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I have been trying to find a equation for the equipotential surface of a dipole , so I started with a simpler system of a singular charged particle , here are few things I know about the equipotential surface

The divergence through the surface should be q(enc.)/ε

$$\iiint\nabla\cdot\vec EdV=q/\epsilon $$

The equipotential surface is always perpendicular to the electric field But I have been struggling to extract the information about the surface from just these criteria.

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    $\begingroup$ Hint: You need to find the electric potential. $\endgroup$ Commented Apr 15 at 8:47

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An equipotential surface is by definition the surface $S=\{(x,y,z)|\phi(x,y,z)=k\}$. In this sense the equipotential surface depends on the exact form of the potential, however, it will still be a function of the spatial coordinates $x,y,z$. Given a relation like $\phi(x,y,z)=k$, one can solve for a coordinate, say $z$, and get an equation of surface: $z=f(x,y)$. As an example, consider the dipole with moment $\vec p=q\vec l$, where the expression for the electric potential is given by: $$\phi(\vec r)={q\over 4\pi\epsilon_0}\bigg[{1\over|\vec r-\vec r^\prime-\vec l|}-{1\over|\vec r-\vec r^\prime|}\bigg].$$ To find the equipotential surface, set $\phi(\vec r)=k$, where $k$ is some constant that defines the level surface; the equaiton of the surface is then: $$k={q\over 4\pi\epsilon_0}\bigg[{1\over|\vec r-\vec r^\prime-\vec l|}-{1\over|\vec r-\vec r^\prime|}\bigg].$$

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