# Work done moving a test charge into conducting shell

• KleZMeR
Therefore, the potential can still be nonzero even if the electric field is zero. In summary, the problem asks to find the work done in moving a test charge from infinity to the center of a spherical shell with given parameters, passing through an infinitesimal hole. By using the equations for work and assuming a zero electric field inside the shell, the final result is that the total work done is zero. However, this result may seem trivial and may not take into account the potential being nonzero even if the electric field is zero.
KleZMeR

## Homework Statement

Ok, so I've read many of the threads on here and they all say the same thing. I think I understand the Gauss Law and the theory behind the spherical shell.

The question is this:

Find the WORK done bringing a test charge q from infinity to the center of a spherical shell of thickness T, radius R, and surface charge $\sigma$. Assume the charge passes through an infinitesimal hole in the shell.

## Homework Equations

$W = q[\phi(inf)-\phi(r)]$

## The Attempt at a Solution

$W_{\infty, R+T} = q[\phi(\infty)-\phi(R+T)] = -\phi(R+T)$
and
$W_{R+T,0} = q[\phi(R+T)-\phi(0)] = \phi(R+T)$

My second equation takes the 'fact' that $E(r) = 0$ inside the shell, resulting in a zero potential.
The final result is that the total WORK = 0
This is for a graduate class, and this result seems somewhat trivial. My other assumption is that the test charge induces an electric field inside the shell, but I do not think work can be done by moving the test charge through its own electric field? I could be totally wrong, and that's why I'm asking this question.

Any help clarifying my result would be greatly appreciated.

The electric field being zero inside the shell does not mean the potential is zero. Recall that electric field is the negative gradient of potential.

Last edited:

## 1. What is the definition of "work done" when moving a test charge into a conducting shell?

The work done is the amount of energy required to move a test charge from one point to another, against the force applied by the electric field of the conducting shell.

## 2. How is the work done calculated when moving a test charge into a conducting shell?

The work done is calculated by multiplying the magnitude of the electric field of the conducting shell by the distance the test charge is moved into the shell.

## 3. Is the work done positive or negative when moving a test charge into a conducting shell?

The work done is negative when moving a test charge into a conducting shell. This is because the electric field of the conducting shell opposes the movement of the test charge, and therefore, work must be done to overcome this force.

## 4. Does the work done depend on the size of the test charge being moved?

No, the work done does not depend on the size of the test charge being moved. It only depends on the magnitude of the electric field and the distance the test charge is moved.

## 5. How does the work done change if the conducting shell is hollow versus solid?

If the conducting shell is hollow, the work done is calculated by using the electric field at the surface of the shell. However, if the shell is solid, the work done is calculated by using the electric field at the center of the shell. This is because the electric field inside a solid conducting shell is zero, while the electric field at the surface is non-zero.

Replies
2
Views
2K
Replies
1
Views
906
Replies
6
Views
1K
Replies
0
Views
577
Replies
10
Views
1K
Replies
6
Views
3K
• Electromagnetism
Replies
4
Views
762
Replies
2
Views
930