Electrostatic Potential and Capacitance Class 12 Physics Important Questions 2026-27

Class 12 Physics • Chapter 2

Electrostatic Potential and Capacitance — Important Questions with Answers

A deep, syllabus-aligned practice set for CBSE Class 12 Physics 2026–27 covering electric potential, potential energy, equipotential surfaces, conductors, dielectrics, capacitance, capacitor combinations and stored energy.

Important note: These are original, exam-oriented questions prepared for practice. They are not presented as leaked, predicted, or copied CBSE board questions. Use the official CBSE curriculum and current Sample Question Paper/Marking Scheme as the final reference for the examination.

What This Chapter Covers in CBSE 2026–27

The current CBSE Class XII Physics syllabus places Electric Charges and Fields and Electrostatic Potential and Capacitance together in Unit I: Electrostatics, which carries 16 marks at the unit level. CBSE assigns marks to the unit rather than giving a fixed public chapter-wise weightage for Chapter 2 alone.

• Electric potential and potential difference
• Potential due to a point charge, dipole and system of charges
• Equipotential surfaces
• Potential energy of two charges and of a dipole in an electrostatic field
• Conductors, insulators, free charges and bound charges
• Dielectrics and electric polarisation
• Capacitors and capacitance
• Capacitors in series and parallel
• Parallel-plate capacitor with and without dielectric
• Energy stored in a capacitor — formulae only in the current syllabus

How to Use These Important Questions

Do not treat an “important question” list as a prediction of the board paper. Use it as a structured coverage check. First solve without looking at the answer, then compare your method, units, signs and final expression. For numerical questions, write the governing formula before substitution.

Section A — 1-Mark Conceptual Questions

1. Define electric potential at a point.

Answer: Electric potential at a point is the work done by an external agent in bringing a unit positive test charge from infinity to that point, without acceleration.

Formula: V = W/q0. Its SI unit is volt (V), or J C⁻¹.

2. Is electric potential a scalar or a vector quantity?

Answer: Electric potential is a scalar quantity. Therefore, potentials due to several charges are added algebraically, with their signs.

3. What is the electric potential due to an isolated point charge q at distance r?

Answer: V = (1/4πε0) q/r, taking the potential at infinity as zero.

4. What is the potential difference between two points on an equipotential surface?

Answer: Zero, because every point on an equipotential surface has the same electric potential.

5. What is the SI unit of capacitance?

Answer: Farad (F).

6. Write the definition of capacitance.

Answer: Capacitance is the charge stored per unit potential difference across a capacitor:

C = Q/V.

7. What happens to the capacitance of a completely filled parallel-plate capacitor when a dielectric of relative permittivity K is inserted?

Answer: The capacitance becomes C' = KC, where C is the capacitance before insertion.

8. Why can no work be done in moving a test charge along an equipotential surface?

Answer: The potential difference is zero, so W = qΔV = 0.

9. What is the electrostatic potential energy of a charge q placed at a point of potential V?

Answer: U = qV, when V represents the potential produced by the other charges responsible for the field.

10. State the relation between electric field and potential in one dimension.

Answer: E = −dV/dx. The negative sign shows that the electric field points in the direction of decreasing potential.

Section B — 2-Mark Questions

11. Why is electric potential the same throughout a conductor in electrostatic equilibrium?

Answer: In electrostatic equilibrium, the electric field inside the conducting material is zero. Hence the potential difference between any two points inside the conductor is zero. Therefore, the conductor is an equipotential body, including its surface.

12. Distinguish between electric potential and electric potential energy.

Answer:

• Electric potential (V): work done per unit positive test charge at a point; it is a property of the electric field configuration.
• Potential energy (U): energy associated with a charge or system of charges in the field.

For a charge q at potential V, U = qV.

13. Why are electric field lines perpendicular to an equipotential surface?

Answer: If the electric field had a tangential component along an equipotential surface, moving a charge tangentially would involve non-zero work. But the potential difference along an equipotential surface is zero. Hence the tangential component must be zero, so the electric field is normal to the surface.

14. Two points A and B have potentials 80 V and 20 V. Find the work done by the electric field in moving a charge of 2 C from A to B.

Solution:

Wfield = q(VA − VB)

= 2(80 − 20) = 120 J.

Answer: 120 J.

15. Write the equivalent capacitance of two capacitors C₁ and C₂ connected in parallel and in series.

Answer:

Parallel: Ceq = C1 + C2

Series: 1/Ceq = 1/C1 + 1/C2, or Ceq = C1C2/(C1+C2).

16. A capacitor remains connected to a battery while a dielectric completely fills the gap. State what happens to C, Q and V.

Answer: Because the battery remains connected, V remains constant. The dielectric increases capacitance to C' = KC. Since Q = CV, the charge becomes Q' = KQ.

17. An isolated charged capacitor receives a dielectric completely between its plates. State what happens to C, Q and V.

Answer: The capacitor is isolated, so Q remains constant. Capacitance becomes C' = KC. Since V = Q/C, the new potential difference is V' = V/K.

18. Why does a charged conductor have no electric field inside its conducting material in electrostatic equilibrium?

Answer: If a non-zero electric field existed inside the conductor, free charges would experience a force and continue moving. Electrostatic equilibrium requires no such macroscopic charge motion. Therefore the net electric field inside the conducting material is zero.

Section C — 3-Mark Questions

19. Obtain the expression for the electric potential due to a point charge.

Solution: For a point charge q, the electric field at distance r is

E = (1/4πε0) q/r2.

Taking potential at infinity as zero, the potential at r is obtained from

V(r) − V(∞) = −∫∞r E·dr.

Therefore,

V = (1/4πε0) q/r.

The sign of V follows the sign of q.

20. Explain why the potential at the equatorial point of an ideal electric dipole is zero.

Answer: Consider a dipole with charges +q and −q. At an equatorial point, the point is equidistant from both charges. The potentials due to the two charges have equal magnitudes but opposite signs:

V = kq/r + k(−q)/r = 0.

Thus, the electric potential at every point on the equatorial plane of an ideal dipole is zero.

21. A system contains charges q₁, q₂ and q₃ at distances r₁, r₂ and r₃ from a point P. Write the potential at P.

Answer: Since potential is scalar, add the contributions algebraically:

V = (1/4πε0)(q1/r1 + q2/r2 + q3/r3).

No vector addition is required for potential.

22. Derive the potential energy of two point charges q₁ and q₂ separated by distance r.

Solution: Bring q₁ from infinity first. No work is required because no other charge is present initially. At the location of q₂, the potential due to q₁ is V = kq1/r. Hence the work required to bring q₂ from infinity is

U = q2V = kq1q2/r.

Therefore, U = (1/4πε0)q1q2/r.

23. A capacitor of capacitance 6 μF is charged to 20 V. Find its charge and stored energy.

Solution:

Q = CV = 6×10−6 × 20 = 120 μC.

U = ½CV2 = ½ × 6×10−6 × 202

= 1.2×10−3 J.

Answer: Q = 120 μC; U = 1.2 mJ.

24. Two capacitors of 4 μF and 6 μF are connected in parallel across a 10 V battery. Find the equivalent capacitance and total charge supplied by the battery.

Solution:

Ceq = 4 + 6 = 10 μF.

Q = CeqV = 10 μF × 10 V = 100 μC.

Answer: Equivalent capacitance = 10 μF; total charge = 100 μC.

25. Two capacitors of 3 μF and 6 μF are connected in series across 18 V. Find the equivalent capacitance and charge on each capacitor.

Solution:

Ceq = (3×6)/(3+6) = 2 μF.

In series, the magnitude of charge on each capacitor is the same:

Q = CeqV = 2 μF × 18 V = 36 μC.

Answer: C_eq = 2 μF and each capacitor carries 36 μC.

26. Explain the effect of inserting a dielectric in an isolated charged capacitor.

Answer: For an isolated capacitor, no charge can flow to or from the plates, so Q remains constant. If a dielectric of relative permittivity K completely fills the gap:

C' = KC, V' = V/K.

Using U = Q2/(2C), the stored energy becomes U' = U/K. Thus capacitance increases while voltage and stored energy decrease.

27. Explain the difference between an isolated capacitor and a battery-connected capacitor when a dielectric is inserted.

Answer:

Quantity	Battery connected	Battery disconnected
C	KC	KC
Q	KQ	Q
V	V	V/K

28. A charge of 4 μC is moved from a point at 50 V to a point at 120 V. Find the change in potential energy and state whether the external agent must supply or remove energy if the charge is moved slowly.

Solution:

ΔU = q(VB − VA)

= 4×10−6(120−50) = 2.8×10−4 J.

The potential energy increases, so a slow displacement against the electrostatic tendency requires positive work by the external agent.

Section D — 5-Mark / Long-Answer Practice

29. Derive the capacitance of a parallel-plate capacitor with air or vacuum between the plates.

Solution outline: Consider two large parallel conducting plates of area A separated by distance d, carrying charges +Q and −Q. Neglecting edge effects, the electric field between the plates is

E = σ/ε0 = Q/(ε0A).

The potential difference is

V = Ed = Qd/(ε0A).

Using C = Q/V,

C = ε0A/d.

Final result: C0 = ε0A/d.

Exam presentation: define geometry, state the field assumption, obtain V, substitute into C = Q/V, and write the final expression with SI units.

30. Derive the equivalent capacitance of capacitors connected in series and explain the charge and potential distribution.

Solution: For capacitors C₁, C₂, … connected in series, the magnitude of charge on each capacitor is the same, Q. The total potential difference is the sum:

V = V1 + V2 + …

Since Vi = Q/Ci,

V = Q(1/C1 + 1/C2 + …).

But V = Q/Ceq. Hence

1/Ceq = 1/C1 + 1/C2 + ….

For two capacitors, Ceq = C1C2/(C1+C2).

31. Derive the equivalent capacitance of capacitors connected in parallel and explain the charge distribution.

Solution: In parallel, every capacitor is connected across the same two nodes, so the potential difference V is the same for all. The total charge is

Q = Q1 + Q2 + ….

Using Qi = CiV,

Q = (C1 + C2 + …)V.

Since Q = CeqV,

Ceq = C1 + C2 + ….

32. A capacitor is charged to potential V and then its geometry or connection condition is changed. Explain the three equivalent expressions for stored electrostatic energy and when each is convenient.

Answer: The stored energy can be written as

U = ½CV2 = ½QV = Q2/(2C).

• Use ½CV² when C and V are known or when V is held constant by a battery.
• Use ½QV when Q and V are directly available.
• Use Q²/(2C) when the capacitor is isolated and charge is fixed.

Exam trap: Never assume V or Q is constant until you identify whether the capacitor remains connected to a battery.

Mixed Higher-Order Practice

33. A capacitor of capacitance C is charged to V and disconnected from the battery. Its plate separation is increased to twice its original value. Neglect fringing. Find the new capacitance, potential difference and stored energy.

Solution: For a parallel-plate capacitor, C = ε0A/d. Doubling d gives C' = C/2. Since the capacitor is isolated, Q remains constant. Therefore

V' = Q/C' = 2V.

Using U = Q²/(2C),

U' = 2U.

Answer: C' = C/2, V' = 2V, U' = 2U.

34. The same capacitor remains connected to a battery while its plate separation is doubled. State the changes in C, Q, V and stored energy.

Solution: Since C ∝ 1/d, doubling d gives C' = C/2. The battery keeps V constant, so V' = V. From Q = CV, Q' = Q/2. From U = ½CV², with V fixed, U' = U/2.

35. A point charge produces a potential of 900 V at a point 0.20 m away in vacuum. Find the charge.

Solution: From V = kq/r,

q = Vr/k = (900)(0.20)/(9×109)

= 2.0×10−8 C.

Answer: q = 20 nC.

36. Two capacitors of 2 μF and 4 μF are connected in series to a 12 V source. Find the voltage across each capacitor.

Solution: The equivalent capacitance is

Ceq = (2×4)/(2+4) = 4/3 μF.

Charge in series is the same:

Q = CeqV = (4/3)×12 = 16 μC.

Therefore

V2μF = Q/C = 16/2 = 8 V,

V4μF = 16/4 = 4 V.

The voltages add to 12 V, as required.

Answer-Audit Checklist for This Chapter

• Potential is treated as a scalar; electric field as a vector.
• Potential differences use the correct sign convention.
• Series capacitors carry the same magnitude of charge.
• Parallel capacitors have the same potential difference.
• Battery-connected conditions keep V fixed.
• Isolated-capacitor conditions keep Q fixed.
• Dielectric insertion uses C' = KC for a completely filled dielectric.
• Energy expressions are mutually consistent: ½CV² = ½QV = Q²/(2C).
• The current 2026–27 syllabus requirement that capacitor energy be formula-based, without derivation, is respected.
• Numerical answers include units and the relevant physical condition.

Chapter 2 Resource Path

Use the resources in sequence rather than opening isolated pages:

1. Electrostatic Potential and Capacitance — Chapter Hub
2. Complete Notes
3. This page: Important Questions with Answers

Official Curriculum & Exam References

For syllabus scope and the current examination framework, verify against the official CBSE documents:

• CBSE Class XII Physics Curriculum 2026–27
• CBSE Class XII Sample Question Papers & Marking Schemes 2026–27

Next practice step: After mastering these questions, move to chapter MCQs, numericals, case-based questions, assertion–reason practice, PYQs/PYQ-style questions, formula revision and the chapter test.