Potential of a grounded spherical shell inside of a metal one

[ohannuks]

Picture:
http://img263.imageshack.us/img263/7361/ongelma.jpg

Homework Statement

Solve the potential energy of a charged sphere-shaped metal shell. What happens when you place a smaller, grounded sphere-shaped metal shell inside of that shell? What is the potential energy then? How about the electric potential?

R is the radius of the bigger metal shell
Q is the total charge of the bigger metal shell
W is potential energy
c=kQ/R

Homework Equations

(1) V(r)=k*∫(dQ)/|r-r'|
Where V(r) is electric potential, k is a constant and dQ is a very small charge

(2) W=∫F*dr
Where W is potential energy and F is force

(3) W=q*∫E*dr
Where q is a charge and E is the electric field

(4) W=q*V(r)
-

(5) E=-(nabla)*V(r)

(6) W=1/2*V(r)*∫dQ

The Attempt at a Solution

Okay so the first one is easy. From the equation (1) it can be calculated that the electric potential inside the metal shell is constant, V=kQ/R.
Let's say c=kQ/R

The potential energy can be calculated from equation (4)

This is equation (6)
W=1/2*V(r)*∫dq

The half is there because we are calculating the potential of the spherical shell from its own electric potential, which means charge a's potential will be calculated in comparison to charge b and charge b's potential will be calculated in comparison to charge a, so everything gets calculated twice. We don't want that.

Now we place the grounded sphere-shaped shell with no charge inside of the metal ball.

So we can calculate V(r) again:

V=c, still! Because the grounded ball is neutral.

So W=1/2*V(r)*∫dq, still!

But here's the catch:

If we have a sphere-shaped shell with no charge inside its electric potential is:
V=0, this is obvious because it has no charge.So if we place it inside, we can calculate that when it is inside the charge is in fact not 0, but c=kQ/R.

So what happens is that the electric potential inside changes.

It should follow that some sort of electric field must have been created by the grounded metal shell, as stated by equation (5). So E=-(nabla)*V(r). If V(r) changes, (nabla)*V(r) should not be zero, at some point anyway.

So now let's look at equation (3):
W=q*∫E*dr

Because we have electric field when the grounded metal shell moves, it MUST have done work, let's call this work W'

So now we have contradicting results.

W should be: (6)
W=1/2*V(r)*∫dq = 1/2*c*Q

But at the same time we have:

W= 1/2*c*Q + W'

Where W' is not zero

So where did I go wrong?

 

[mark726]

Thank you for your question regarding the potential energy of a charged sphere-shaped metal shell. I would like to address your confusion and provide a clearer explanation.

Firstly, let's clarify the setup of the problem. We have a larger metal shell with radius R, total charge Q, and electric potential V=kQ/R. Inside this shell, we place a smaller, grounded metal shell with no charge. We want to calculate the potential energy and electric potential in this system.

Your approach of using equations (6) and (4) to calculate the potential energy is correct. However, there seems to be a misunderstanding in your interpretation of the electric potential and electric field inside the metal shell.

The electric potential inside the metal shell is not constant, as you have stated. It varies depending on the distance from the center of the shell. This is because the electric potential is determined by the charge distribution, which is not uniform in this case. The charge is concentrated at the surface of the shell, so the electric potential will be higher near the surface and lower near the center.

Now, when we place the smaller, grounded metal shell inside the larger shell, the electric potential inside the larger shell will change. This is because the grounded shell will redistribute the charge inside the larger shell. As you correctly pointed out, there will be an electric field created by the grounded shell, which will change the electric potential inside the larger shell. This can be calculated using equation (1).

As for the potential energy, it will also change due to the presence of the grounded shell. However, this change is not simply given by equation (6). This equation assumes that the potential energy is solely determined by the electric potential, which is not the case in this situation. There is also the contribution of the work done by the electric field created by the grounded shell, as given by equation (3). Therefore, the correct expression for the potential energy in this case is:

W=q*∫E*dr + 1/2*V(r)*∫dq

Where q is the charge of the larger shell and E is the electric field created by the grounded shell. This expression takes into account both the contribution of the electric potential and the work done by the electric field.

In summary, the potential energy and electric potential inside the larger metal shell will change when we place a smaller, grounded shell inside it. This is due to the redistribution of charge and the creation of