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Problem An insulating sphere with radius R contains a total non-uniform charge (i.e. Hydrogen atom) Q such that its volume charge density is 3B p= p312 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 DE dÃ= qenc / epsilono We will choose a symmetric Gaussian surface, which is the surface of a sphere, then evaluate the dot product to obtain E A = qenc / epsilon0 (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 dqenc 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
Transcribed Image Text:Problem An insulating sphere with radius R contains a total non-uniform charge (i.e. Hydrogen atom) Q such that its volume charge density is 3B p= p312 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 DE dÃ= qenc / epsilono We will choose a symmetric Gaussian surface, which is the surface of a sphere, then evaluate the dot product to obtain E A = qenc / epsilon0 (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 dqenc 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
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 = , 3B dv 3/2 where dV is the infinitesimal volume. By evaluating the integral and simplifying, we obtain the following Q = 8 IT B R 3/2 for the limits from 0 to R Thus, B can then be expressed as B =Q/ 8piR3/2 Now we are ready to solve for the charged enclosed by the Gaussian surface. We apply the same definition of volume charge density, to obtain the integral 3B dv 9enc p312 the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain denc 3/2 I R3/2 %3D By substitution to Equation 1 above, then using A = and simplifying, we obtain E = (1/ 4piepsilon0 (Qr -1/2 IR 3/2
Transcribed Image Text: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 = , 3B dv 3/2 where dV is the infinitesimal volume. By evaluating the integral and simplifying, we obtain the following Q = 8 IT B R 3/2 for the limits from 0 to R Thus, B can then be expressed as B =Q/ 8piR3/2 Now we are ready to solve for the charged enclosed by the Gaussian surface. We apply the same definition of volume charge density, to obtain the integral 3B dv 9enc p312 the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain denc 3/2 I R3/2 %3D By substitution to Equation 1 above, then using A = and simplifying, we obtain E = (1/ 4piepsilon0 (Qr -1/2 IR 3/2
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