Question

How much work is required to be done on a dipole in turning it from its position of stable equilibrium to its position of unstable equilibrium in uniform electric field e?

Answer 1

As we know, dW= pEsin theta d theta

Also, theta= 0 when stable equilibrium and 180 when in unstable equilibrium.

So integrating dW=pE (sin theta d theta)

dW= pE(cos theta1– costheta2)= pE (cos 180-cos 0)

dW= pE( -1–(1))=-2pE

Putting the values of E and P, we get

W= - 2×10^-8×10^3 =-2× 10^-5Nm

Answer 2

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The condition of stable equilibrium is achieved when the direction of extrernal electric field is parallel to the direction of the dipole moment.

Hence, Potential Energy = -PEcos?

Since ?=0, cos ?=1

Thereferoe, potential energy=PE=maximum

In unstable equilibirum, the direction of the electric field is antiparallel to the direction of dipole moment.

therefore, ?=180 degrees. cos?= -1

hence, Potential enegy = -PE

Hence, total work done = Initial energy - final energy = PE - (-PE) = 2PE