Obtain the equation of electric potential energy of a dipole from equation of potential energy of a system of two electric charges.

Explanation

The equation of potential energy of a system a two charges is,
`U (theta) = q_(1)V(r_(1))+q_(2)V(r_(2))+(kq_(1)q_(2))/(r_(12))`
For dipole `q_(1)=+q,q_(2)=-q` and their positior vectors are `vecr_(1)` and `vecr_(2)` .
`:. U.(theta)=q[V(r_(1))-V(r_(2))]-(kq_(1)q_(2))/(r_(12))`
The potential difference between positions `r_(1)`and `r_(2)` equals the work done in bringing a unit positive charge against field from `r_(2)` to `r_(1)`. The displacement parallel to the force is 2acose `theta` .
`:. V(r_(1))-V(r_(2))=-Exx2a cos theta[ because `W=Fd]
`=- p E costheta[because` p =q(2a)=2a]
`:. U.(theta)=-pEcostheta-(kq^(2))/(2a)`
`:.U.theta=-(vecp.vecE)-(kq^(2))/(2a)`
For a given dipole U(6) differs from U(`theta`) only by a constant.
If `theta_(0)=(pi)/(2) ` be taken then `q[V(r_(1))-V(r_(2))]=0`
If we drop the second term in equation (2) then,
`U.(theta)=(vecP.vecE) ` is like `U =- (vecP.vecE).`