Chapter-2 Electrostatic Potential and Capacitance — Online MCQ Test
PHYSICS · CLASS 12 SECOND PUC · Karnataka State Board
Practice Chapter-2 Electrostatic Potential and Capacitance with a free chapter-wise online MCQ test.
This chapter covers: electric potential - equipotential surfaces - capacitance - parallel plate capacitor - capacitor combinations - electric dipole potential.
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter-2 Electrostatic Potential and Capacitance — Important Questions & Answers
Electric potential at a point is defined as the work done by an external agent in bringing a unit positive charge from infinity to that point. What is the SI unit of electric potential?
- A. Joule per Coulomb (J/C)
- B. Newton per Coulomb (N/C)
- C. Coulomb per Joule (C/J)
- D. Ampere per meter (A/m)
Answer: A. Joule per Coulomb (J/C)
Electric potential is work per unit charge, so its SI unit is Joule per Coulomb, which is also called Volt (V).
Electric potential is work per unit charge, so its SI unit is Joule per Coulomb, which is also called Volt (V).
What is the electric potential due to a point charge q at a distance r from it?
- A. V = kq²/r
- B. V = kq/r
- C. V = kr/q
- D. V = kq/r²
Answer: B. V = kq/r
The electric potential due to a point charge is directly proportional to the charge and inversely proportional to the distance: V = kq/r, where k = 9 × 10⁹ N⋅m²/C².
The electric potential due to a point charge is directly proportional to the charge and inversely proportional to the distance: V = kq/r, where k = 9 × 10⁹ N⋅m²/C².
The work done by the electric field in moving a charge q between two points in an electric field is related to the potential difference. Which statement is true?
- A. W = q(V₂ - V₁)
- B. W = q(V₁ - V₂)
- C. W = -q(V₁ - V₂)
- D. W = q²(V₁ - V₂)
Answer: B. W = q(V₁ - V₂)
Work done by electric field W = q(V₁ - V₂), where V₁ and V₂ are potentials at initial and final points respectively.
Work done by electric field W = q(V₁ - V₂), where V₁ and V₂ are potentials at initial and final points respectively.
The electric field inside a parallel plate capacitor is uniform and equal to E = σ/ε₀, where σ is surface charge density. If the separation between plates is doubled while keeping the charge constant, what happens to the energy stored?
- A. Energy doubles
- B. Energy becomes half
- C. Energy remains constant
- D. Energy quadruples
Answer: A. Energy doubles
With constant charge, U = Q²/(2C) = Q²d/(2ε₀A). When d doubles, energy doubles because energy is directly proportional to plate separation when charge is constant.
With constant charge, U = Q²/(2C) = Q²d/(2ε₀A). When d doubles, energy doubles because energy is directly proportional to plate separation when charge is constant.
Two capacitors with capacitances C and 2C are connected in series and then connected to a battery of voltage V. The voltage across the capacitor with capacitance C is:
- A. V/3
- B. 2V/3
- C. V
- D. 3V/2
Answer: B. 2V/3
In series, both capacitors have equal charge Q. For series capacitors connected to voltage V: V = V_C + V_2C. Since Q = CV_C = 2C·V_2C, we get V_2C = V_C/2. Therefore: V_C + V_C/2 = V, so V_C = 2V/3.
In series, both capacitors have equal charge Q. For series capacitors connected to voltage V: V = V_C + V_2C. Since Q = CV_C = 2C·V_2C, we get V_2C = V_C/2. Therefore: V_C + V_C/2 = V, so V_C = 2V/3.