Class 12 Physics Test Paper Chapter 2 Electric Potential and Capacitance | CBSE Board | Detailed Solutions & Marking Scheme
VEDANT IGNITE TEST SERIES
Class: 12
Subject: Physics
Chapter: Electric Potential and Capacitance
Board: CBSE
Time: 1.5 Hours | Marks: 30
Section A (7 Marks)
(5 MCQs + 2 Assertion-Reason, 1 mark each)
Q1. A charge of 2 × 10⁻⁶ C is placed in vacuum. Find the electric potential at a point 10 cm away.
Options:
(A) 1.8 × 10⁵ V (B) 3.6 × 10⁵ V (C) 9 × 10⁵ V (D) 1.8 × 10⁶ V
Q2. A student observes that when a dielectric slab is inserted between capacitor plates (battery disconnected), potential difference decreases. What is the correct reason?
Options: (A) Charge increases (B) Capacitance decreases (C) Capacitance increases (D) Energy increases
Q3. In a laboratory setup, two capacitors 3µF and 6µF are connected in series. The equivalent capacitance is:
Options: (A) 9µF (B) 2µF (C) 3µF (D) 1µF
Q4. A charged capacitor stores energy. If voltage is doubled, the energy becomes:
Options: (A) same (B) double (C) four times (D) half
Q5. A metal sphere is given charge. Where is potential maximum?
Options: (A) centre (B) surface (C) outside (D) zero everywhere
Assertion-Reason
Q6.
Assertion (A): Work done in moving a charge along an equipotential surface is zero.
Reason (R): Electric field is perpendicular to equipotential surface.
Options: (A) Both true & R explains A (B) Both true but R not explanation (C) A true R false (D) A false R true
Q7.
Assertion (A): Capacitance increases when dielectric is inserted.
Reason (R): Dielectric reduces effective electric field inside capacitor.
Options: (A) Both true & R explains A (B) Both true but R not explanation (C) A true R false (D) A false R true
Section B (8 Marks)
(4 Questions × 2 marks each)
Q8. A student moves a charge of (1\mu C) between two points of equal potential. Calculate work done. Explain significance.
Q9. In a physics lab, two capacitors (5µF and 10µF) are connected in parallel across 12V supply. Find total capacitance and total charge stored.
Q10. A parallel plate capacitor has plate area (0.02 m^2) and separation 1 mm. Calculate capacitance.
Q11. A student observes that electric field inside a conductor is zero. Explain using potential concept.
Section C (6 Marks)
(2 Questions × 3 marks each)
Q12. (Case-based Numerical)
A capacitor of 5µF is charged to 100V. It is then connected to another uncharged capacitor of 10µF.
(i) Find final potential
(ii) Find charge redistribution
(iii) State energy loss reason
(CBSE PYQ concept)
Q13. (Story-based Concept + Numerical)
In an experiment, a dielectric slab (K = 5) is inserted into a charged capacitor (battery disconnected). Initial capacitance = 10µF, voltage = 100V.
(i) Find new capacitance
(ii) Find new voltage
(iii) Explain physical reason
Section D (5 Marks)
Q14. (Competency-Based Long Question)
A physics student designs a capacitor network using 2µF, 3µF, and 6µF capacitors connected in series across 12V battery.
(i) Find equivalent capacitance
(ii) Find total charge stored
(iii) Find potential across each capacitor
(iv) Calculate total energy stored
(v) What happens to energy if connected in parallel instead?
(PYQ inspired)
Section E (4 Marks)
Case Study (1+1+2)
Q15.
A student is studying energy storage in capacitors for use in flash cameras. He uses a capacitor of 100µF connected to a 12V battery.
(i) Calculate energy stored
(ii) If voltage is doubled, what happens to energy?
(iii) Explain why capacitors are used in energy storage devices with reasoning
Here is a complete CBSE Board-style solution set with:
✔ Step-by-step solutions (exam-ready format)
✔ Proper marking scheme (CBSE pattern)
✔ Easy-to-copy math format
✔ Paper analysis
✔ Pro tips for full marks
✔ Common mistakes sheet
🌟 VEDANT IGNITE TEST SERIES
Class 12 Physics
Chapter: Electric Potential and Capacitance
Total Marks: 30 | Time: 1.5 Hours
🟢 SECTION A (7 × 1 = 7 Marks)
Q1. Electric Potential
Given:
q = 2 × 10⁻⁶ C
r = 10 cm = 0.1 m
k = 9 × 10⁹
Formula:
V = kq / r
Solution:
V = (9 × 10⁹ × 2 × 10⁻⁶) / 0.1
V = (18 × 10³) / 0.1
V = 18 × 10⁴
V = 1.8 × 10⁵ V
✅ Answer: (A) 1.8 × 10⁵ V
🟡 Marks:
Formula (0.5) + substitution (0.25) + answer (0.25)
Q2. Dielectric in Capacitor
Battery disconnected → Q constant
Dielectric increases capacitance: C ↑
So V = Q/C → V decreases
✅ Answer: (C) Capacitance increases
Q3. Series Capacitor
Formula:
1/Ceq = 1/C1 + 1/C2
1/Ceq = 1/3 + 1/6 = (2+1)/6 = 3/6
Ceq = 2 µF
✅ Answer: (B) 2 µF
Q4. Energy of Capacitor
U = ½ CV²
V doubled → U ∝ V² → 4 times
✅ Answer: (C) four times
Q5. Conducting Sphere
Electric potential is constant inside conductor and maximum at surface
✅ Answer: (B) surface
Q6. Assertion-Reason
A: True
R: True
R explains A (since E ⟂ equipotential → no work)
✅ Answer: (A)
Q7. Assertion-Reason
A: True (C increases)
R: True (E decreases)
R explains A
✅ Answer: (A)
🟢 SECTION B (8 Marks)
Q8. Work Done in Equipotential Surface
W = qΔV
ΔV = 0
W = 1 × 10⁻⁶ × 0 = 0 J
Explanation:
No work is done because potential difference is zero
🟡 Marks:
Formula (1) + substitution (0.5) + explanation (0.5)
Q9. Parallel Capacitors
C1 = 5 µF, C2 = 10 µF
V = 12 V
Equivalent:
Ceq = C1 + C2 = 15 µF
Charge:
Q = CV
Q = 15 × 10⁻⁶ × 12
Q = 180 × 10⁻⁶ C
Q = 180 µC
🟡 Marks:
Ceq (1) + Q formula (1)
Q10. Parallel Plate Capacitor
C = ε₀A/d
Given:
A = 0.02 m²
d = 1 mm = 10⁻³ m
ε₀ = 8.85 × 10⁻¹²
Calculation:
C = (8.85 × 10⁻¹² × 0.02) / 10⁻³
C = 1.77 × 10⁻¹³ F
✅ Answer: 1.77 × 10⁻¹³ F
Q11. Electric Field in Conductor
E = 0 inside conductor
Since E = -dV/dx
If E = 0 → V constant
Explanation:
No potential difference inside conductor → no electric field
🟢 SECTION C (6 Marks)
Q12. Capacitor Sharing Charge
C1 = 5 µF, V1 = 100 V
C2 = 10 µF (uncharged)
Step 1: Initial charge
Q = CV = 5 × 100 = 500 µC
Step 2: Final voltage
Total C = 5 + 10 = 15 µF
Vf = Q / C
Vf = 500 / 15 = 33.33 V
Step 3: Charge distribution
Q1 = 5 × 33.33 = 166.65 µC
Q2 = 10 × 33.33 = 333.3 µC
Step 4: Energy loss reason
Energy lost as heat due to redistribution of charge (resistance present)
Q13. Dielectric Problem
K = 5
C = 10 µF
V = 100 V
(i) New capacitance
C' = KC = 5 × 10 = 50 µF
(ii) New voltage (battery disconnected)
Q constant
Q = CV = 10 × 100 = 1000 µC
V' = Q / C'
V' = 1000 / 50 = 20 V
(iii) Reason
Dielectric reduces electric field → increases capacitance → voltage decreases
🟢 SECTION D (5 Marks)
Q14. Series Capacitor Network
C1 = 2 µF, C2 = 3 µF, C3 = 6 µF
V = 12 V
(i) Equivalent capacitance
1/Ceq = 1/2 + 1/3 + 1/6
LCM = 6
= 3/6 + 2/6 + 1/6 = 6/6
Ceq = 1 µF
(ii) Charge
Q = CV = 1 × 12 = 12 µC
(iii) Potential difference
V1 = Q/C1 = 12/2 = 6 V
V2 = 12/3 = 4 V
V3 = 12/6 = 2 V
(iv) Energy
U = ½ CV²
U = ½ × 1 × 144
U = 72 µJ
(v) If parallel
C increases → energy increases significantly (U = ½ CV²)
🟢 SECTION E (4 Marks)
Q15. Energy in Capacitor
C = 100 µF = 100 × 10⁻⁶
V = 12 V
(i) Energy
U = ½ CV²
U = ½ × 100 × 10⁻⁶ × 144
U = 7200 × 10⁻⁶
U = 7.2 × 10⁻³ J
(ii) If voltage doubles
U ∝ V² → energy becomes 4 times
(iii) Reason
Capacitors:
store energy quickly
release energy instantly
used in flash cameras due to rapid discharge
📊 PAPER ANALYSIS (CBSE BOARD LEVEL)
Difficulty:
Easy: 35%
Moderate: 45%
Hard: 20%
Key Competency Areas:
✔ Capacitor combinations
✔ Energy formula
✔ Dielectric concept
✔ Potential concept reasoning
Most Important Board Topic:
👉 Series/Parallel combination + energy relations
🚀 PRO TIPS FOR FULL MARKS
✔ Always write formula first
✔ Convert units properly (mm → m, µF → F)
✔ Use stepwise charge conservation
✔ Highlight final answer clearly
✔ Draw simple capacitor diagrams if possible
✔ Use Q = CV everywhere in redistribution questions
⚠️ COMMON MISTAKE SHEET
❌ Forgetting unit conversion (mm → m)
❌ Mixing series & parallel formulas
❌ Using voltage constant in dielectric problem (wrong if battery disconnected)
❌ Not conserving charge in capacitor sharing
❌ Writing final answer without steps
❌ Confusing energy formula (U = CV² vs ½CV²)
🎯 FINAL BOARD STRATEGY
✔ Section A → 5–6 min
✔ Section B → 20 min
✔ Section C → 25 min
✔ Section D → 20 min
✔ Section E → 10 min
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