# Behaviour of electric field when an electric dipole is placed inside a conducting shell

Suppose there is a spherical shell made of a perfectly conducting material with inner radius $$R_{1}$$ and outer radius $$R_{2}$$. A dipole with charges $$+q$$ and $$-q$$ separated by a distance $$a$$ ($$a) is placed inside the shell.

What would the electric field be inside and outside the shell? What would the charge induced look like on both the inside surface and the outside surface?

This is a standard problem. I will assume that the dipole is very small and located at the origin, but the method can be generalized to physical dipoles with finite separations at arbitrary locations. The key point is that there must be zero field inside the thick conducting shell ($$\vec{E}=0$$ always inside a conductor), so there must be an surface charge at the inner surface that will cancel all the electric field of the dipole at radii $$r>R_{1}$$.

The potential of the dipole itself (with dipole moment $$\vec{p}=qa\hat{z}$$; the $$z$$-direction is chosen to point in the direction the dipole—that is, from the negative charge to the positive charge) at the origin is (in MKS units) $$\Phi_{\mathrm{dip}}=\frac{p\cos\theta}{4\pi\epsilon_{0}r^{2}}.$$ Except at $$r=0$$, this is a special case of the fact that the electrostatic potential in a vacuum region, for a system with rotational symmetry around the $$z$$-axis, may be written in the separation-of-variables form $$\Phi(r,\theta)=\sum_{\ell=0}^{\infty}\left(A_{\ell}r^{\ell}+B_{\ell}r^{-\ell-1}\right)P_{\ell}(\cos\theta),$$ where the $$P_{\ell}(\xi)$$ are the Legendre polynomials. Note that, since $$P_{1}(\xi)=\xi$$, the dipole field $$\Phi_{\mathrm{dip}}$$ has this form with only the $$B_{1}$$ coefficient nonzero.

The combined $$\Phi$$ of the dipole $$\vec{p}$$ and the surface charge layer $$\sigma_{1}$$ on the inner surface of the conductor must have the separable form in between $$r=0$$ and $$r=R_{1}$$, since that region is vacuum. As $$r\rightarrow0$$, it should approach the (singular) form $$\Phi_{\mathrm{dip}}$$. At $$r=R_{1}$$, the combined potential must be a constant, since the conductor is an equipotential. [We can take its potential to be zero; if it has a different value $$V_{0}$$, you just add that $$V_{0}$$—which is a $$A_{0}$$ term in the expansion, since $$P_{0}(\xi)=1$$—to the whole expression, raising of lowering the potential everywhere in space by a constant.] To satisfy these boundary conditions, we need a combination of $$A_{1}$$ and $$B_{1}$$ terms, $$\Phi(r

This turns out to be the only region where the field is nonzero. Since the total charge contained within the conductor was zero, there was no net charge to accumulate on the outer surface. If there were a total nonzero $$Q$$ inside the hollow conductor, the total of the surface charge $$\sigma_{1}$$ needed to cancel it at the inner surface $$R_{1}$$ would be $$-Q$$. Then, in order that the conducting mass itself should remain overall charge neutral, there must be a compensating surface charge density $$\sigma_{2}$$ located at the outer surface $$R_{2}$$. This $$\sigma_{2}$$ is always spread out uniformly over the surface, because the charges interior to it exert no net field outside the hollow hole in the conductor. So the distribution of $$\sigma_{2}$$ is determined just by the self-repulsion of the total charge $$Q$$, which spreads it uniformly over the spherical surface. This is a special case of a famous and somewhat surprising fact about conductors: The electrostatic field outside a conductor is determined entirely by the total charge located anywhere inside the conductor, whether that charge is contained in insulating hollows or just dropped on the surface. In this case, since the total charge of the interior dipole is $$Q=0$$, the field everywhere outside the conductor is also vanishing.

This qualitative result is unchanged if we relax the assumptions I mentioned in the first paragraph. So long as the total charge in the hollow is zero, the potential everywhere outside the grounded inner surface of the conductor is also zero. Moreover, the exact field even inside the spherical hollow is known analytically for arbitrary arrangements of pointlike charges and dipoles. It may be found using the method of images, but the expressions are not particularly illuminating.

• Thank you @Buzz , that was a very thorough answer. Just to make sure I've got the point, for a distance r which is less than R1, the field would be non zero, meanwhile for any distance r which is in between R1 and R2 and also for a distance greater than R2, the field would be zero. Is that correct ? Also, I noticed the equation that you used for the field due to the dipole itself but isn't that the equation for potential ? Commented Jan 3 at 6:35
• @wonderingwhy If the total charge inside the hollow is zero, then there is no net field anywhere outside the inner surface of the conductor. That is true regardless of the shape of the surrounding conductor. As to the "field" terminology, physicists tend to refer to the potential as the field (since the potential is more important in quantum mechanics and such, and it is, mathematically speaking, a field). However, I have edited that to make it clearer.
– Buzz
Commented Jan 4 at 4:17