# Uniform Electric Field in Cavity of Solid Sphere

• mrlebowski

## Homework Statement

A charge q is uniformly distributed over the volume of a solid sphere of radius R. A spherical cavity is cut out of this solid sphere, and the material and its charges are discarded. Show that the electric field in the cavity will then be uniform, of magnitude (1/4*pi*epsilon knot)(qd/R^3), where d is the distance between the center of the original sphere and the center of the spherical cavity.

## Homework Equations

p(ro)=q/v (charge density)

electric flux= E*A=q(enclosed)/epsilon knot (E is the electric field, and A is the surface area of the Gaussian surface used to find flux)

electrical flux of spherical gaussian surface = E *4pi*r^2

## The Attempt at a Solution

I'm having a really hard time wrapping my mind around this one, and I am loathe to begin calculating before I know what is going on conceptually. However, I have been trying to the electric field just before and after the cavity. So if the radius of the cavity is H, finding the field at (d-H) and (d+H) might be able to tell me something about the field in between those points (inside the cavity)? But in calculating the field at the outer edge of the sphere, where the radius is (d+H), I had a hard time accounting for the loss of volume and charge and surface area in calculating the field at this point. The point of the problem is just to show that inside the sphere the field is uniform and its magnitude is as given above. If someone could just explain how that answer came to be, and why it is the case, I would be most grateful. Thanks!

There is a trick for such problems. You can make a cavity free of charge by inserting a sphere of opposite charge density there.

ehild