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Electric Potential due to conducting sphere and conducting shell

1. The problem statement, all variables and given/known data
A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the same two surfaces is????
Answer is V.Let R be the radius of the outer surface of shell and r of inner surface and let the radius of sphere be X.
Consider epsilon not as 'ε'

2. Relevant equations
potential at a point =Q/4∏ε.


3. The attempt at a solution
If Q charge is on the surface of sphere then it will induce -Q charge on the inner surface of the shell and Q on the outer surface of shell.Potential at the surface of sphere due to charge inside sphere is Q/4∏εX but I am unable to calculate NET potential at surface of sphere(potential due to charge outside and inside sphere).
I know that potential on the outer surface of shell is Q/4∏εR.
 

ehild

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I know that potential on the outer surface of shell is Q/4∏εR.
Yes, it is true. And you also know the electric field between the sphere and the inner wall of the shell (think of Gauss' Law). How do you get the potential difference then?

ehild
 
E at any point between sphere and shell is kQ/a^2 where a is the distance of a point from the center of sphere.
since,dV/dr=-E. so v=kQ/a.I do not know whether it is right or wrong.
 
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ehild

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You get V, the potential difference between the sphere (a=X) and the inner surface of the shell (a=r) by integrating -E between these boundaries. You also know the potential at the outer surface of the shell is Q/(4∏εR), and that a conductor has the same potential everywhere. Your formula misses the integration constant that you need to match to the potential values at a=X and a=r.

ehild
 

SammyS

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1. The problem statement, all variables and given/known data
A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the same two surfaces is????
Answer is V.Let R be the radius of the outer surface of shell and r of inner surface and let the radius of sphere be X.
Consider epsilon not as 'ε'
BTW: It's epsilon naught (ε0) .

(Homonyms)
2. Relevant equations
potential at a point =Q/4∏ε.


3. The attempt at a solution
If Q charge is on the surface of sphere then it will induce -Q charge on the inner surface of the shell and Q on the outer surface of shell.Potential at the surface of sphere due to charge inside sphere is Q/4∏εX but I am unable to calculate NET potential at surface of sphere(potential due to charge outside and inside sphere).
I know that potential on the outer surface of shell is Q/4∏εR.
 
You get V, the potential difference between the sphere (a=X) and the inner surface of the shell (a=r) by integrating -E between these boundaries. You also know the potential at the outer surface of the shell is Q/(4∏εR), and that a conductor has the same potential everywhere. Your formula misses the integration constant that you need to match to the potential values at a=X and a=r.

ehild
So potential on the outer surface of sphere is kQ/R+kQ/r-kQ/X.so,potential difference between the two surfaces is kQ/r-kQ/X.Now if shell is given the charge -3Q then net charge on it will be -2Q(as initially it has Q charge).The shell will induce -3Q charge on the surface of sphere.So,net charge on the sphere will be -2Q. Now if we calculate potential difference we will get P.D=k(-2Q)/r-k(-2Q)/X.This value is not the same as previous one but the answer says that it should be same as previous one.
 

ehild

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So potential on the outer surface of sphere is kQ/R+kQ/r-kQ/X
.
No, the inner and outer surface of the conducting shell are at the same potential.

so,potential difference between the two surfaces is kQ/r-kQ/X.
That is true, but what is the sign? What is V(sphere)-V(shell)?

Now if shell is given the charge -3Q then net charge on it will be -2Q(as initially it has Q charge).
There was zero net charge on the shell, -Q inside and Q outside. Adding the shell -3Q charge, the net charge will be -3Q. How is it shared between the inner and outer surfaces?

The shell will induce -3Q charge on the surface of sphere.So,net charge on the sphere will be -2Q. Now if we calculate potential difference we will get P.D=k(-2Q)/r-k(-2Q)/X.This value is not the same as previous one but the answer says that it should be same as previous one.
No, the charge on the shell will not induce any charge inside. Does the electric field change between the sphere and the shell if you add charge to the outer shell? Think of Gauss' Law.


ehild
 
In reply to ehild
No, the inner and outer surface of the conducting shell are at the same potential.

As the electric field inside conducting shell is 0.so,the inner and outer surface of the conducting shell are at the same potential.


That is true, but what is the sign? What is V(sphere)-V(shell)?

I do not understand what is meant by sign in this case.


There was zero net charge on the shell, -Q inside and Q outside. Adding the shell -3Q charge, the net charge will be -3Q. How is it shared between the inner and outer surfaces?

Please explain the sharing of charge between inner and outer surface.As initially inner surface of shell has charge -Q so as to get net charge -3Q outer surface of charge should have charge -2Q. I do not know if it is right or wrong.So please explain.


No, the charge on the shell will not induce any charge inside. Does the electric field change between the sphere and the shell if you add charge to the outer shell? Think of Gauss' Law.

That was my mistake.Yes,while calculating E using gauss law we do not consider charge outside the gaussian surface.So E between shell and sphere will not change.
I was not able to quote so I replied in this way.
 
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ehild

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Yes,while calculating E using gauss law we do not consider charge outside the gaussian surface.So E between shell and sphere will not change.
I was not able to quote so I replied in this way.
Use the quote button under the post. If you want to quote only some sentences, write quote in front in [], and end the quotation with /quote also between [].


ehild
 
Use the quote button under the post. If you want to quote only some sentences, write quote in front in [], and end the quotation with /quote also between [].
I got it .Thank you for your help.
 
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No, the inner and outer surface of the conducting shell are at the same potential. ehild
As the electric field inside conducting shell is 0.so,the inner and outer surface of the conducting shell are at the same potential.


.That is true, but what is the sign? What is V(sphere)-V(shell)?ehild
kQ/R+kQ/r-kQ/X- kQ/R.


.There was zero net charge on the shell, -Q inside and Q outside. Adding the shell -3Q charge, the net charge will be -3Q. How is it shared between the inner and outer surfaces?ehild
Please explain the sharing of charge between inner and outer surface.As initially inner surface of shell has charge -Q so as to get net charge -3Q outer surface of charge should have charge -2Q. I do not know if it is right or wrong.So please explain.


.No, the charge on the shell will not induce any charge inside. Does the electric field change between the sphere and the shell if you add charge to the outer shell? Think of Gauss' Law. ehild
That was my mistake.Yes,while calculating E using gauss law we do not consider charge outside the gaussian surface.So E between shell and sphere will not change.
 

ehild

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As the electric field inside conducting shell is 0.so,the inner and outer surface of the conducting shell are at the same potential.
What is the potential at the outer surface of the shell?

ehild
 
What is the potential at the outer surface of the shell?

ehild
As I have mentioned above,potential at the outer surface of shell is kQ/R.In second case I think that it is k(-2Q)/R as the inner surface of shell has charge -Q as as to get net charge -3Q on shell,outer surface of shell should have charge -2Q.I do not know if it is right or not.
 
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ehild

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The outer surface of the shell is at distance R from the centre. The electric potential is the same as the whole charge of the system was in the centre. In the first case, it is Q, and the potential both at the outer and the inner surface of the shell is kQ/R. In the second case, the net charge is -2Q. The potential of the shell - on both sides- is -2kQ/R.

At the same time, the electric field between the sphere and the shell does not change. Nor does the potential difference.

ehild
 
The outer surface of the shell is at distance R from the centre. The electric potential is the same as the whole charge of the system was in the centre. In the first case, it is Q, and the potential both at the outer and the inner surface of the shell is kQ/R. In the second case, the net charge is -2Q. The potential of the shell - on both sides- is -2kQ/R.

At the same time, the electric field between the sphere and the shell does not change. Nor does the potential difference.

ehild
So the charge on the outer surface of shell in second case is -2Q and the potential difference in two cases will remain same . THANK YOU sir for answering.Now I understood it completely.
 

ehild

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I have found the answer of this question in a google book as
ANSWER:Because the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere.Since the two values have not changed therefore potential difference does not change.
I want to ask

that the answer mentions the potential difference between solid sphere and hollow shell
but the question asks potential difference between
surface of the solid sphere and that of the outer surface of the hollow shell
Are both same?
 
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And the answer also mentions
:Because the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere.Since the two values have not changed therefore potential difference does not change.
Which two spheres?

I think there is shell and sphere in the question but the answer mentions two spheres?
 

ehild

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In the answer of the OP
i.e
Because the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere.Since the two values have not changed therefore potential difference does not change.
Here charge on the inner sphere means charge on solid conducting sphere ?
 

ehild

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Charge of the inner sphere is the charge of the conducting sphere inside.
 
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And please answer my other two questions POST #18 & #19 also.Please!
 

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