Electrostatics — Previous Year Questions (Class 12 Physics)
Electrostatics covers electric charge, Coulomb's law, electric field, potential, and capacitance. It provides the foundation for understanding electrical p
TL;DR: Electrostatics covers electric charge, Coulomb's law, electric field, potential, and capacitance. It provides the foundation for understanding electri…
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Electrostatics covers electric charge, Coulomb's law, electric field, potential, and capacitance. It provides the foundation for understanding electrical p
Electrostatics — Previous Year Questions with Solutions
Q (2023, 3 marks): Two charges Q1 = 2 × 10^-6 C and Q2 = 3 × 10^-6 C are separated by 30 cm. Find the force between them. (k = 9 × 10^9 N·m²/C²)
Answer: Using Coulomb's law: F = k|Q1||Q2|/r²
where k = 9 × 10^9 N·m²/C², Q1 = 2 × 10^-6 C, Q2 = 3 × 10^-6 C, r = 0.3 m
F = (9 × 10^9 × 2 × 10^-6 × 3 × 10^-6) / (0.3)²
F = (9 × 10^9 × 6 × 10^-12) / 0.09
F = (54 × 10^-3) / 0.09
F = 0.054 / 0.09
F = 0.6 N
The force is repulsive (both charges are positive).
Q (2022, 3 marks): Define electric field intensity. Calculate field intensity at a point 10 cm from a charge of 2 × 10^-7 C.
Answer: Electric field intensity (E) is the force per unit charge: E = F/q = kQ/r²
Given: Q = 2 × 10^-7 C, r = 0.1 m, k = 9 × 10^9 N·m²/C²
E = (9 × 10^9 × 2 × 10^-7) / (0.1)²
E = (9 × 10^9 × 2 × 10^-7) / 0.01
E = (18 × 10^2) / 0.01
E = 1800 / 0.01
E = 1.8 × 10^5 N/C
Direction: Radially away from the charge (since it's positive)
Q (2023, 2 marks): Calculate the potential at a distance of 20 cm from a charge of 5 × 10^-9 C.
Answer: Electric potential: V = kQ/r
where k = 9 × 10^9 N·m²/C², Q = 5 × 10^-9 C, r = 0.2 m
V = (9 × 10^9 × 5 × 10^-9) / 0.2
V = 45 / 0.2
V = 225 V
Q (2021, 2 marks): A parallel plate capacitor has area 100 cm² and plate separation 2 mm. Find its capacitance. (ε0 = 8.85 × 10^-12 F/m)
Answer: Capacitance of parallel plate capacitor: C = ε0 × A / d
where ε0 = 8.85 × 10^-12 F/m, A = 100 × 10^-4 m² = 0.01 m², d = 2 × 10^-3 m
C = (8.85 × 10^-12 × 0.01) / (2 × 10^-3)
C = (8.85 × 10^-14) / (2 × 10^-3)
C = 4.425 × 10^-11 F
C = 44.25 pF
Q (2022, 3 marks): Two point charges 4 μC and -2 μC are separated by 30 cm. Find the point on the line joining them where electric field is zero.
Answer: Let point P be at distance x from 4 μC charge
At zero field point: E1 = E2 (magnitudes equal, opposite directions)
k(4 × 10^-6)/x² = k(2 × 10^-6)/(30-x)²
4/x² = 2/(30-x)²
4(30-x)² = 2x²
2(30-x)² = x²
sqrt(2)(30-x) = x
30*sqrt(2) - x*sqrt(2) = x
30*sqrt(2) = x(1 + sqrt(2))
x = 30*sqrt(2) / (1 + sqrt(2)) = 30*sqrt(2) × (sqrt(2) - 1) / ((1 + sqrt(2))(sqrt(2) - 1))
x = 30*sqrt(2) × (sqrt(2) - 1) / (2 - 1) = 30*sqrt(2)(sqrt(2) - 1)
x = 30(2 - sqrt(2)) = 60 - 30*sqrt(2) ≈ 17.6 cm from 4 μC charge
Q (2023, 2 marks): A capacitor is charged to 100 V and stores 4 × 10^-6 J of energy. Find its capacitance.
Answer: Energy stored in capacitor: U = (1/2)CV²
where U = 4 × 10^-6 J, V = 100 V
4 × 10^-6 = (1/2) × C × (100)²
4 × 10^-6 = (1/2) × C × 10000
4 × 10^-6 = 5000C
C = (4 × 10^-6) / 5000
C = 8 × 10^-10 F = 0.8 nF
Frequently Asked Questions
What is the difference between electric potential and electric potential energy?
Electric potential (V) is the potential energy per unit charge at a point: V = U/q, measured in volts. Electric potential energy (U) is the total energy a charge possesses in an electric field, measured in joules.
Why is the electric field zero inside a conductor in electrostatic equilibrium?
In electrostatic equilibrium, free charges in the conductor redistribute until the electric field inside becomes zero. This prevents further motion of charges, achieving equilibrium.
More Class 12 Physics PYQs
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- Electromagnetic Induction
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- Atoms and Nuclei
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