Irodov Problem 3.29

The solution to this problem exactly the same as that in Problem 3.28, except that we now have a cylinder instead of a sphere - i.e. using superposition principle. The cylinder with the cavity is same as the superposition of a solid cylinder with charge density and a smaller cylinder corresponding to the cavity of density.

First let us compute the electric field at a distance r from the axis inside a solid uniformly charged cylinder. Similar to in problem 3.28 we consider a cylindrical Gaussian surface of radius r and a very large length L about the axis of the cylinder. The total charge in the cylinder will be,



Then from Gauss's Law we have,


















If we consider as a planar vector in the horizontal plane from the axis as shown in figure 2, then,





Let be the planar vector drawn from the axis of the cylinder to the axis of the cavity, then as in Problem 3.28, the net electric field is given by,

Irodov Problem 3.28

This problem is an excellent illustration of the superposition principle of electrostatic. The basic idea is to understand that the electric field due to the sphere with a displaced cavity can be represented as the sum of electric fields due to a solid sphere of positive charge density and another displaced sphere with negative charge density of (the negative and positive charge densities cancel each other in the volume of the displaced sphere thus creating the displaced cavity).

Now let us compute the electric filed inside a uniformly charged sphere at a distance r from the center. As in case of problems 3.25-3.27, we can compute this electric field using a Spherical Gaussian surface concentric with the original sphere. The total charge contained within the Gaussian surface will be,



Now using Gauss Law we can write,






Further, since the electric field vector will in a direction radially outwards (2) can be written in vector notation as,




























A point located at will be at a location as seen from the center of the sphere displaced such that its center lies at with respect to the center of the original sphere. Consequently, the electric fields due to each of the spheres based on (2a) will be,








Hence, the net electric field inside the cavity will be,