Problem An insulating sphere with radius R contains a total non-uniform charge (i.e. Hydrogen atom) Q such that volume charge density is 3B p= p3/2 where B is a constant and r is the distance from the center of the sphere. What is electric field at any point inside the sphere? Solution To find the electric field inside the non-uniformly charged sphere, we may apply integration method or the Gauss's Law method. Here, let us use the Gauss's law which is expressed as %3D We will choose a symmetric Gaussian surface, which is the surface of a sphere, then evaluate the dot product to obtain A = (Equation 1) The issue however is how much charge does the Gaussian surface encloses? Since, our sphere is an insulating material, charges will get distributed non-uniformly within the volume of the object. So, we look into the definition of volume charge density to find the enclosed charge. So, we have dq p = dV Based on the given problem, we can also say that dd enc p= 3B 13/2 dV Let us first solve for B. Our enclosed charge would have limits from 0 to Q. Then r would have the limits 0 to R. Thus, the equation above becomes R Q = 13/2 Ap- where dV is the infinitesimal volume. By evaluating the integral and simplifying, we obtain the following

Problem An insulating sphere with radius R contains a total non-uniform charge (i.e. Hydrogen atom) Q such that volume charge density is 3B p= p3/2 where B is a constant and r is the distance from the center of the sphere. What is electric field at any point inside the sphere? Solution To find the electric field inside the non-uniformly charged sphere, we may apply integration method or the Gauss's Law method. Here, let us use the Gauss's law which is expressed as %3D We will choose a symmetric Gaussian surface, which is the surface of a sphere, then evaluate the dot product to obtain A = (Equation 1) The issue however is how much charge does the Gaussian surface encloses? Since, our sphere is an insulating material, charges will get distributed non-uniformly within the volume of the object. So, we look into the definition of volume charge density to find the enclosed charge. So, we have dq p = dV Based on the given problem, we can also say that dd enc p= 3B 13/2 dV Let us first solve for B. Our enclosed charge would have limits from 0 to Q. Then r would have the limits 0 to R. Thus, the equation above becomes R Q = 13/2 Ap- where dV is the infinitesimal volume. By evaluating the integral and simplifying, we obtain the following