Solve this:
Q. An electric dipole of length 2 cm is placed with its axis making an angle of 30$°$ to a uniform electric field ${10}^5$ N/C. If it experiences a torque of $10\sqrt{3}$ Nm, then potential energy of the dipole
(1) - 10 J
(2) - 20 J
(3) - 30 J
(4) - 40 J

Dear student,
The torque on an electric dipole placed in uniform electric field is given as

τ = q2dE sinθ

or magnitude of charge will be

q = τ / 2dEsinθ

here

τ = 17.32 Nm

d = 2cm = 0.02m

E = 10⁵ N/C

θ = 30 degrees

so,

q = 866 x 10^-5 C

now,

The potential energy of an electric dipole placed in uniform electric field is given as

U = -pEcosθ = -(q2d)Ecosθ

so,

U =- 866 x 10^-5 x 2x0.02 x 10⁵ x cos30

thus,

U =- 30J
Regards