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 3/2 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 PĒ - dà = 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 AP Based on the given problem, we can also say that dqenc 3B 3/2 %3D dV Let us first solve for 6

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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 =
3/2
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
PĒ - dÃ=
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
dV
Based on the given problem, we can also say that
dq 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
3B
O =
where dV is the infinitesimal volume.
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 = 3/2 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 PĒ - dÃ= 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 dV Based on the given problem, we can also say that dq 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 3B O = where dV is the infinitesimal volume.
where dV is the infinitesimal volume.
By evaluating the integral and simplifying, we obtain the following
Q =
for the limits from 0 to R
Thus, B can then be expressed as
B =Q/
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
denc
%3D
3/2
the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain
denc =
By substitution to Equation 1 above, then using A =
, and simplifying, we obtain
E = (1/
(Qr
/R
Transcribed Image Text:where dV is the infinitesimal volume. By evaluating the integral and simplifying, we obtain the following Q = for the limits from 0 to R Thus, B can then be expressed as B =Q/ 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 denc %3D 3/2 the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain denc = By substitution to Equation 1 above, then using A = , and simplifying, we obtain E = (1/ (Qr /R
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