# Electric flux through a sphere

1. Feb 13, 2015

### PosteMortem

1. The problem statement, all variables and given/known data
An uncharged nonconductive hollow sphere of radius 12.0 cm surrounds a 11.0 µC charge located at the origin of a cartesian coordinate system. A drill with a radius of 1.00 mm is aligned along the z axis, and a hole is drilled in the sphere. Calculate the electric flux through the hole.

2. Relevant equations
Flux=EA

3. The attempt at a solution
I've got the solution for this problem. I found the flux for the sphere to be equal to (kq/r^2)*(pi*r^2)=k*q*pi, and after multiplying that value by the ratio of the smaller radius squared to the larger radius squared, I got the answer. However, I was reading the textbook, and it shows the flux through a sphere to be equal to 4*k*q*pi, the reason being that integrating over the entire surface will give us the surface area of a sphere. I'm confused as to how these two answers could be different. Shouldn't it be the same in both cases?

2. Feb 13, 2015

### Staff: Mentor

Check the units of each. One is a flux, the other a field strength.

What motivated you multiply the electric field strength by $\pi r^2$? It provided a correction factor, but it seems geometrically unmotivated at first glance (at least to me). A bit of algebra shows that it will work if you use the ratio of radii squared as a multiplier to select the amount of flux, but one is the radius of a volume and the other a radius of an area. Not exactly intuitive. Nothing succeeds like success, I suppose! But what would you have done if the hole happened to be a square punched in the surface?

When you start with the total flux over a spherical surface and want to know the flux that comes through some small aperture at the surface you multiply the total flux by the ratio of the total surface area to the area of the aperture.