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Q.
The electric potential due to an extremely short dipole at a distance r from it is proportional to
Electrostatic Potential and Capacitance
Solution:
Potential at point $P =\frac{ KV }{ r _{1}}-\frac{ KV }{ r _{2}}$
$V = KV \left(\frac{ r _{2}- r _{1}}{ r _{1} r _{2}}\right)$
we are taking extremely short dipole $r_{1} \cong r_{2}=r$
$r _{2}- r _{1} \cong d \cos \theta$
$V =\frac{ KV d \cos \text { theta }}{ r ^{2}}$
$V =\frac{ K P \cos \text { theta }}{ r ^{2}} P = V d$,
where $P$ is the dipole moment.
$\therefore V \propto \frac{1}{ r ^{2}}$
