A short dipole $\mathrm{p}\hat{\mathrm{i}}$ is placed at origin in x-y plane. The locus of equipotential surface around the dipole if a constant electric field $\mathrm{E}\hat{\mathrm{i}}$ also exists through out the space is $\left(\mathrm{k}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\left({\mathrm{V}}_{\text{origin }}=0\right)$

A equipotential surface will be formed when potential due to dipole + potential due to electric field =0 

Let the position of point be given by $\overrightarrow{\mathrm{a}}$

Potential due to dipole is given by $\frac{\mathrm{k} \overrightarrow{\mathrm{P}}.\overrightarrow{\mathrm{a}}}{{\mathrm{a}}^3}$

Potential due to the electric field is given by $-\overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{a}}$

$\Rightarrow \frac{\mathrm{k} \mathrm{Pacos\theta }}{{\mathrm{a}}^3}+\left(-\mathrm{Eacos\theta }\right)=0$

$\Rightarrow \frac{\mathrm{kP}}{{\mathrm{a}}^2}=\mathrm{Ea}\Rightarrow {\mathrm{a}}^3=\frac{\mathrm{kP}}{\mathrm{E}}\Rightarrow \mathrm{a}={\left(\frac{\mathrm{kP}}{\mathrm{E}}\right)}^{1/3}$

$\begin{array}{l}\Rightarrow \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}={\left(\frac{\mathrm{kP}}{\mathrm{E}}\right)}^{1/3}\\ {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2={\left(\frac{\mathrm{kP}}{\mathrm{E}}\right)}^{2/3}\end{array}$