An electric dipole of length 2 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 8√3 Nm. Calculate the potential energy of the dipole, if it has a charge of ±4 nC.

Torque, τ = PEsin θ = 2Qlsin θ
Here l is the length of the dipole, Q is the charge and E is the electric field.
Potential energy, U = −PEcos θ = −2Qlcos θ
τ = 8√3 Nm
θ = 60°
2l = 2cm = 0.02m
Q = 4nC = 4 × 10⁻⁹C
τ = 2QlEsin θ
8√3 = 2 × 4 × 10⁻⁹ × 0.01 × E × sin 60°
E = 8√3 / (8 × 10⁻¹¹ × √3/2) = 2 × 10¹¹ N/C
U = −2QlEcos θ = −2 × 4 × 10⁻⁹ × 0.01 × 2 × 10¹¹ × cos 60°
U = −8 × 10⁻⁸ × 2 × 10¹¹ × 1/2 = −8 × 10³ J
U = -8000 J