# Point Charge in the presence of charged, insulated, conducting sphere

1. Sep 27, 2013

### M. next

So here we are talking about solving this problem by method of images.
The approach taken by most of electrodynamics textbooks is as follows:

"If we wish to consider the problem of an insulated conducting sphere with total charge Q in the presence of a point charge q, we can build up the solution for the potential by linear superposition. IN an operational sense, we can imagine that we start with the grounded conducting sphere (with its charge q' distributed over its surface). We then disconnect the ground wire and add to the sphere an amount of charge (-q'). This brings the total charge on the sphere up to Q. To find the potential we merely note that the added charge (Q-q') will distribute itself uniformly over the surface, since the electrostatic forces due to the point charge q are already balanced by the charge q'. Hence the potential due to the added charge (Q-q') will be the same as if a point charge of that magnitude were at the origin, at least for points outside the sphere." (Chapter 2, p. 61) from Electrodynamics by J. D. Jackson.

Then to find the potential, the author uses the superposition principle (the author superposes the potential due to q, q' and (Q-q').

My question here is: Can we find the electric field of this charges sphere directly from Gauss's law and then find the potential due to it as a whole & then superpose it with the charge q outside the sphere? (i.e, without the referring to q') If not, why?

2. Sep 27, 2013

### WannabeNewton

Yes you can; consider for example the case wherein we want the potential of the system exterior to the conducting sphere of radius $R$. Suppose the conducting sphere is grounded i.e. $\varphi(R) = 0$; place the origin of the spherical coordinates on the center of the sphere. We can always orient our coordinate system so that the point charge lies on the polar axis at some $a > R$. There will be some induced potential $\varphi_{i}$ associated with the sphere of charge. By superposition, the net potential is then $\varphi = \varphi_i + \varphi_p$ where $\varphi_p$ is the potential due to the point charge.

By Gauss's law, we are then solving Laplace's equation $\nabla^{2}\varphi_i = 0$ for the induced potential in the region outside the sphere (excluding the point charge delta function source of course) with the boundary condition $\varphi_i (R) = -\varphi_p(R)$. Because of the aziumuthal symmetry of the system, the general solution is $\varphi_{i} = \sum (A_{l}r^{l}P_{l}(cos\theta) + B_{l}r^{-l-1}P_{l}(cos\theta))$ where $P_{l}$ are the Legendre polynomials and $A_l,B_l$ are expansion coefficients. We can immediately conclude that $A_{l} = 0$ for all $l$ because $r^{l}$ blows up at $r = \infty$ which cannot happen for this system (there are no point charges at infinity) hence we are just left with $\varphi_{i} = \sum_{l}B_{l}r^{-l-1}P_{l}(cos\theta)$.

To find the $B_l$ we just apply the boundary condition stated above and we are done.