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 = <b>√3 × 10⁻² C ≈ 1.73 × 10⁻² C</b>

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 = <b>−30 J</b>

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