
A. Directly proportional to r
B. Inversely proportional to r
C. Inversely proportional to r²
D. Independent of r
Answer: B. Inversely proportional to r
Explanation: The potential due to a single point charge is given by V = q / (4πε₀r), showing that V ∝ 1/r.
6. What is the electric potential at any point on the equatorial plane of an electric dipole?
A. Maximum
B. Minimum
C. Zero
D. Infinite
Answer: C. Zero
Explanation: Every point on the equatorial plane is equidistant from both +q and -q charges, so the net potential V = V_+ + V_- = 0.
7. The relation between electric field 'E' and electric potential 'V' is given by:
A. E = -dV/dr
B. E = dV/dr
C. V = -dE/dr
D. E = V × r
Answer: A. E = -dV/dr
Explanation: Electric field is equal to the negative potential gradient (E = -dV/dr), indicating that field points in the direction of decreasing potential.
8. What is the electric potential inside a hollow charged spherical conductor of radius R?
A. Zero
B. Same as on its surface
C. Half of surface potential
D. Infinite
Answer: B. Same as on its surface
Explanation: Since electric field inside a hollow charged conductor is zero (E = 0), dV/dr = 0, meaning potential V remains constant and equal to its surface value everywhere inside.
9. Electrostatic potential energy of a system of two point charges q₁ and q₂ separated by distance 'r' is:
A. U = (1 / 4πε₀) * (q₁q₂ / r)
B. U = (1 / 4πε₀) * (q₁q₂ / r²)
C. U = (1 / 4πε₀) * (q₁ + q₂ / r)
D. U = zero
Answer: A. U = (1 / 4πε₀) * (q₁q₂ / r)
Explanation: Potential energy U is equal to work done in bringing charges from infinity to distance r, given by U = (q₁q₂) / (4πε₀r).
10. What is the potential energy of an electric dipole placed parallel (θ = 0°) to a uniform electric field?
A. U = +pE
B. U = -pE
C. U = 0
D. U = pE / 2
Answer: B. U = -pE
Explanation: Dipole potential energy U = -pE cosθ. For θ = 0°, cos 0° = 1, giving minimum potential energy U = -pE (stable equilibrium).
11. An electric dipole is in most unstable equilibrium in a uniform electric field when angle θ between 'p' and 'E' is:
A. 0°
B. 90°
C. 180°
D. 360°
Answer: C. 180°
Explanation: At θ = 180°, potential energy U = -pE cos(180°) = +pE, which is maximum, making it the most unstable equilibrium state.
12. What is the SI unit of Electrical Capacitance?
A. Volt
B. Farad
C. Ohm
D. Henry
Answer: B. Farad
Explanation: Capacitance C = Q/V has the SI unit Coulomb per Volt, which is defined as Farad (F).
13. The capacitance of an isolated spherical conductor of radius 'R' is given by:
A. C = 4πε₀ R
B. C = 4πε₀ / R
C. C = R / (4πε₀)
D. C = 2πε₀ R
Answer: A. C = 4πε₀ R
Explanation: Since V = Q / (4πε₀R) on a sphere, capacitance C = Q/V = 4πε₀ R, showing capacitance is directly proportional to radius.
14. The capacitance of a parallel plate capacitor with plate area 'A' and separation 'd' in vacuum is:
A. C = ε₀ A / d
B. C = ε₀ d / A
C. C = A d / ε₀
D. C = 2ε₀ A / d
Answer: A. C = ε₀ A / d
Explanation: Capacitance of a parallel plate capacitor in vacuum is C₀ = ε₀ A / d.
15. When a dielectric medium of dielectric constant 'K' is completely filled between the plates of a capacitor, its capacitance:
A. Decreases K times
B. Increases K times
C. Remains unchanged
D. Becomes zero
Answer: B. Increases K times
Explanation: Inserting a dielectric increases the capacitance to C = K × C₀, where K is the dielectric constant of the material.
16. Three capacitors each of capacitance 'C' are connected in series. Their equivalent capacitance is:
A. 3C
B. C / 3
C. C / 2
D. 9C
Answer: B. C / 3
Explanation: In series, 1/C_eq = 1/C + 1/C + 1/C = 3/C, which gives C_eq = C / 3.
17. Three capacitors each of capacitance 'C' are connected in parallel. Their equivalent capacitance is:
A. 3C
B. C / 3
C. C
D. C / 9
Answer: A. 3C
Explanation: In parallel connection, capacitances simply add up: C_eq = C₁ + C₂ + C₃ = C + C + C = 3C.
18. The energy stored in a charged capacitor of capacitance 'C' carrying charge 'Q' and potential 'V' is given by:
A. U = 1/2 C V²
B. U = 1/2 Q V
C. U = Q² / (2C)
D. All of the above
Answer: D. All of the above
Explanation: Stored energy formulas are equivalent: U = 1/2 C V² = 1/2 Q V = Q² / (2C).
19. Energy density (energy stored per unit volume) in an electric field 'E' in free space is:
A. u = 1/2 ε₀ E²
B. u = ε₀ E²
C. u = 1/2 ε₀ E
D. u = 2 ε₀ E²
Answer: A. u = 1/2 ε₀ E²
Explanation: The energy stored per unit volume in an electric field is given by u = 1/2 ε₀ E².
20. What happens to the energy stored in a capacitor when the distance between its plates is doubled keeping battery connected?
A. Doubles
B. Halves
C. Quadruples
D. Remains same
Answer: B. Halves
Explanation: Since battery is connected, V is constant. C = ε₀A/d, so doubling d halves C. Energy U = 1/2 C V² therefore becomes half.
21. The Earth is assumed to be a spherical conductor of radius 6400 km. Its capacitance is approximately:
A. 711 μF
B. 1 Farad
C. 9 × 10⁹ F
D. 6400 μF
Answer: A. 711 μF
Explanation: C = 4πε₀ R = (6.4 × 10⁶ m) / (9 × 10⁹) ≈ 7.11 × 10^-4 F = 711 μF.
22. During the sharing of charges between two charged conductors, there is always a loss of:
A. Charge
B. Electrostatic Energy
C. Mass
D. Both Charge and Energy
Answer: B. Electrostatic Energy
Explanation: Total charge remains conserved during charge sharing, but some electrostatic energy is always lost as heat and light during current flow.
23. Dielectric strength of a dielectric medium is defined as:
A. Maximum electric field it can withstand without breakdown
B. Minimum potential applied to break it
C. Amount of charge stored per unit area
D. Its capacitance value
Answer: A. Maximum electric field it can withstand without breakdown
Explanation: Dielectric strength is the maximum electric field an insulating material can tolerate before breakdown and becoming conducting.
24. What is the capacitance of a parallel plate capacitor if a thin conducting plate of same area is introduced between plates?
A. Decreases
B. Increases
C. Remains constant
D. Becomes zero
Answer: B. Increases
Explanation: Introducing a conducting slab effectively reduces the effective distance 'd' between plates, thereby increasing capacitance (C = ε₀A / (d - t)).
25. A 10 μF capacitor is charged to 50 V. The charge on the capacitor is:
A. 5 × 10^-4 C
B. 500 C
C. 5 C
D. 5 × 10^-3 C
Answer: A. 5 × 10^-4 C
Explanation: Q = C × V = (10 × 10^-6 F) × (50 V) = 500 × 10^-6 C = 5 × 10^-4 C.
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