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
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
Related questions
Question
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
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 4 steps
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
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 = ___
Solution
by Bartleby ExpertKnowledge 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
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON