Question

A dipole of length 2 cm is placed with its axis making an angle of 600 to a uniform electric field experiences a torque of 4 √3 Nm. Calculate the (i) magnitude of electric field, (ii) potential energy of dipole. The dipole has charges ±8 nC

Answer 1

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/U = 2Qlsinθ/−2Qlcoθ (where P=qd)

=−tanθ

τ/U = – tanθ

U = τ/ – tan60° = –8√3/√3 = – 8 joule