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 N/C. If it experiences a torque of 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