Electrostatic Potential and Capacitance Class 12 Physics Case Based Questions 2026-27
Class 12 Physics • Chapter 2 • Competency Practice
Electrostatic Potential and Capacitance Class 12 Physics Case-Based Questions with Answers
Original competency-focused case studies that require students to read a situation, identify the governing principle, use data correctly and justify the answer.
Why Case-Based Practice Matters
The current CBSE Class XII examination framework includes objective, short-answer, longer-answer and case-study/competency-oriented assessment formats. The official 2026–27 SQP/MS page provides the current subject sample papers and marking schemes.
Chapter 2 is part of Unit I: Electrostatics, which carries 16 marks together with Chapter 1 in the current Physics curriculum. The Chapter 2 scope includes potential, potential energy, equipotential surfaces, conductors, dielectrics, capacitance, combinations of capacitors, parallel-plate capacitors and capacitor energy formulae.
How to Solve a Case-Based Question
- Read the situation before looking at the sub-questions.
- Underline quantities that remain constant, especially Q or V.
- Identify whether the capacitor is connected to a battery or isolated.
- Choose the formula only after identifying the physical condition.
- For conceptual parts, explain the principle rather than writing only a formula.
- For numerical parts, include units and check the result.
Case Study 1 — Equipotential Surfaces and Work
A student maps the potential around a charged conducting object and finds that several points on a particular surface have the same potential. The student moves a small positive test charge from point A to point B along that surface. The electric field at the surface is directed normal to it.
1(i). What is the potential difference between A and B?
Answer: Zero.
Reason: A and B lie on the same equipotential surface.
1(ii). What is the work done by the electrostatic field in moving charge q from A to B?
Answer: Zero, because Wfield = q(VA−VB) = 0.
1(iii). Why is the electric field perpendicular to the surface?
Answer: A tangential component of E would produce non-zero work along the surface. Since the surface is equipotential, the tangential component must be zero.
1(iv). If the charge is moved to a different surface of lower potential, in which direction does the electric field point?
Answer: The electric field points toward decreasing potential.
Case Study 2 — Capacitor Connected to a Battery
A parallel-plate capacitor of capacitance 4 μF is connected to an ideal 100 V battery. A dielectric with relative permittivity K = 5 is then inserted completely between the plates while the battery remains connected.
2(i). What is the new capacitance?
Answer: C' = KC = 5×4 = 20 μF.
2(ii). What happens to the potential difference?
Answer: It remains 100 V because the capacitor is still connected to the ideal battery.
2(iii). Find the new charge on the capacitor.
Solution: Q' = C'V = 20 μF × 100 V = 2000 μC = 2 mC.
Answer: 2 mC.
2(iv). By what factor does the stored energy change?
Answer: It becomes 5 times the original value because V is constant and U = ½CV², while C becomes 5C.
Case Study 3 — Isolated Capacitor and Dielectric
A charged 6 μF capacitor is disconnected from its battery. It initially has a potential difference of 50 V. A dielectric of relative permittivity 3 is inserted completely between its plates.
3(i). Which quantity remains constant during dielectric insertion?
Answer: Charge Q remains constant because the capacitor is isolated.
3(ii). Find the new capacitance.
Solution: C' = KC = 3×6 = 18 μF.
3(iii). Find the new potential difference.
Solution: Since Q is constant, V' = V/K = 50/3 ≈ 16.67 V.
Answer: Approximately 16.7 V.
3(iv). What happens to the stored energy?
Answer: It decreases by a factor of 3. For fixed Q, U = Q²/(2C), so increasing C by 3 makes U one-third.
Case Study 4 — Series Capacitors
Two capacitors of 3 μF and 6 μF are connected in series across an 18 V source. A student initially assumes that each capacitor gets 9 V because the source voltage is 18 V.
4(i). Is the student's assumption necessarily correct?
Answer: No. In series, the charge magnitude is the same, but the voltage across each capacitor depends on its capacitance.
4(ii). Find the equivalent capacitance.
Solution: Ceq = (3×6)/(3+6) = 2 μF.
4(iii). Find the charge on each capacitor.
Solution: Q = CeqV = 2 μF × 18 V = 36 μC.
Answer: 36 μC on each capacitor in magnitude.
4(iv). Find the voltage across each capacitor.
Solution:
V3μF = 36/3 = 12 V
V6μF = 36/6 = 6 V
Answer: 12 V across 3 μF and 6 V across 6 μF.
Case Study 5 — Parallel Capacitors
A 4 μF and a 6 μF capacitor are connected in parallel across a 10 V battery. Because both capacitors are connected between the same two nodes, they have the same potential difference.
5(i). Find the equivalent capacitance.
Answer: Ceq = 4+6 = 10 μF.
5(ii). Find the charge on the 4 μF capacitor.
Answer: Q = CV = 4×10 = 40 μC.
5(iii). Find the charge on the 6 μF capacitor.
Answer: Q = 6×10 = 60 μC.
5(iv). Find the total charge supplied by the battery.
Answer: Qtotal = 40+60 = 100 μC.
Case Study 6 — Changing Plate Separation
A parallel-plate capacitor initially has capacitance 12 μF and is charged to 50 V. In one experiment the battery remains connected while the plate separation is doubled. In another experiment the capacitor is first disconnected from the battery and then the same mechanical change is made.
6(i). In the battery-connected experiment, what is the new capacitance?
Answer: Since C ∝ 1/d, doubling d gives C' = 6 μF.
6(ii). In the battery-connected experiment, what happens to Q and V?
Answer: V remains 50 V and Q becomes half because Q = CV.
6(iii). In the isolated experiment, what happens to V?
Answer: Q remains constant while C halves, so V doubles: V' = 100 V.
6(iv). State the energy change in the two experiments.
Answer: With the battery connected, V is constant and C halves, so U becomes half. With the capacitor isolated, Q is constant and C halves, so U = Q²/(2C) becomes twice the initial energy.
Case Study 7 — Electric Potential from Multiple Charges
At a point P, two charges +4 μC and −2 μC are both 0.50 m away. A student tries to add electric fields as vectors even though the question asks for potential.
7(i). Should potential contributions be added as vectors?
Answer: No. Electric potential is scalar, so contributions are added algebraically.
7(ii). Find the potential at P.
Solution:
V = k(q1+q2)/r
= (9×109)(2×10−6)/0.50
= 3.6×104 V.
Answer: 36 kV.
7(iii). What would change if the question asked for electric field instead?
Answer: Electric field is a vector, so its magnitude and direction must be combined vectorially.
7(iv). Can electric field be non-zero at a point where the potential is zero?
Answer: Yes. V = 0 at a point does not require the spatial gradient of V to be zero.
Case Study 8 — Energy Stored in a Capacitor
A 10 μF capacitor is charged to 100 V. A student uses different forms of the capacitor-energy equation depending on the information available.
8(i). Find the charge on the capacitor.
Solution: Q = CV = 10×10−6}×100 = 1.0×10−3 C.
Answer: 1 mC.
8(ii). Find the stored energy.
Solution: U = ½CV² = ½(10×10−6})(100²) = 0.05 J.
Answer: 0.05 J.
8(iii). Which energy expression is convenient if Q and V are known?
Answer: U = ½QV.
8(iv). Which expression is most convenient when Q remains fixed and C changes?
Answer: U = Q²/(2C).
Case Study 9 — Dipole in a Uniform Electric Field
An electric dipole with dipole moment 4×10−8 C m is placed in a uniform electric field of 5×104 N/C. Its orientation makes an angle of 60° with the field.
9(i). Find the potential energy of the dipole.
Solution: U = −pE cosθ.
U = −(4×10−8})(5×104})(1/2) = −1.0×10−3 J.
Answer: −1.0 mJ.
9(ii). At what orientation is the dipole in stable equilibrium?
Answer: θ = 0°, with p parallel to E.
9(iii). At what orientation is the potential energy maximum?
Answer: θ = 180°, when p is antiparallel to E.
9(iv). Why is θ = 0° stable?
Answer: At θ = 0°, U = −pE, the minimum possible value of the dipole's potential energy.
Case Study 10 — Dielectric, Capacitance and Charge
A parallel-plate capacitor has capacitance 8 μF in vacuum. A dielectric with relative permittivity 4 completely fills the space between the plates. Consider two separate experiments: in Experiment A the capacitor stays connected to a battery of 20 V; in Experiment B the capacitor is charged to 20 V and then isolated before the dielectric is inserted.
10(i). Find the new capacitance.
Answer: C' = KC = 4×8 = 32 μF.
10(ii). In Experiment A, find the new charge.
Solution: V remains 20 V.
Q' = C'V = 32×20 = 640 μC.
Answer: 640 μC.
10(iii). In Experiment B, find the new voltage.
Solution: Q is fixed, so V' = V/K = 20/4 = 5 V.
Answer: 5 V.
10(iv). Compare the energy change in the two experiments.
Answer: In Experiment A, V is fixed, so U becomes 4U. In Experiment B, Q is fixed, so U becomes U/4.
Competency Traps to Remember
- Same potential does not mean same electric field.
- Zero potential does not automatically mean zero field.
- Series ≠ same voltage: series capacitors have the same charge magnitude.
- Parallel ≠ same charge: parallel capacitors have the same voltage.
- Battery connected: ideal battery fixes V.
- Battery disconnected: isolated capacitor keeps Q constant.
- Dielectric insertion: always identify the electrical condition before deciding what happens to energy.
- Potential is scalar: do not use vector addition for potential.
Chapter 2 Resource Path
- Electrostatic Potential and Capacitance — Chapter Hub
- Complete Notes
- Important Questions with Answers
- MCQs with Answers
- Numericals with Step-by-Step Solutions
- This page: Case-Based Questions
Official Curriculum & Exam References
CBSE Class XII Physics Curriculum 2026–27
CBSE Class XII Sample Question Papers & Marking Schemes 2026–27
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