# Total Charge on Conducting Hollow Sphere

1. Mar 24, 2013

### *FaerieLight*

I've just worked out using the method of images that the total induced charge on a grounded hollow conducting sphere in the presence of a dipole outside the sphere pointing in the radial direction is non-zero. I can't think of an intuitive explanation as to why a dipole outside would induce a non-zero charge on the sphere.

I think the problem goes back to why a point charge outside a hollow conducting sphere would induce a net charge on the sphere that is not equal in magnitude to the point charge, which seems to be the case also, according to the method of images. I suppose there is no reason for the induced charge to be equal to the point charge in magnitude, but then for a grounded conducting plane, with a point charge above it, the induced charge on the plane is the same in magnitude to the point charge! I don't get why it should be equal in magnitude for a plane, and why it isn't for a sphere. I know they have different geometries but I can't see what specific difference in their geometries leads to the difference in induced charge magnitude.

Thanks a lot!

2. Mar 24, 2013

### Staff: Mentor

Consider an extreme dipole: One charge close to the sphere and the other charge at "infinite distance": You have a net charge close to the sphere, which will attract the opposite charge towards the sphere.

Right. They are equal for an infinite sheet only.
Only in the limit of negligible distance to the sphere (or infinite radius of the sphere).

3. Mar 24, 2013

### stevendaryl

Staff Emeritus
There's no particular reason for the real charge and the image charge to be equal.

The significance of a "grounded conducting sphere" is that the potential $V$ must be everywhere $0$ on the surface of the sphere. That's what "grounded" means.

The method of images is a guess as to how to make the potential zero on the surface of the sphere by placing imaginary charges inside the sphere.

Let's work in the x-y plane, with a sphere of radius $R$ centered at the origin, and a point charge $Q$ on the x-axis at a distance $X$ from the origin. We guess that the charges on the sphere have the same effect outside the sphere as if there were a point charge $Q'$ on the x-axis at a distance $X'$ from the origin. An arbitrary point $\vec{r}$ on the sphere in the x-y plane will have coordinates

$\vec{r} = R cos(\theta) \hat{x} + R sin(\theta) \hat{y}$

The distance of this point from $Q$ is
$D = \sqrt{R^2 - 2RX cos(\theta) + X^2}$

The distance of this point from $Q'$ is
$D' = \sqrt{R^2 - 2RX' cos(\theta) + X'^2}$

For the potential at the point on the sphere to be zero, it must be that:

$\dfrac{Q}{D} + \dfrac{Q'}{D'} = 0$

This equation must be true for all values of $\theta$. That determines $Q'$ and $X'$, and there's no particular reason for $Q'$ to be equal in magnitude to $Q$.

Last edited: Mar 24, 2013