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= 1312 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?

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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
3B
dV
9enc
'o p3/2
the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain
lenc =
By substitution to Equation 1 above, then using A =
and simplifying, we obtain
E = (1/
(Qr
/R
Transcribed Image Text: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 3B dV 9enc 'o p3/2 the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain lenc = By substitution to Equation 1 above, then using A = and simplifying, we obtain E = (1/ (Qr /R
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
fE - dÃ= Q
/ epsilono
%|
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
dqenc
3B
p=
dV
1312
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
38
dV
p3/2
where dV is the infinitesimal volume.
By evaluating the integral and simplifying, we obtain the following
R
for the limits from 0 to R
%3D
Transcribed Image Text: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 fE - dÃ= Q / epsilono %| 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 dqenc 3B p= dV 1312 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 38 dV p3/2 where dV is the infinitesimal volume. By evaluating the integral and simplifying, we obtain the following R for the limits from 0 to R %3D
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