Calculating Electric Field and Induced Charge in a Charged Sphere System

AI Thread Summary
The discussion focuses on calculating the electric field and induced charge in a system of a charged insulating sphere and a concentric uncharged conducting sphere. Participants apply Gauss's Law to determine the electric field in different regions: inside the insulating sphere, between the insulating and conducting spheres, inside the conducting sphere, and outside the conducting sphere. Key points include that the electric field inside a conductor is zero and that the charge enclosed must be calculated correctly for each Gaussian surface. The correct expressions for the electric field in each region are derived, and clarification is provided on the relationships between the regions and their respective electric fields. The conversation emphasizes the importance of accurately applying Gauss's Law to solve the problem.
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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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