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Given two point charges of opposite sign I need to prove that inside the electric field they create there is an equipotential sphere.

I'm very positive that this is more geometry than anything else and I really question why I've been given this exercise. Here's my thoughts anyways and everything I can remember from my background in math and geometry.

The potential for a point charge is given by $$V=\frac{q}{4\pi ε_0r}$$

Let's assume the charges are correlated by the following ratio $$\frac{q_1}{q_2}=-a$$ What we want is $$V_1+V_2=0=>\frac{a}{r_1}=\frac{1}{r_2}=>\\r_1=ar_2$$

The distances $r$ are the distance from the charge to any point in the equipotential surface.

The equation of a sphere is the following: $$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2$$

I can also write $$r_1=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2+(z_1-z_0)^2}\\r_2=\sqrt{(x_2-x_0)^2+(y_2-y_0)^2+(z_2-z_0)^2}$$and then use r1=ar2. But then I really get lost.

Is there another way? And if not how do I show that this is a sphere?

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The electric field around the charges will have rotational symmetry about the line joining them, so the problem can be reduced to 2D and the task of proving that the locus is a circle.

The only physics here is getting $r_1=ar_2$. The rest is geometry. One proof relates to the Apollonian Circles Theorem.

Using co-ordinate geometry :

Suppose the charges are located at A(0,0) and B(d,0) where d=AB is the fixed distance between them. The distances AP, BP of some point P(x,y) from A,B are given by
$AP^2=x^2+y^2$
$AB^2=(d-x)^2+y^2$.
Set $a^2AP^2=BP^2$ to get an equation for the locus of points P. This has the form
$x^2+y^2-2fx-2gy+c=0$
which is the equation of a circle.

This problem is the inverse of finding the image charge A of a point charge B in a grounded conducting sphere. See Griffiths 2007, Problem 3.7.

I think Icchyamoy is wrong : the non-zero potential surfaces are not spherical. However, if a 3rd charge of suitable magnitude and position is introduced, the surface with any non-zero potential can be made spherical by a suitable choice of the magnitude and position of the 3rd charge. See A point charge near a conducting sphere.

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I guess this can be done if we use polar coordinates. In polar coordinates, the equation of a sphere looks like $R=C_0,$ where $C_0$ is a constant. Then the thing can be dealt in this way:

EDIT: $\mathbf {Z= r_0-r_1}$

Solution

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  • $\begingroup$ @sammy seems to be correct regarding nonzero equipotential surfaces not being spherical $\endgroup$ – Icchyamoy Jul 30 '17 at 9:07

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