Conducting sphere within a conducting shell

  • Thread starter Beatrix
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Homework Statement


A hollow spherical shell (B) with inner radius R2 and esternal radius R3 is negatively charged with Q.
A spherical conducter (A) with radius R1 is placed within the the shell. A is charged with Q.
The centers of both shells coincide.
Then a negative point charge q is placed at R3.
Tasks:
1. Calculate the electric field genrated by the spheres at any point.
2. Calculate the force on the charge q
3.Calculate the work done by an external agent to bring q to infinity

2. Homework Equations

Gauß-Law

The Attempt at a Solution


(Non-native English speaker here but my lectures are in English so excuse any grammar or spelling mistakes, ok?)
First of all I tried to imagine how the charges on the spherical shell would redistribute.
The positive charge Q on the surface of A can't exactly escape so it has to stay there^^
I then thought that the charge -Q of B would arange along R2 on the inner surface of the shell.
Then there is no charge left on the outside surface of B.

So the object is seen as neutral when looked at from a distance r>R3, or isn't it?

1. According to Gauß law there is no E-field at a distance r<R1 since no charge is enclosed.
At R2<r<R1 the E-field should be
equal to ## \frac{1}{4\pi\epsilon}*\frac{Q}{r^2} ##
At r<R2 I think I can use the super position principle, adding the E-Field of both spheres
but since the charges have the same value (one negative one positive)
My E-field should be zero here....

but if this is true I don't get the second question...
since the force should be ## F=E*q ##
and the Field is equal to zero at R3 the force should be zero as well?

I really don't understand this task so please help....
 

Answers and Replies

  • #2
BvU
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Hello Beatrix, welcome to PF :smile: !

I'm non-native english speaker too, and I have a hard time interpreting your exercise:
A hollow spherical shell (B) with inner radius R2 and esternal radius R3 is negatively charged with Q.
A spherical conductor (A) with radius R1 is placed within the the shell. A is charged with Q.
So for B I don't know if it is conducting or not. You assume it is.
And it's negatively charged with Q. Does that mean that Q is a negative number of Coulombs ?
Because then A and B have the same charge and the rest of the exercise makes a bit more sense !

Further on, "placing q at R3" probably means on the surface but not in electrical contact ? You assume so and I agree.
 
  • #3
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B is conducring (excersise is called two conducting spheres foorgot to mention this)
Q has a positive value (Q=3*10^-7) (the excersice contains values but since that doesn't help me much when trying to understand the problem I left them out)
I mean I can calculate it if they had the same charge but they don't....
It's an excersise for exam preparation so I'll ask my professor about it.
But thanks, I really was going crazy thinking I could't even understand the basics of this course
 
  • #4
BvU
Science Advisor
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Well, then I agree with your results (barring the replacement of < by > here and there :wink: ). So you're good: you understand the situation, you can do the exercise both ways: the almost trivial -Q + Q and the more sensible -Q -Q probably just as well. After all the actual answer isn't that important; the understanding is).
And prof will be more than happy to help out such an eager student !
 

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