Demonstrate that the potential of the field generated by a dipole with the electric moment `p` (fig) may be represented as `varphi = pr//4pi epsilon_(0) r^(3)`, where `r` is the redius vector. Using this expression, find the magnitude of the electric strength vector as a funcition of `r` and `theta`.

Explanation

Let two charges `+q` and `-q` be separated by a distance `l`. Then electric potential at a point at a distance `r gt gt l` from this dipole,
`varphi(r ) = (+q)/(4pi epsilon_(0) r_(+)) + (-q)/(4pi epsilon_(0) r_(-)) = (q)/(4pi epsilon_(0)) ((r_(-) - r_(+))/(r_(+) r_(-)))` (1)
But `r_(-) - r_(+) = l cos theta` and `r_(+) r_(-) ~= r^(2)`
From Eqs. (1) and (2).
`varphi(r) = (q l cos theta)/(4pi epsilon_(0) r^(2)) = (p cos theta)/(4pi epsilon_(0) r^(2)) varphi = (vec(p). vec(r))/(4pi epsilon_(0) r^(3))`
where `p` is magnitude of electric moment vector
Now, `E_(r) = - (del varphi)/(del r) = (2 p cos theta)/(4pi epsilon_(0) r^(3))`
and `E_(theta) = - (del varphi)/(r del theta) = (p cos theta)/(4pi epsilon_(0) r^(3))`
So `E = sqrt(E_(r)^(2) + E_(0)^(2)) = (p)/(4pi epsilon_(0) r^(3)) sqrt(4 cos^(2) theta + sin^(2) + theta)`