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My question relates to the example given by Griffiths’ Electrodynamics regarding a sphere in a constant electric field.

Specifically, in Example 3.8 of Griffiths’ Electrodynamics, there is a metal sphere of radius $R$. It is placed in a uniform electric field $\textbf{E} = E_{0}\hat{x}$.

Now, Griffiths is able to obtain a formula for the potential in spherical coordinates and shows that $$V(r, \theta) = -E_{0}\left(r - \frac{R^3}{r^2}\right)\cos\theta$$

We can then just find the electric field by taking the negative gradient of $V$ (this is, of course, for outside the sphere).

However, this example involves an empty metallic sphere. The question I have is: suppose we now have the same situation, except place a point charge with charge $Q$ inside of the sphere, and specifically place it at a position of $(x,y,z)$, where this position vector is contained within the sphere (hence, $Q$ need not be at the centre of the sphere but must be contained inside).

My question is, what is the electric field outside the sphere?

I'm not entirely sure how to even go about this problem. I feel like maybe we could use superposition, but I'm not sure if that's allowed in this case. Ideally, I'd like to solve this problem just as Griffiths does in Example 3.8; by considering the boundary values of the problem and finding potential, then taking the negative gradient to obtain the field.

$\textbf{EDIT:}$ The issue here is that I would like to solve this using the method Griffiths' uses, regarding spherical coordinates. I understand intuitively why superposition might work, but I want to know how to use this method from the book (3.3) specifically.

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  • $\begingroup$ The answer is actually just a superposition of examples 3.8 and 2.10 in Griffiths' book. $\endgroup$
    – Buzz
    Sep 20, 2022 at 19:33
  • $\begingroup$ Thanks for your comment! I suppose I'm not understanding why this is true. In particular, I'm curious as to how we'd solve for the potential (which then gives us the electric field) using Griffiths' method of azimuthal symmetry. What I have now is that the boundary conditions are $V(R) = 0$ and $V$approaches $-E_{0}r\cos(\theta) + Q/4\pi\varepsilon$ in the limit that $r$ gets larger. However, when I try and solve given these boundary conditions, I don't get the superposition result. $\endgroup$ Sep 21, 2022 at 21:52
  • $\begingroup$ And thus I'm wondering how we apply the method shown in Section 3.3 of the book (azimuthal symmetry & spherical coordinates) to prove that the answer must be superposition. $\endgroup$ Sep 21, 2022 at 21:54
  • $\begingroup$ The exterior potential is just $-E_{0}(r-R^{3}/r^{2})\cos\theta+Q/4\pi\epsilon_{0}r$. $\endgroup$
    – Buzz
    Sep 22, 2022 at 0:30

2 Answers 2

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The spherical shell must have some nonzero thickness, which we can treat as arbitrarily close to zero, but positive. The effect of the external electric field is to induce a distribution of charge solely on the outer surface of the shell such that the external field is cancelled everywhere inside this surface (i.e. at all deeper points within the shell and in the empty interior of the sphere).

(Why? Suppose we had a solid metal ball. There would be no inner surface for any induced charge to reside on, and there couldn't be any nonzero charge density anywhere in the interior of the ball, since it would violate Gauss's law. So all induced charge would lie on the surface. Since the induced charge on the outer surface would be sufficient to cancel the external field everywhere inside the ball, without any help from any internal charge density, we can now imagine hollowing out the ball, leaving an inner surface. This wouldn't change the charge distribution anywhere, so the electric field would still be zero everywhere inside the shell except on the outer surface. This implies that there would be nothing to induce any charge on the inner surface.)

If we place a point charge $Q$ anywhere inside the sphere, the inner surface of the shell will acquire a charge $-Q$ that will cancel the point charge's field everywhere outside the inner surface. We know this is possible because Gauss's law requires it.

To maintain neutrality, this charge of $-Q$ on the inner surface must be balanced by a charge of $+Q$ that distributes itself somehow over the outer surface. We know that this induced charge $+Q$ won't feel any tangential force from the sum of the applied field and the charge that the applied field induced on the outer surface (since, if there were any such force, the latter would not have been at equilibrium in the first place). So it will distribute itself in the same way that it would in the absence of the applied field and the applied field's induced charge: that is, uniformly.

The total field outside the sphere will therefore be the sum of the previously calculated field (i.e. due to the combination of the applied field and the charge distribution induced by the applied field) and the field due to a uniformly charged spherical shell with total charge $Q$.

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Well Electric field will be the part you have plus simple culomb potential. The 1st part is for '0' solutions of poisson equation i.e charge density is zero part which will give you that part and next part just potential from center as independent of location of charge.. as conductor equipotential surface

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Sep 20, 2022 at 19:04

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