Chapter Exam

Chapter-2 Electrostatic Potential and Capacitance — Online MCQ Test

Q: 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?

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).

Q: What is the electric potential due to a point charge q at a distance r from it?

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.

Q: 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?

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.

Q: 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?

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.

Q: 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:

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.

PHYSICS · CLASS 12 SECOND PUC · Karnataka State Board

Practice Chapter-2 Electrostatic Potential and Capacitance with a free chapter-wise online MCQ test for Karnataka State Board CLASS 12 SECOND PUC PHYSICS. 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.

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Chapter-2 Electrostatic Potential and Capacitance — Important Questions & Answers (FAQ)

Frequently asked questions from Karnataka State Board CLASS 12 SECOND PUC PHYSICS — Chapter-2 Electrostatic Potential and Capacitance, with answers and explanations. These are sample questions; the exam has its own separate question set.

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).

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 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.

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.

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.