The region R lessthanorequalto a contains a uniform charge d

The region R lessthanorequalto a contains a uniform charge density rho_v, while for R > a the charge density is zero. Find the electric field in all regions. In the region 0

Solution

1 ) The charge distribution is spherically symmetric,,,,,   2 ) Since +Q is uniformly distributed throughout the volume, the electric field E must be radially symmetric and directed outward. The magnitude of the electric field is constant on spherical surfaces of radius . r,,,,,,,     3 ) The charge density of the sphere is uniform and given by

        where V is the volume of the sphere. The charge distribution divides space into two regions,   

1. Ra      and 2. . R>a

Consider the first case where R.a,        4a) We choose our Gaussian surface to be a sphere of radius R.a , as shown in Figure

   The flux through the Gaussian surface is

6 ) The charge distribution is uniform. Therefore, the charge enclosed in the Gaussian sphere of radius    R is ,

we can now apply Gauss’s Law and equate ,   

,,, We can now solve for the magnitude of the electric field in the region R >a,      ,     We now repeat steps 4 through 7 for the second region, We choose our Gaussian surface to be a sphere of radius, as shown in previous Figure      ,     : The flux through the Gaussian surface is, , ,,,,The charge distribution is uniform. Since the radius of the Gaussian surface is greater than the radius of the sphere all the charge is enclosed in our Gaussian surface, ,   ,   : Upon applying Gauss’s Law, we obtain,           We can now solve for the magnitude of the electric field in the region R .> a, , The qualitative behavior of R as a function of E is plotted in Figure