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Electrostatics Solved Examples (Class 12 Physics)

Electrostatics studies electric charges at rest and the electric fields and forces they create. These examples cover Coulomb's law, electric potential, cap

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TL;DR: Electrostatics studies electric charges at rest and the electric fields and forces they create. These examples cover Coulomb's law, electric potential…

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Electrostatics studies electric charges at rest and the electric fields and forces they create. These examples cover Coulomb's law, electric potential, cap

Electrostatics — Solved Numerical Examples (Step by Step)

Example 1: Two point charges of 4 μC and 8 μC are separated by 2 m. Calculate the force between them. (k = 9 × 10⁹ N m²/C²)

Solution: Given: q1 = 4 μC = 4 × 10⁻⁶ C, q2 = 8 μC = 8 × 10⁻⁶ C, r = 2 m, k = 9 × 10⁹ N m²/C²

Using Coulomb's law: F = k × q1 × q2 / r²
F = (9 × 10⁹) × (4 × 10⁻⁶) × (8 × 10⁻⁶) / 2²
F = (9 × 10⁹) × (32 × 10⁻¹²) / 4
F = (288 × 10⁻³) / 4
F = 72 × 10⁻³ = 0.072 N = 72 mN

Example 2: A uniform electric field of magnitude 1000 N/C points from south to north. Find the force on an electron placed in this field. (e = 1.6 × 10⁻¹⁹ C)

Solution: Given: Electric field E = 1000 N/C (pointing south to north), charge on electron q = -1.6 × 10⁻¹⁹ C

Force on charge: F = qE
F = (-1.6 × 10⁻¹⁹) × 1000
F = -1.6 × 10⁻¹⁶ N

Magnitude = 1.6 × 10⁻¹⁶ N
Direction: opposite to field (south), since electron is negatively charged

Example 3: Find the electric potential due to a point charge of 5 × 10⁻⁹ C at a distance of 10 cm. (k = 9 × 10⁹ N m²/C²)

Solution: Given: Charge q = 5 × 10⁻⁹ C, distance r = 10 cm = 0.1 m, k = 9 × 10⁹ N m²/C²

Electric potential: V = kq/r
V = (9 × 10⁹) × (5 × 10⁻⁹) / 0.1
V = (45) / 0.1
V = 450 V

Example 4: A parallel plate capacitor has plates of area 100 cm² separated by 2 mm. Find its capacitance. (ε₀ = 8.85 × 10⁻¹² F/m)

Solution: Given: Area A = 100 cm² = 100 × 10⁻⁴ m² = 0.01 m², separation d = 2 mm = 2 × 10⁻³ m, ε₀ = 8.85 × 10⁻¹² F/m

For parallel plate capacitor: C = ε₀A/d
C = (8.85 × 10⁻¹²) × 0.01 / (2 × 10⁻³)
C = (8.85 × 10⁻¹⁴) / (2 × 10⁻³)
C = 4.425 × 10⁻¹¹ F = 44.25 × 10⁻¹² F = 44.25 pF

Example 5: A capacitor of 5 μF is connected to a 12 V battery. Calculate the charge stored and energy stored.

Solution: Given: Capacitance C = 5 μF = 5 × 10⁻⁶ F, voltage V = 12 V

Charge: Q = CV = (5 × 10⁻⁶) × 12 = 60 × 10⁻⁶ C = 60 μC

Energy stored: U = (1/2)CV² = (1/2) × (5 × 10⁻⁶) × 12²
U = (1/2) × (5 × 10⁻⁶) × 144
U = 360 × 10⁻⁶ J = 360 μJ

Alternatively: U = (1/2)QV = (1/2) × (60 × 10⁻⁶) × 12 = 360 × 10⁻⁶ J

Example 6: A small sphere of mass 2 g and charge 4 × 10⁻⁸ C hangs from a string in a horizontal uniform electric field of 1000 N/C. Find the angle the string makes with vertical at equilibrium. (g = 10 m/s²)

Solution: Given: mass m = 2 g = 0.002 kg, charge q = 4 × 10⁻⁸ C, electric field E = 1000 N/C, g = 10 m/s²

Weight: W = mg = 0.002 × 10 = 0.02 N
Electric force: F_E = qE = (4 × 10⁻⁸) × 1000 = 4 × 10⁻⁵ N

At equilibrium, the string makes angle θ with vertical:
tan θ = F_E / W = (4 × 10⁻⁵) / 0.02 = 4 × 10⁻⁵ / (2 × 10⁻²) = 2 × 10⁻³
θ = arctan(2 × 10⁻³) ≈ 0.115° (very small angle)

Tips

  • In Coulomb's law, the force can be attractive or repulsive depending on the signs of charges (same sign = repulsive, opposite = attractive).
  • Electric potential is a scalar quantity, so it adds algebraically (with sign) when due to multiple charges.
  • The electric field points from positive to negative charges, while the force on a positive charge is in the direction of the field.
  • Energy stored in a capacitor increases with the square of voltage, so even small voltage increases can store significantly more energy.

Frequently Asked Questions

What is the relationship between electric field and electric potential?

Electric field and electric potential are related by E = -dV/dr, where E is the electric field and V is the potential. In simpler terms, the electric field points in the direction of decreasing potential, and its magnitude tells us how rapidly the potential changes with distance.

Can electric potential be negative?

Yes, electric potential can be negative. The sign depends on the type of charge creating the potential. A positive charge creates positive potential, while a negative charge creates negative potential. The potential is always measured relative to infinity, which is taken as zero reference.

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