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??

  • 1
    $\begingroup$ 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

2 Answers 2


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 $C = \frac{2}{Kq}$ :



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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.