Effects of Charges on Conducting Sphere

Most recent answer: 03/19/2011

Q:
I've been told that if two point charges of unequal magnitude q1 and q2 are fixed symmetrically about a solid conducting sphere of zero net charge each at a distance x away from the surface and diametrically opposite, then the conducting sphere acquires a tendency of expansion due to electrostatic forces on induced charges on its surface irrespective of its sign. I can't reason out why this is true when both charges have the same sign because then each will not be able to induce an opposite charge on the surface closest to it because the sphere must remain neutral. Please explain where I'm going wrong.
- Niharika (age 18)
India
A:
Think of it this way. Let's say that the only way the sphere could polarize was by a dipole distribution of charges. So let's see what those external charges do. They consist of a symmetrical average charge (0.5(q1+q2)) and an antisymmetric difference (0.5(q1-q2)). True, the symmetric part has no dipole moment about the center of the sphere, but the antisymmetric part does. So it induces a dipole on the sphere, lowering the energy. If the sphere expands, it can lower the energy more. In other words, ignore the average charge, and just think of the difference.

Now what if q1=q2, so there's no difference? There are still induced charges on the sphere, mainly a quadrupole distribution. Say q1 and q2 were positive. There would be negative patches near them and a positive band around the middle between them. Again, this lowers the electrostatic energy. The energy can still be lowered more if the sphere expands.

Mike W

(published on 03/19/2011)