Charge inside conducting shell

In summary, the problem involves a conducting shell with a negative charge q at a distance a from the center M and brought up to a potential V. The goal is to find the surface charge on the inside and outside of the conductor. The method of mirror charges is suggested, but the expert suggests using the method of applying Gauss' law to find the total charge on both surfaces of the shell. The electric field inside the shell is zero and the potential of a spherical shell with charge Q at distance R from the center is V = Q / (4 * pi * epsilon_0 * R). The expert guides the conversation towards the solution of finding the surface charge on the outer surface as |q| and on the inner surface as -q, resulting
  • #1
Shukie
95
0

Homework Statement


Inside a metal conducting shell of radius [itex]R[/itex] there is a negative charge [itex]q[/itex] at a distance [itex]a[/itex] from the center [itex]M[/itex]. The shell is brought up to a potential [itex]V[/itex]. What is the surface charge on the inside and outside of the conductor?

The Attempt at a Solution



I'm not sure if this is the way to solve this problem, but I know that in the case of a grounded conducting shell this system can be replaced by the original charge [itex]q[/itex] at distance [itex]a[/itex] from M and some imaginary charge [itex]\frac{R}{a}q[/itex] outside the shell, at distance [itex]a' = \frac{R^2}{a}[/itex] from [itex]M[/itex]. Does the added potential change anything about this 'imaginary' setup?
 
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  • #2
The method of mirror charges is used to calculate the field in case of a metal surface and a point charge. Here the surface charge is the question. Apply Gauss' Law.

ehild
 
  • #3
I really don't know how to proceed. Do I add the potential of the negative charge and [itex]V[/itex] and calculate the electric field from that and then use Gauss' law to find [itex]Q = \sigma A[/itex]?
 
  • #4
"The shell is brought up to a potential V." That should mean that the potential of the conducting shell is V (with respect to infinity, as no grounding is mentioned).

ehild
 
  • #5
So we have:

[tex]V = - \int_{\infty}^R \mathbf{E} \cdot d\mathbf{r}[/tex]

I don't know how to continue...
 
  • #6
What is the the potential of a spherical shell with charge Q at distance R of the centre?

ehild
 
  • #7
Do I need to use image charges to solve this? Because otherwise I have no idea.
 
  • #8
Well, try the image charges. I do not see how you get the surface charge on the shell with that method, show me please. Note that not the surface charge density is asked, but the total charge both on the outer surface of a spherical conducting shell and on the inner one.

I would apply Gauss" law. You know it, do not you? ehild
 
  • #9
Yes, [itex]\int \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\epsilon_0}[/itex], but I don't know how to apply it here. The potential [itex]V[/itex] will it affect it right?
 
  • #10
Forget the charge inside the shell for a moment. There is a conducting spherical surface with charge Q. What is its electric field at distance r from the centre?

ehild
 
  • #11
It would be 0 for [itex]r < R[/itex] and [itex]\frac{Q}{4 \pi \epsilon_0 r^2}[/itex] for [itex]r \geq R[/itex].
 
  • #12
OK. The electric field is the same as if all the charge were concentrated in the centre. What is the potential of such charged conducting sphere in terms of r?

ehild
 
  • #13
That's [itex]\frac{Q}{4 \pi \epsilon_0 r}[/itex] for [itex]r \geq R[/itex].
 
  • #14
You know the potential at r=R. It is V. What is the charge then on the outer surface of the shell?
The position of the inner charge does not influence the field outside, as the metal wall screens it. The inner charge induces a charge distribution on the inner surface of the metal shell, and if you want to determine it, you can use the method of image charges. But you need only the total surface charge inside. You know that electric field is zero inside the metal wall. Taking a Gaussian surface inside the wall, the integral of E is zero on it, that is, the enclosed charge is zero. So what is the inner surface charge?

ehild
 
Last edited:
  • #15
So [itex]Q = V 4 \pi \epsilon_0 R[/itex] and so [itex]\sigma_{out} = \frac{Q}{A} = \frac{V 4 \pi \epsilon_0 R}{4 \pi R^2} = \frac{V \epsilon_0}{R}[/itex].

On the inner surface the charge is simply [itex]q[/itex], so [itex]\sigma_{in} = \frac{q}{4 \pi R^2}[/itex].
 
  • #16
No, σin, the inner surface charge density depends on the distance from the negative charge placed inside the shell. But the total surface charge is +|q|. If I understand the problem correctly, it asks only the surface charge and not its density.

ehild
 
  • #17
Yes, you're right, so on the outer surfaceit's just [itex]Q = V 4 \pi \epsilon_0 R[/itex] and the inner surface [itex]q[/itex].
 
  • #18
Take care, the sign or the inner charge is opposite to that of q. It was said "q negative charge". The inner surface charge has opposite sign. So answer |q|.

ehild
 
  • #19
Thanks a lot!
 
  • #20
You are welcome:smile:

ehild
 

Related to Charge inside conducting shell

1. What is a conducting shell?

A conducting shell is a hollow, metallic object that allows charges to move freely on its surface. Examples of conducting shells include metal spheres, cylinders, and boxes.

2. How does charge distribute inside a conducting shell?

Inside a conducting shell, the charge distributes in such a way that the electric field inside the shell is zero. This is due to the fact that charges on the surface of a conductor will rearrange themselves in order to minimize their potential energy.

3. How does the charge inside a conducting shell affect the electric field outside the shell?

The charge inside a conducting shell does not affect the electric field outside the shell. This is because the charges on the surface of the conductor rearrange themselves in such a way that the electric field inside the shell is zero, and therefore there is no contribution to the electric field outside the shell.

4. Can the charge inside a conducting shell be changed?

Yes, the charge inside a conducting shell can be changed by adding or removing charges from its surface. However, as long as the conducting shell remains closed, the total charge inside the shell will remain constant and the electric field inside the shell will remain zero.

5. What is the purpose of a conducting shell?

The purpose of a conducting shell is to shield any charges inside it from external electric fields. This is due to the fact that charges on the surface of a conductor rearrange themselves in order to minimize their potential energy, resulting in a zero electric field inside the shell. This is why conducting shells are often used in electronic devices to protect sensitive components from external electric fields.

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