Electrostatic Potential and Capacitance Class 12 Physics Numericals 2026-27

Class 12 Physics • Chapter 2 • Numericals

Electrostatic Potential and Capacitance Class 12 Physics Numericals with Solutions

A carefully audited set of original numerical problems covering electric potential, potential energy, dipoles, capacitance, capacitor combinations, dielectric effects and stored energy for CBSE 2026–27.

How to use this page: Attempt each problem yourself before opening the solution. Write the formula, substitute SI units, calculate, and check whether the final answer has the correct unit and physical sign.

Numerical Coverage

This page is not a collection of random formula substitutions. The problems are arranged to cover the numerical skills students repeatedly need in this chapter: scaling, signs, combinations, conditions, and multi-step reasoning.

• Potential due to a point charge
• Potential difference and work
• Potential energy of charge systems
• Dipole potential and energy
• Parallel-plate capacitance
• Series and parallel combinations
• Charge and voltage distribution
• Dielectric insertion
• Battery-connected versus isolated capacitors
• Energy stored in capacitors

Constants Used

• k = 1/(4πε0) ≈ 9×109 N m² C−2
• ε0 ≈ 8.85×10−12 F m−1
• 1 μC = 10−6 C
• 1 nC = 10−9 C
• 1 μF = 10−6 F

Section A — Electric Potential and Potential Difference

1. A point charge of 2 μC is placed in vacuum. Find the potential at a point 0.30 m away.

Solution: Use V = kq/r.

V = (9×109)(2×10−6)/0.30

V = 6.0×104 V.

Answer: 60 kV.

2. A point charge produces a potential of 900 V at a distance of 0.20 m. Find the charge.

Solution: q = Vr/k.

q = (900)(0.20)/(9×109) = 2.0×10−8 C.

Answer: 20 nC.

3. A charge of −3 μC is placed in vacuum. Find the potential at a point 0.50 m away.

Solution:

V = kq/r = (9×109)(−3×10−6)/0.50

V = −5.4×104 V.

Answer: −54 kV. The negative sign comes from the negative source charge.

4. A charge of 5 μC is moved from a point at 100 V to a point at 400 V. Find the change in its electrostatic potential energy.

Solution:

ΔU = q(VB − VA)

= 5×10−6(400−100)

= 1.5×10−3 J.

Answer: 1.5 mJ increase.

5. A charge of 2 μC moves from 500 V to 200 V. Find the work done by the electrostatic field.

Solution:

Wfield = q(VA − VB)

= 2×10−6(500−200)

= 6.0×10−4 J.

Answer: 6.0×10⁻⁴ J.

6. Two charges +4 μC and −2 μC are each 0.50 m from a point P. Find the potential at P.

Solution: Potential is scalar, so add algebraically:

V = k(q1+q2)/r

= (9×109)(2×10−6)/0.50

= 3.6×104 V.

Answer: 36 kV.

7. Two equal and opposite charges of magnitude 3 μC are separated by 0.20 m. Find the potential at the midpoint.

Solution: The midpoint is equidistant from both charges. Their potentials are equal in magnitude and opposite in sign:

V = k(3 μC)/0.10 + k(−3 μC)/0.10 = 0.

Answer: 0 V.

Section B — Potential Energy

8. Two charges +2 μC and +3 μC are separated by 0.30 m in vacuum. Find their electrostatic potential energy.

Solution:

U = kq1q2/r

= (9×109)(2×10−6)(3×10−6)/0.30

= 0.18 J.

Answer: +0.18 J.

9. Two charges +2 μC and −3 μC are separated by 0.30 m. Find their potential energy.

Solution:

U = (9×109)(2×10−6)(−3×10−6)/0.30

U = −0.18 J.

Answer: −0.18 J.

10. A charge of 4 μC is placed at a point where the potential due to other charges is 250 V. Find its potential energy.

Solution:

U = qV = (4×10−6)(250)

= 1.0×10−3 J.

Answer: 1.0 mJ.

11. An electric dipole of dipole moment 4×10⁻⁸ C m is placed in a uniform electric field of 5×10⁴ N/C at 60°. Find its potential energy.

Solution:

U = −pE cosθ

= −(4×10−8)(5×104)(1/2)

= −1.0×10−3 J.

Answer: −1.0 mJ.

12. An electric dipole of moment 2×10⁻⁸ C m is rotated in a uniform field of 3×10⁵ N/C from 0° to 90°. Find the increase in potential energy.

Solution:

U1 = −pE cos0° = −pE

U2 = −pE cos90° = 0

Therefore, ΔU = pE = (2×10−8)(3×105) = 6×10−3 J.

Answer: 6 mJ.

Section C — Capacitance and Parallel-Plate Capacitor

13. A parallel-plate capacitor has plate area 2×10⁻² m² and separation 1 mm. Find its capacitance in vacuum. Take ε₀ = 8.85×10⁻¹² F/m.

Solution:

C = ε0A/d

= (8.85×10−12)(2×10−2)/(1×10−3)

= 1.77×10−10 F.

Answer: 177 pF.

14. A capacitor has capacitance 5 μF and is connected to a 12 V battery. Find the charge stored.

Solution:

Q = CV = 5×10−6×12

= 60×10−6 C.

Answer: 60 μC.

15. A capacitor stores 80 μC of charge at 20 V. Find its capacitance.

Solution:

C = Q/V = 80×10−6/20

= 4×10−6 F.

Answer: 4 μF.

16. Two capacitors of 4 μF and 6 μF are connected in parallel across 10 V. Find the equivalent capacitance and total charge.

Solution:

Ceq = 4+6 = 10 μF.

Q = CeqV = 10×10 = 100 μC.

Answer: C_eq = 10 μF; Q = 100 μC.

17. Capacitors of 3 μF and 6 μF are connected in series across 18 V. Find the equivalent capacitance and charge on each.

Solution:

Ceq = (3×6)/(3+6) = 2 μF.

Q = CeqV = 2×18 = 36 μC.

In series, the charge magnitude is the same on each capacitor.

Answer: C_eq = 2 μF; Q = 36 μC on each.

18. For the capacitors in Question 17, find the potential difference across each capacitor.

Solution:

V3μF = Q/C = 36/3 = 12 V.

V6μF = 36/6 = 6 V.

12 + 6 = 18 V, confirming the result.

Answer: 12 V across 3 μF and 6 V across 6 μF.

19. Three capacitors of 2 μF, 3 μF and 6 μF are connected in parallel across 20 V. Find the total charge supplied by the battery.

Solution:

Ceq = 2+3+6 = 11 μF.

Q = 11×20 = 220 μC.

Answer: 220 μC.

20. Three capacitors of 2 μF, 3 μF and 6 μF are connected in series. Find the equivalent capacitance.

Solution:

1/Ceq = 1/2 + 1/3 + 1/6 = 1 (in μF⁻¹).

Therefore, Ceq = 1 μF.

Answer: 1 μF.

Section D — Capacitor Energy

21. A 6 μF capacitor is charged to 20 V. Find the energy stored.

Solution:

U = ½CV²

= ½(6×10−6})(20²)

= 1.2×10−3 J.

Answer: 1.2 mJ.

22. A capacitor stores 100 μC of charge at a potential difference of 50 V. Find the stored energy.

Solution:

U = ½QV = ½(100×10−6})(50)

= 2.5×10−3 J.

Answer: 2.5 mJ.

23. A capacitor of capacitance 4 μF stores 0.72 mJ of energy. Find the potential difference across it.

Solution:

U = ½CV², so V = √(2U/C).

V = √[(2×0.72×10−3)/(4×10−6)]

= √360 = 18.97 V.

Answer: ≈19.0 V.

24. A capacitor of capacitance 10 μF is charged to 100 V. Find the charge and stored energy.

Solution:

Q = CV = 10×10−6×100 = 1.0×10−3 C.

U = ½CV² = ½(10×10−6)(100²) = 0.05 J.

Answer: Q = 1 mC; U = 0.05 J.

Section E — Dielectric and Condition-Based Numericals

25. A capacitor has capacitance 4 μF. A dielectric of relative permittivity 5 completely fills the gap. Find the new capacitance.

Solution:

C' = KC = 5×4 = 20 μF.

Answer: 20 μF.

26. An isolated capacitor has capacitance 4 μF and is charged to 100 V. A dielectric of constant 5 is inserted completely. Find the new capacitance and voltage.

Solution:

C' = KC = 5×4 = 20 μF.

Because the capacitor is isolated, Q remains constant.

Initial charge: Q = CV = 4×100 = 400 μC.

New voltage: V' = Q/C' = 400/20 = 20 V.

Answer: C' = 20 μF; V' = 20 V.

27. In Question 26, find the initial and final stored energies.

Solution:

Initial:

U = ½CV² = ½(4×10−6})(100²) = 0.02 J.

Final, using constant Q:

U' = Q²/(2C') = (400×10−6})²/(2×20×10−6}) = 0.004 J.

Answer: Initial U = 0.020 J; final U = 0.004 J.

28. A 4 μF capacitor remains connected to a 100 V battery while a dielectric of constant 5 is inserted fully. Find the new charge and stored energy.

Solution:

New capacitance: C' = 5×4 = 20 μF.

Battery keeps V = 100 V.

Q' = C'V = 20×100 = 2000 μC = 2 mC.

U' = ½C'V² = ½(20×10−6})(100²) = 0.10 J.

Answer: Q' = 2 mC; U' = 0.10 J.

29. An isolated parallel-plate capacitor has plate separation 2 mm. The separation is increased to 4 mm. If its initial capacitance is 8 μF and initial voltage is 10 V, find the new capacitance and voltage.

Solution: Since C ∝ 1/d, doubling d halves C:

C' = 4 μF.

The capacitor is isolated, so Q remains constant. Therefore V doubles:

V' = 20 V.

Answer: C' = 4 μF; V' = 20 V.

30. A parallel-plate capacitor remains connected to a battery. Its plate separation is doubled. If initially C = 12 μF and V = 50 V, find the new capacitance and charge.

Solution: Doubling d gives C' = 6 μF. Battery keeps V = 50 V.

Q' = C'V = 6×50 = 300 μC.

Answer: C' = 6 μF; Q' = 300 μC.

Section F — Higher-Order Multi-Step Numericals

31. Two capacitors 6 μF and 3 μF are first connected in parallel to a 12 V battery. They are then disconnected from the battery and connected in series with each other, with opposite polarity avoided. Find the charge on each capacitor in the final series arrangement.

Solution: Initially in parallel:

Cinitial = 6+3 = 9 μF.

Total charge supplied initially is Qtotal = 9×12 = 108 μC.

For the stated isolated reconnection, the final series configuration requires a circuit-specific charge redistribution analysis. The total charge available on the isolated two-capacitor system is not by itself enough to determine the final voltage unless the exact polarity and node connections are specified.

Answer: Do not assign a unique numerical value from the information given. This is an intentional exam-skill check: identify when a capacitor-network problem is under-specified rather than inventing a result.

32. A 2 μF capacitor is charged to 100 V and disconnected. It is connected in parallel to an uncharged 6 μF capacitor. Find the final common voltage and the final total electrostatic energy.

Solution: Initial charge is

Qinitial = 2×100 = 200 μC.

Total capacitance after parallel connection:

Ctotal = 2+6 = 8 μF.

Charge is conserved:

Vfinal = 200/8 = 25 V.

Final energy:

Ufinal = ½CtotalV² = ½(8×10−6})(25²) = 2.5×10−3 J.

Answer: V_final = 25 V; U_final = 2.5 mJ.

33. A 5 μF capacitor charged to 20 V is connected in parallel with an uncharged 15 μF capacitor. Find the final voltage and the charge on each capacitor.

Solution: Initial charge = 5×20 = 100 μC.

Total capacitance = 20 μF.

Vfinal = 100/20 = 5 V.

Charges:

Q5μF = 5×5 = 25 μC.

Q15μF = 15×5 = 75 μC.

Answer: V = 5 V; charges are 25 μC and 75 μC.

34. A capacitor of 8 μF is charged to 50 V. Find the charge and energy. If the voltage is doubled while capacitance remains unchanged, by what factor does the energy change?

Solution:

Q = CV = 8×50 = 400 μC.

U = ½CV² = ½(8×10−6})(50²) = 0.010 J.

Since U ∝ V² for fixed C, doubling V makes energy four times.

Answer: Q = 400 μC; U = 0.010 J; energy becomes 4 times.

35. A capacitor of capacitance C is charged to voltage V. Its capacitance is increased to 3C while the battery remains connected. Find the ratios Q'/Q and U'/U.

Solution: With the battery connected, V is constant.

Q = CV, so Q'/Q = 3.

U = ½CV², so U'/U = 3.

Answer: Q'/Q = 3; U'/U = 3.

36. A capacitor of capacitance C is charged and then isolated. Its capacitance is increased to 3C. Find the ratios V'/V and U'/U.

Solution: Isolation keeps Q constant.

Since V = Q/C, V'/V = 1/3.

Since U = Q²/(2C), U'/U = 1/3.

Answer: V'/V = 1/3; U'/U = 1/3.

Answer-Audit and Numerical-Solving Checklist

• Convert μC, nC, μF, mm and other quantities to SI units before substitution.
• For potential, remember V ∝ q/r, not q/r².
• Potential is scalar, so signs must be handled algebraically.
• For potential energy of two charges, the sign of q1q2 matters.
• For series capacitors, charge magnitude is the same.
• For parallel capacitors, voltage is the same.
• When a battery remains connected, V is fixed by the ideal battery.
• When a capacitor is isolated, Q is fixed.
• For a completely inserted dielectric, C' = KC.
• Use ½CV², ½QV or Q²/(2C) according to the quantities known and the physical condition.
• Always check whether the question supplies enough information for a unique answer. Question 31 intentionally demonstrates this.

Chapter 2 Resource Path

1. Electrostatic Potential and Capacitance — Chapter Hub
2. Complete Notes
3. Important Questions with Answers
4. MCQs with Answers
5. This page: Numericals with Step-by-Step Solutions

Official Curriculum & Exam References

The current CBSE 2026–27 Physics curriculum places Chapter 2 under Unit I: Electrostatics, which carries 16 marks together with Chapter 1. The current CBSE Class XII SQP/MS page now provides the 2026–27 Physics Sample Question Paper and Marking Scheme. Use these official documents as the final authority for the syllabus and examination framework.

CBSE Class XII Physics Curriculum 2026–27

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

Next step: After numerical practice, move to Chapter 2 case-based questions to test whether you can extract data and apply the correct electrostatic or capacitor principle from a real-world-style passage.