How Does Charging a Conducting Shell Affect Potential Difference with a Sphere?

[cupid.callin]

Solved
stupid mistake!

i forget to consider potential on sphere due to shell ... my Bad!
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Homework Statement

Suppose there is a conducting sphere surrounded by a concentric conducting shell. the sphere is given a charge Q. Let the potential difference b/w the sphere and shell be V.
Now the shell is given a charge -3Q, Find their new potential difference in terms of V.

Homework Equations

The Attempt at a Solution

now assuming radius of shell and sphere be R and r
Well for first part of course the inner surface of shell will get charge -Q and outer Q.

So,, V(shell) - V(sphere)= (KQ/R - KQ/R + KQ/R) - (KQ/r) = KQ/R - KQ/r = V

After giving charge -3Q, inner surface of shell will have -Q harge and outer -2Q.

V(shell) - V(sphere) = (-KQ/R - 2KQ/R + KQ/R) - (KQ/r) = -2KQ/R - KQ/r = ?

*** and also one query ... will the potential of sphere be KQ/r alone? or this plus sometheing

 

[anathema]

due to that shell?

Hello there,

Thank you for sharing your attempt at solving this problem. It seems like you have made a small mistake in your calculation. The potential difference between the shell and the sphere should be V, not -V. Therefore, the equation should be:

V(shell) - V(sphere) = (K(-Q)/R + K(-2Q)/R + K(Q)/R) - (KQ/r) = -3KQ/R - KQ/r = -4KQ/R

So, the new potential difference between the shell and the sphere will be -4V.

As for your query, the potential of the sphere will indeed be KQ/r alone, as the potential due to the shell will be cancelled out by the potential due to the sphere itself.

I hope this helps and keep up the good work! Remember, even scientists make mistakes sometimes, what's important is to learn from them. Keep exploring and asking questions!