An electric dipole of length 2 cm is placed with its axis making 30...

Question

An electric dipole of length 2 cm is placed with its axis making 30° to a uniform electric field of 10⁵ N/C. It experiences a torque of 10√3 N·m. Find (i) magnitude of charge on dipole and (ii) potential energy of the dipole.

Accepted Answer

Given:
• Length of dipole, 2l = 2 cm = 2 × 10⁻² m → l = 10⁻² m
• Electric field, E = 10⁵ N/C
• Torque, τ = 10√3 N·m
• Angle, θ = 30°

Part (i) — Finding Charge q:

Formula: τ = pE sinθ (where p = q × 2l)
10√3 = p × 10⁵ × sin 30°
10√3 = p × 10⁵ × 0.5
p = 10√3 / (0.5 × 10⁵) = 20√3 × 10⁻⁵ C·m
p = 2√3 × 10⁻⁴ C·m

Now, p = q × 2l:
q = p / 2l = (2√3 × 10⁻⁴) / (2 × 10⁻²)
q = √3 × 10⁻² C ≈ 1.73 × 10⁻² C

Part (ii) — Potential Energy:

Formula: U = −pE cosθ
U = −(2√3 × 10⁻⁴) × 10⁵ × cos 30°
U = −2√3 × 10 × (√3/2)
U = −2√3 × 10 × 0.866
U = −30 J

∴ Charge q = √3 × 10⁻² C and Potential Energy = −30 J

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