Charged shpere inside grounded shell

[Terocamo]

Homework Statement

There is a Charged metal sphere with radius b placed inside the central of a ground metallic shell with a radius a. What will be the work done in bring a small positive charge q from infinity to the surface of the inner sphere?

Homework Equations

$$\emph{V=}\frac{qQ}{4r\pi\epsilon}$$

The Attempt at a Solution

What i think is that if the outer shell is earthed, than i will have 0V potential, and an infinite object also have 0V so there is no work done to bring a charged object q from infinite far to the outer shell, so what left is the potential of the inner sphere which is
$$\emph{V=}\frac{qQ}{4b\pi\epsilon}$$
However, the solution manual mentioned there is induced charge induced on the outher shell and thus, its produce a potential $$\emph{V=}\frac{-qQ}{4a\pi\epsilon}$$.

What bothers me is if the shell is grounded, how can there be any potential?

 

[ehild]

The potential is zero outside the big sphere and on it, but differs from zero inside.

ehild

 

[Terocamo]

The potential is zero outside the big sphere and on it, but differs from zero inside.

ehild

I see, so it is not the outer shell which produce a negative potential but shielding the space outside from being affected by the field produced by the inner sphere is that right?

 

[ehild]

The outer shell is at the same potential as the ground. There is no work done when a charge moves from infinity to the outer shell, but there is work while it moves from the outer shell to the inner sphere.
Find the electric field in the space between the sphere and shell. The charge on the inner sphere induces an equal but opposite surface charge (-q) on the inner surface of the shell. If the shell were not grounded, its outer surface would carry q charge. But it is grounded so this charge has flown off, to the ground.

I don't understand the formula you cited for the potential, as it is not potential but work. It is true that you can choose the zero point of potential anywhere, if it is at the surface of the inner sphere then the outer shell together with the whole world outside are at a constant negative potential. ehild

 

[rwisz]

As a scientist, it is important to consider all factors and variables in a given situation. In this case, while the outer shell may be grounded, it is still important to take into account the presence of the charged metal sphere inside it. This sphere will have its own potential, which can induce a potential on the outer shell through the process of induction. This potential on the outer shell can be calculated using the equation V = (qQ)/(4aπε), where a is the radius of the outer shell and Q is the charge on the sphere. This potential may be negative due to the opposite charge on the sphere, but it still exists and must be taken into consideration when calculating the work done in bringing a small positive charge q from infinity to the surface of the inner sphere. Therefore, the correct calculation for the work done would be to take into account both the potential of the inner sphere and the induced potential on the outer shell.