Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform electric field E, as shown in Fig.
As we know that in a uniform electric field, the dipole experiences no net force; but experiences a torque τ given by,
τ = p×E
which will tend to rotate it (unless p is parallel or antiparallel to E).
Suppose an external torque τ(ext) is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle θo to angle θ1 at an infinitesimal angular speed and without angular acceleration. The amount of work done by the external torque will be given by
This work is stored as the potential energy of the system.
We can then associate potential energy U(θ) with an inclination θ of the dipole.
Similar to other potential energies, there is a freedom in choosing the angle wherethe potential energy U is taken to be zero.
A natural choice is to take θo= π / 2. We can then write,
U(θ) = - pE cos θ = - p . E
This expression can alternately be understood also from Eq.
U = q1 V(r1) + q2 V(r2) + q1q2 / 4πε0 r12
We apply above equation to the present system of two charges +q and –q. The potential energy expression then reads
U' (θ)= q[V(r1) - V(r2)] + q^2 / 4πε0 r12
Here, r1 and r2 denote the position vectors of +q and –q.
Now, the potential difference between positions r1 and r2 equals the work done in bringing a unit positive charge against field from r2 to r1.
Thee displacement parallel to the force is 2a cosθ. Thus, [V(r1)–V (r2)] =–E × 2a cosθ. We thus obtain,
U' (θ) = –qE × 2a cosθ + q^2 / 4πε0 r12
U' (θ) = –pE cosθ + q^2 / 4πε0 r12
( Since p = q2a)
U' (θ) = –p.E + q^2 / 4πε0 r12
