Derive an expression for electrostatic potential energy of the dipole in a uniform electric field .

Explanation

Electrostatic potential energy of a dipole in a uniform electric Field: place in the uniform electric field `overlineE`.A dipole experiences a torque when kept ina uniform electric field `overline E`.This torque rotates the dipole to align it with the direction of the electric field.To rotate the dipole (at constant angular velocity ) fromits initial angle `theta` another angle `theta` against the torque exerted by the electric field ,an equal and opposite external torque must be applies on the dipole.
The work done by the ecternal torque to rotate the diplole from the angle `theta` to `theta` at constant angular velocity is `W=int_(theta)^(theta)tau_(ext)d theta`
Since `tau_(ext)` is equal and opposite to `tau_E=overlinePxxoverlineE`,we have `|overlinetau_(ext)|=|overlinetau_E|=|overlinePxxoverlineE|`
Subsituting equation (2) in equation (1) we get, `W=int_(theta)^(theta)pEsin thetad theta=pE(costheta.-costheta)`
This work done is equal to the potential energy difference between the angular potions `theta` and `theta U(theta)-(U theta)=triangleU=-pE cos theta +pE cos theta`
If the initial aangle is `=theta.=90^(@)` and is taken as reference point ,then `U(theta.)+pE cos 90^(@)=0`.The potential energy stored in the system of dipole kept in the uniform electric field is given by `U=-pEcos theta=-overlinep .overlineE`
In addition to P and E ,the potential energy also depends on the orientation `theta` of the electric dipole with respect to the external electric field
The potential energy is maxmimum when the dipole is aligned anti-parallel `(theta=pi)` to the external electric field and maximum when th dipole is aligned parallel `(theta=0)` to the external electric field.