Calculating Electric Field and Induced Charge in a Charged Sphere System

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Homework Help Overview

The problem involves calculating the electric field and induced charge in a system consisting of a solid insulating sphere with a uniform charge density and a concentric uncharged conducting hollow sphere. The discussion focuses on determining the electric field in various regions defined by the radii of the spheres and the induced charge on the surfaces of the hollow sphere.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss applying Gauss's Law in different regions of the sphere system, questioning the charge enclosed in each Gaussian surface. There is exploration of the electric field expressions for various regions and attempts to clarify the conditions inside and outside the conducting sphere.

Discussion Status

Some participants have provided guidance on applying Gauss's Law correctly and have pointed out potential misunderstandings regarding the charge enclosed in certain regions. There is an ongoing exploration of the relationships between the electric fields in different regions, with some participants questioning their calculations and assumptions.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the information they can share or the methods they can use. There is a noted confusion regarding the interpretation of the regions and the application of Gauss's Law.

tag16
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Homework Statement


A solid insulating sphere of radius a has a uniform charge density p and a total charge Q. Concentric with this sphere is an uncharged, conducting, hollow sphere whose inner and outer radii are b and c. (a) Find the magnitude of the electric field in the regions r<a, a<r<b, b<r<c, and r<c. (b) Determine the induced charge per unit area on the inner and outer surfaces of the hollow sphere.


Homework Equations



Gauss's Law= S E dA= Q/E_0

The Attempt at a Solution



S E dA= Q/E_0
= E(4pir^2)= Q/E_0

I am doing this right so far? if so not sure what to do next, if not, not sure what to do.
 
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You are doing fine so far. Next you need to apply Gauss' Law by choosing different gaussian surfaces in each of the regions of interest

First at r<a
Second at a<r<b
Third at b<r<c
Fourth at r>c (You say r<c, but I assume you meant r>c)

Apply Gauss' Law as you have stated it. Make sure that you calculate the charge enclosed by each surface correctly and remember that inside a conductor the electric field is zero.
 
ok this is what I have so far...

for r>a

S E dA= Q/E_0
= E(4pir^2)= Q/E_0
E=Q/4piE_0r^2= kQ/r^2

for a<r<b

q(internal)=pV'=p(4/3pir^3)
S E dA = E S dA= E(4pir^2)= q(internal)/E_0
E=qin/4piE_0r^2=p(4/3pir^3)/4piE_0r^2=pr/3E_0
E= (Q/(4/3)pia^3)r/3(1/4pik)= kr(Q/a^3)

for r>c
would it be the same as the 1st one?

for b<r<c
would it be 0?

Did I do any of these right? and for the wrong ones what do I need to do?
 
tag16 said:
ok this is what I have so far...

for r>a

S E dA= Q/E_0
= E(4pir^2)= Q/E_0
E=Q/4piE_0r^2= kQ/r^2
I assume you mean r < a. In this case not all the charge Q is enclosed by the Gaussian surface. Only a fraction of it. Can you figure out what that fraction is?

for a<r<b

q(internal)=pV'=p(4/3pir^3)
S E dA = E S dA= E(4pir^2)= q(internal)/E_0
E=qin/4piE_0r^2=p(4/3pir^3)/4piE_0r^2=pr/3E_0
E= (Q/(4/3)pia^3)r/3(1/4pik)= kr(Q/a^3)
Actually this is the answer to the previous part and the answer to the previous part is the answer to this one. Can you see why?

for r>c
would it be the same as the 1st one?
It would be the same as in the region a<r<b (corrected).
for b<r<c
would it be 0?
Yes it would.
Did I do any of these right? and for the wrong ones what do I need to do?
See above.
 
Thanks...I think I just copied it down wrong in here from what I wrote down.
 

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