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= 312 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= 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 dgenc p= 3B 3/2 dv Let us first solve for B. Our enclosed charge would have limits from 0 to Q. Then rwould have the limits 0 to R. Thus, the equation above becomes 3B 312 dv 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 B=a 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- 312 genc = 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
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= 312 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= 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 dgenc p= 3B 3/2 dv Let us first solve for B. Our enclosed charge would have limits from 0 to Q. Then rwould have the limits 0 to R. Thus, the equation above becomes 3B 312 dv 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 B=a 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- 312 genc = 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
Chapter6: Gauss's Law
Section: Chapter Questions
Problem 70P: A point charge q=5.01012 C is placed at the center of a spherical conducting shell of inner radius...
Related questions
Question
100%
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=
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?
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
da
p=
dv
Based on the given problem, we can also say that
dgenc
p=
3B
3/2
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
3B
dv
Q =
312 0
where dV is the infinitesimal volume.
By evaluating the integral and simplifying, we obtain the following
Q =
TT
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
3B
dv
genc =
312
Ap-
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
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you
Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning
Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning