Suppose two charged conducting spheres of different radii $r_1$ and $r_2 > r_1$ are very far apart but connected by a conducting wire. Why does the sphere with the smaller radius $(r_1)$ have the greater charge density? So why don't you have two spheres with equal charge density?

I thought that when the system is in equilibrium, then because the two spheres are very far apart, their fields are assumed not to interfere with themselves and since the system has the same potential everywhere (equilibrium), the charge density of each sphere would be same.

  • $\begingroup$ A nice example for charge concentration is St. Elm's fire. From the sharp of a wooden (!) mast goes plasma "to heaven" in time of thunderstorms. This is due to a very high electric potential and the one potential side is wide spreaded and on the other end ( the mast) the potential will be concentrated in the sharp or tip. $\endgroup$ Commented May 10, 2015 at 11:03

2 Answers 2


The potential at the surface of a charged sphere or radius $r$ is:

$$ V = k\frac{Q}{r} \tag{1} $$

Since the area of the sphere is $4\pi r^2$, the charge density is:

$$ \rho = \frac{Q}{4\pi r^2} $$

and rearranging gives:

$$ Q = \rho 4\pi r^2 $$

and substituting for $Q$ in equation (1) we get:$$ V = 4\pi k\rho r $$

Can you take it from here?

  • $\begingroup$ I'm sorry, could you elaborate on what you trying to say? $\endgroup$
    – AReeves
    Commented May 10, 2015 at 12:46
  • $\begingroup$ I'm saying that because $V_1 = V_2$, $4\pi k\rho_1 r_1 = 4\pi k\rho_2 r_2$, so $\rho_1/\rho_2 = r_2/r_1$. $\endgroup$ Commented May 10, 2015 at 15:23

Because they should have the same electric potential and electric potentials of them depends on charge not charge density.


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