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= where Bis a constant and ris 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 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 dqanc 38 p= dV 3/2 Let us first solve for B. Our enclosed charge would have limits from 0 to Q. Then r would have the limits O to R. Thus, the equation above becomes 38 Q= where dV is the infinitesimal volume. By evaluating the integral and simplifying, we obtain the following Q= R for the limits from 0 to R Thus. B can then be expressed as 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 Tenc = /38 the difference is now, the limits of r will be from O 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
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= where Bis a constant and ris 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 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 dqanc 38 p= dV 3/2 Let us first solve for B. Our enclosed charge would have limits from 0 to Q. Then r would have the limits O to R. Thus, the equation above becomes 38 Q= where dV is the infinitesimal volume. By evaluating the integral and simplifying, we obtain the following Q= R for the limits from 0 to R Thus. B can then be expressed as 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 Tenc = /38 the difference is now, the limits of r will be from O 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
Chapter6: Gauss's Law
Section: Chapter Questions
Problem 86AP: Two non-conducting spheres of radii R1 and R2 are uniformly charged with charge densities p1 and p2...
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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
38
p=
3/2
where Bis a constant and ris 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=
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=
AP
dV
Based on the given problem, we can also say that
dg enc
p= dv
38
312
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
38
5dv
Q=
o 3/2
where dV is the infinitesimal volume.
By evaluating the integral and simplifying, we obtain the following
Q =
R
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
3B
AP
denc =
the difference is now, the limits ofr will be from 0 to r. Evaluating the integral and simplifying, we obtain
| and
denc =
By substitution to Equation 1 above, then using A =
and simplifying, we obtain
E = (1/
(Qr
/R
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Hello! May I ask what should I input on the following boxes?
1.) By evaluating the integral and simplifying, we obtain the following
Q = __ pi__R__for the limits from 0 to R
2. "Thus, beta can be expressed as
beta=Q/___ ??"
3.)
the difference is now, the limits of r will be from 0 to r. Evaluating the integral and simplifying, we obtain
qenc =__ r__/__
and
By substitution to Equation 1 above, then using A = ___
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