Class 12 Physics Test Paper Chapter 2 Electric Potential and Capacitance | GSEB Board | Detailed Solutions & Marking Scheme
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Date: 11/04/2026
Class: 12
Subject: Physics
Chapter: Electric Potential and Capacitance
Board: GSEB
Marks: 30
VEDANT IGNITE TEST SERIES
Section A – MCQs (1 mark each) [7 Marks]
Q1. Two charges (+2\mu C) and (-2\mu C) are placed 10 cm apart. Potential at midpoint is:
(a) 0 V (b) 1.8×10⁵ V (c) 3.6×10⁵ V (d) 9×10⁵ V
Q2. Work done in moving a charge of (5\mu C) through a potential difference of 200 V is:
(a) 1×10⁻³ J (b) 2×10⁻³ J (c) 1×10⁻² J (d) 5×10⁻³ J
Q3. A capacitor of 5 μF is charged to 10 V. Energy stored is:
(a) 2.5×10⁻⁴ J (b) 5×10⁻⁴ J (c) 1×10⁻³ J (d) 2×10⁻³ J
Q4. Two capacitors 4 μF and 6 μF are connected in series. Equivalent capacitance is:
(a) 2.4 μF (b) 10 μF (c) 5 μF (d) 1 μF
Q5. A parallel plate capacitor has plate area 2 m² and separation 1 mm. Capacitance is (ε₀ = 8.85×10⁻¹²):
(a) 1.77×10⁻⁸ F (b) 1.77×10⁻⁹ F (c) 8.85×10⁻⁹ F (d) 3.54×10⁻⁸ F
Q6. Three capacitors 2 μF, 3 μF and 6 μF are in parallel. Equivalent capacitance is:
(a) 11 μF (b) 1 μF (c) 6 μF (d) 3 μF
Q7. A 10 μF capacitor is connected to 100 V battery. Charge stored is:
(a) 1×10⁻³ C (b) 1×10⁻⁴ C (c) 1×10⁻⁵ C (d) 1×10⁻² C
Section B – (2 Marks each) [Attempt Any 3] [6 Marks]
Q8. Two point charges (+4\mu C) and (+6\mu C) are 20 cm apart. Find the potential at a point 10 cm from each charge.
Q9. A charge of (3\mu C) is moved between two points in an electric field. Work done is (6×10^{-4}) J. Find potential difference.
Q10. Two capacitors 3 μF and 6 μF are connected in series across 12 V battery. Find charge on each capacitor.
Q11. A parallel plate capacitor has capacitance 10 μF. If potential difference is increased from 10 V to 20 V, find change in energy stored.
Q12. Find equivalent capacitance of combination: 2 μF and 4 μF in series, connected in parallel with 3 μF.
Section C – (3 Marks each) [Attempt Any 3] [9 Marks]
Q13. Three charges (+2\mu C), (-3\mu C), and (+4\mu C) are placed at vertices of an equilateral triangle of side 10 cm. Find potential at centroid.
Q14. A capacitor of 5 μF is charged to 100 V and then disconnected. A dielectric of constant 2 is inserted. Find new potential and energy.
Q15. Two capacitors 4 μF and 8 μF are connected in parallel and then this combination is connected in series with 6 μF. Find equivalent capacitance.
Q16. A parallel plate capacitor of area 0.5 m² and separation 2 mm is filled with dielectric of constant 4. Find capacitance.
Q17. Calculate energy stored in a capacitor of capacitance 2 μF when charges are 4 μC.
Section D – (4 Marks each) [Attempt Any 2] [8 Marks]
Q18. Three capacitors 2 μF, 3 μF and 6 μF are connected such that 2 μF and 3 μF are in series and this combination is in parallel with 6 μF. The combination is connected to 12 V battery. Find:
(i) Equivalent capacitance
(ii) Total charge
(iii) Charge on each capacitor
Q19. A parallel plate capacitor (area = 1 m², separation = 1 mm) is connected to 100 V battery. A dielectric slab of constant 5 completely fills the space. Find:
(i) Initial capacitance
(ii) Final capacitance
(iii) Change in energy
Q20. Four capacitors each of 2 μF are connected in a square. A battery is connected across opposite corners. Find equivalent capacitance between those corners.
Here is your FULL BOARD-LEVEL DETAILED SOLUTION + MARKING SCHEME + ANALYSIS + PRO TIPS + COMMON MISTAKES SHEET exactly in your preferred easy-to-copy format 👇
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Answer Key (Detailed Solutions + Board Strategy)
Class: 12 | Physics | Chapter: Electric Potential & Capacitance | GSEB
✅ Section A – MCQs (1 mark each)
Q1
Potential at midpoint:
V = k ( q₁/r + q₂/r )
= k ( 2μC/r + (-2μC)/r )
= 0
✔ Answer: 0 V
Marks: 1
Q2
W = qV
= 5×10⁻⁶ × 200
= 1×10⁻³ J
✔ Answer: 1×10⁻³ J
Marks: 1
Q3
U = (1/2) C V²
= (1/2) × 5×10⁻⁶ × (10)²
= 2.5×10⁻⁴ J
✔ Answer: 2.5×10⁻⁴ J
Marks: 1
Q4
1/C = 1/4 + 1/6
= 5/12
C = 12/5 = 2.4 μF
✔ Answer: 2.4 μF
Marks: 1
Q5
C = ε₀ A/d
= 8.85×10⁻¹² × (2 / 10⁻³)
= 1.77×10⁻⁸ F
✔ Answer: 1.77×10⁻⁸ F
Marks: 1
Q6
Parallel:
C = 2 + 3 + 6 = 11 μF
✔ Answer: 11 μF
Marks: 1
Q7
Q = CV
= 10×10⁻⁶ × 100
= 1×10⁻³ C
✔ Answer: 1×10⁻³ C
Marks: 1
✅ Section B – (2 Marks each)
Q8
Given: r = 10 cm = 0.1 m
V = k ( q₁/r + q₂/r )
= 9×10⁹ [ (4×10⁻⁶ / 0.1) + (6×10⁻⁶ / 0.1) ]
= 9×10⁹ × (10×10⁻⁶ / 0.1)
= 9×10⁵ V
✔ Final Answer: 9×10⁵ V
Marking Scheme:
Formula (1) + Calculation (1)
Q9
V = W/q
= (6×10⁻⁴) / (3×10⁻⁶)
= 200 V
✔ Final Answer: 200 V
Marking:
Formula (1) + Answer (1)
Q10
Series:
Ceq = (3×6)/(3+6) = 2 μF
Q = CV = 2×10⁻⁶ × 12
= 24×10⁻⁶ C
✔ Charge on each capacitor = 24 μC
Marking:
Ceq (1) + Charge (1)
Q11
U₁ = (1/2) C V₁²
= (1/2)×10×10⁻⁶×(10)² = 5×10⁻⁴ J
U₂ = (1/2)×10×10⁻⁶×(20)² = 2×10⁻³ J
ΔU = U₂ − U₁
= 1.5×10⁻³ J
✔ Answer: 1.5×10⁻³ J
Marking:
Both energies (1) + Difference (1)
Q12
Series:
C = (2×4)/(2+4) = 1.33 μF
Parallel:
C = 1.33 + 3 = 4.33 μF
✔ Answer: 4.33 μF
Marking:
Series (1) + Final (1)
✅ Section C – (3 Marks each)
Q13
Distance centroid:
r = a/√3 = 0.1/√3
Net charge:
q = 2 − 3 + 4 = 3 μC
V = kq/r
= 9×10⁹ × (3×10⁻⁶) / (0.1/√3)
= 9×10⁹ × 3×10⁻⁶ × √3 / 0.1
≈ 4.67×10⁵ V
✔ Answer: 4.67×10⁵ V
Marking:
Geometry (1) + Formula (1) + Calculation (1)
Q14
Battery removed → Q constant
New potential:
V' = V/K = 100/2 = 50 V
Initial energy:
U = (1/2)×5×10⁻⁶×(100)² = 0.025 J
Final energy:
U' = U/K = 0.0125 J
✔ Answer: V' = 50 V, U' = 0.0125 J
Marking:
Concept (1) + V' (1) + U' (1)
Q15
Parallel:
C = 4 + 8 = 12 μF
Series:
C = (12×6)/(12+6) = 4 μF
✔ Answer: 4 μF
Marking:
Step 1 (1) + Step 2 (2)
Q16
C = K ε₀ A/d
= 4 × 8.85×10⁻¹² × (0.5 / 2×10⁻³)
= 8.85×10⁻⁹ F
✔ Answer: 8.85×10⁻⁹ F
Marking:
Formula (1) + Substitution (1) + Answer (1)
Q17
U = Q² / 2C
= (4×10⁻⁶)² / (2×2×10⁻⁶)
= 4×10⁻⁶ J
✔ Answer: 4×10⁻⁶ J
Marking:
Formula (1) + Calculation (2)
✅ Section D – (4 Marks each)
Q18
Series (2 & 3 μF):
C₁ = (2×3)/(2+3) = 1.2 μF
Parallel with 6 μF:
Ceq = 1.2 + 6 = 7.2 μF
Total charge:
Q = CV = 7.2×10⁻⁶ × 12 = 86.4 μC
Voltage across series branch:
V₁ = Q/C₁ = 86.4 / 1.2 = 72 V
Charge in series capacitors (same):
Q₁ = C₁ V₁ = 1.2×72 = 86.4 μC? ❌ (check carefully)
Better method:
Voltage division:
V₂μF = (3/(2+3)) × 72 = 43.2 V
V₃μF = (2/(2+3)) × 72 = 28.8 V
Charges:
Q₂ = C V = 2×43.2 = 86.4 μC
Q₃ = 3×28.8 = 86.4 μC
✔ Final: All series capacitors have same charge = 86.4 μC
6 μF capacitor:
Q = CV = 6×12 = 72 μC
Marking Scheme:
Ceq (1) + Total Q (1) + Series concept (1) + Final charges (1)
Q19
Initial:
C = ε₀ A/d
= 8.85×10⁻¹² × (1 / 10⁻³)
= 8.85×10⁻⁹ F
Final:
C' = K C = 5×8.85×10⁻⁹
= 44.25×10⁻⁹ F
Energy change (battery connected → V constant):
ΔU = (1/2)(C' − C)V²
= (1/2)(35.4×10⁻⁹)(100²)
= 1.77×10⁻⁴ J
✔ Answer: 1.77×10⁻⁴ J
Q20
Symmetry method:
Equivalent reduces to two parallel branches → each 4 μF
Series of both →
Ceq = (4×4)/(4+4) = 2 μF
✔ Answer: 2 μF
📊 PAPER ANALYSIS (VERY IMPORTANT FOR YOU)
🔥 Difficulty Level:
Section A: Easy NCERT
Section B: Moderate
Section C: Moderate to tricky
Section D: Concept-heavy
👉 Overall Level: 70% Moderate + 30% Tricky
🎯 Chapter Coverage:
Potential: 25%
Capacitance: 50%
Energy: 25%
👉 Perfect board-weight distribution ✔
⚠️ High Scoring Areas:
Series/Parallel capacitors
Energy formulas
Dielectric cases
🚀 PRO TIPS (BOARD EXAM GAME CHANGER)
Always write formula first
→ even if answer wrong → 1 mark savedUse μF → convert carefully
→ biggest mistake areaFor dielectric: remember 3 cases
Battery connected → V constant
Battery removed → Q constant
Series rule:
Charge sameParallel rule:
Voltage same
❌ COMMON MISTAKES SHEET (VERY IMPORTANT)
🔴 Concept Mistakes:
Writing Q different in series ❌
Forgetting dielectric condition ❌
Using wrong energy formula ❌
🔴 Calculation Mistakes:
μF not converted
cm → m conversion missed
Squaring errors in V²
🔴 Exam Mistakes:
Skipping unit → marks cut
Direct answer → no steps
Rough work not shown
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