How Does Charging a Conducting Shell Affect Potential Difference?

[Tanishq Nandan]

Homework Statement

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
A)V B)2V C)4V D)-2V

Homework Equations

▪Potential of a spherical shell of radius R having charge Q=kQ/R
▪Charge on conductors resides only on the surface.

The Attempt at a Solution

First case:
Charge on inner sphere:+Q
Charge on inner surface of outer conducting shell: -Q
Charge on outer surface of outer conducting shell: +Q

According to question,V diff between sphere and outer surface of shell is asked.
Assuming radius of inner sphere to be r1 and outer shell to be r2
So,difference is equal to:kQ(1/r1 - 1/r2)

2nd case:
Charge on inner sphere:+Q
Charge on inner surface of outer conducting shell: -Q
Charge on outer surface of outer conducting shell: -2Q

Same method,p.d comes out to be:kQ(1/r1 + 2/r2)
Now,how do I relate this to the original V??
Answer is V,so I'm most probably writing down the potential difference in the 2nd case wrong...

 

[NFuller]

Same method,p.d comes out to be:kQ(1/r1 - 2/r2)

Q on the outer shell is negative in this case so your minus sign should be a plus sign.

 

[Tanishq Nandan]

Oops,sorry,I'll correct it right away,but that doesn't help the answer

 

[Tanishq Nandan]

Now what?

 

[NFuller]

Well you want an answer in terms of V. You can use the equation for the potential in the first case to find Q in terms of V.
$$ V_{1}=kQ\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right) $$
I'm using $V_{1}$ instead of just $V$ to differentiate this from the second case.

Now use this to substitute $V_{1}$ for Q in the second equation
$$ V_{2}=kQ\left(\frac{1}{r_{1}}+\frac{2}{r_{2}}\right) $$
and simplify.

 

[Tanishq Nandan]

Well,if the answer is V, then in both the cases,the potential difference will have to be the same and I doubt any amount of simplification can make these two terms equal. :(

 

[TSny]

The most direct route to the answer to this question is probably to consider the electric field that exists between the sphere and the inner surface of the shell. How is this field modified when -3Q is added to the shell?

However, if you want to solve it along the lines you posted then consider the following:

The Attempt at a Solution

First case:
Charge on inner sphere:+Q
Charge on inner surface of outer conducting shell: -Q
Charge on outer surface of outer conducting shell: +Q

Yes.

Assuming radius of inner sphere to be r1 and outer shell to be r2
So,difference is equal to:kQ(1/r1 - 1/r2)

This is correct only if you can neglect the thickness of the shell.
You haven't actually shown how you got your result, so it's hard to tell if you are thinking about it correctly.

2nd case:
Charge on inner sphere:+Q
Charge on inner surface of outer conducting shell: -Q
Charge on outer surface of outer conducting shell: -2Q

Yes.

Same method,p.d comes out to be:kQ(1/r1 + 2/r2)

As you suspect, this is not correct. But we need to see your reasoning to find out why.

Note that it is not correct to say that the potential at the surface of the sphere is kQ/r1. The charge on the shell affects the potential of the sphere.

 

[Tanishq Nandan]

The charge on the shell affects the potential of the sphere.

Of course!My bad,got it,Thanks..