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How can we draw the equipotential surfaces (i.e the surface which has constant potential on all its points) for an electric dipole and a system of two like charges? I read it in some book that they are distorted spheres but why??

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    $\begingroup$ This Wolfram demonstration shows both the electric field lines and the equipotentials. Do not worry when you move a slider and see a very strange pattern, calculations are being done to produce the final diagram. demonstrations.wolfram.com/LinesOfForceForTwoPointCharges Very near each of the charges you would expect something which is nearly a circle as in that region one charge dominates. Far way from two charges of the same sign again near circular because they look like one charge. $\endgroup$
    – Farcher
    Commented Feb 25, 2016 at 14:20

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Let's consider two charges $q_1 = q_2$ charged by $q$ placed respectively at (x,y) = (-1,0) and (1,0) in a plane.

Books tells us that the electrical potential $V(M)$ at point $M = (x_m,y_m)$ is given by : $$ V(M) = Kq\left(\frac{1}{\sqrt{(x_m+1)^2+y_m^2}}+\frac{1}{\sqrt{(x_m-1)^2+y_m^2}}\right) $$ Which is simply the sum of the potential created by $q_1$ and the potential created by $q_2$, according to the superposition principle.

To extract an equation of an equipotential surface (which in a 2D plane is a line), we have to find for which $(x_m,y_m)$'s the quantity $V(M)$ remains the same. Let's set a constant $C$ for example, then, the equipotential line equation for $C$ is given by : $$ Kq\left(\frac{1}{\sqrt{(x_m+1)^2+y_m^2}}+\frac{1}{\sqrt{(x_m-1)^2+y_m^2}}\right) = C $$

Here is an example plot with $C = \frac{1}{Kq}$ :

equipo 1

and another with $C = \frac{2}{Kq}$ :

equipo2

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Pretty easy! So let the charges be $-q_1$ and $-q_2$, then we can write potential at any point $P$ as:

$$ V = \frac{-kq_1}{r_1} - \frac{kq_2}{r_2}$$

Where $r_1$ is the distance of the vector which connects $q_1$ to the point of consideration and $r_2$ is the vector which connects $q_2$ to that point. Now, took find equipotential surface, simply fix $V$ like so:

$$ C= - \frac{kq_1}{r_1} - \frac{kq_2}{r_2}$$

By dividing both sides by $-k$, and saying that $ C' = - \frac{C}{k}$ then:

$$ C' = \frac{q_1}{r_1} + \frac{q_2}{r_2} \tag{1}$$

I have made a plot of this by assuming $q_1$ is at origin and $q_2$ is at point (a,b), view my graph here

Examples:

If the negative charges are close and equal in magnitude:

enter image description here

If they are very different in magnitude:

enter image description here

If they are far apart and of different magnitude:

enter image description here

If it's a dipole, simply switch signs of one of the charges:

enter image description here

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