# Electric field near conducting shell

• lys04
lys04
Homework Statement
Electric field near conducting shell
Relevant Equations
E=kQ/r^2
How would I do this question? I am having trouble figuring out what the radius is meant to be, why is it not 3R?

The charge is inside a conducting shell. What is the relevance of that?

PeroK said:
The charge is inside a conducting shell. What is the relevance of that?
Induces a charge in the inner surface of the shell?

lys04 said:
Induces a charge in the inner surface of the shell?
Have you studied that? Or, learned any techniques to find the electric field outside an asymmetric charge configuration?

PeroK said:
Have you studied that? Or, learned any techniques to find the electric field outside an asymmetric charge configuration?
I've learnt Gauss's law, if that's applicable.

lys04 said:
I've learnt Gauss's law, if that's applicable.
Gauss's law is useful. What can you say about the potential on the surface of a conductor?

MatinSAR
lys04 said:
I've learnt Gauss's law, if that's applicable.
In a way: you will have to argue why the required symmetry (*) is present.

spherical symmetry of the field outside the shell

##\ ##

MatinSAR
PeroK said:
Gauss's law is useful. What can you say about the potential on the surface of a conductor?
It's uniform?

BvU said:
In a way: you will have to argue why the required symmetry (*) is present.

spherical symmetry of the field outside the shell

##\ ##
Yeah I'm not sure about that

lys04 said:
It's uniform?
Yes. The surface of the shell must be an equipotential. Now, the potential is fully determined by the boundary conditions (technically this is a uniqueness property of electrostatic solutions). You know that outside the shell a spherically symmetric potential is a solution, so it must be the only solution.

Now, you can apply Gauss's law.

MatinSAR

## What is the electric field outside a conducting shell?

The electric field outside a conducting shell is determined by the charge on the shell and follows the inverse square law, similar to the field produced by a point charge located at the center of the shell. Mathematically, it is given by $$E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}$$, where $$Q$$ is the total charge on the shell, $$r$$ is the distance from the center of the shell, and $$\epsilon_0$$ is the permittivity of free space.

## What is the electric field inside a conducting shell?

The electric field inside a conducting shell is zero. This is because the charges on a conductor rearrange themselves in such a way that they cancel out any internal electric fields. This phenomenon is a consequence of electrostatic equilibrium, where the electric potential is constant throughout the conductor.

## How does the presence of a charge inside a conducting shell affect the electric field outside the shell?

The presence of a charge inside a conducting shell does not affect the electric field outside the shell. The conducting shell redistributes its own charge to cancel the field created by the internal charge within the shell, ensuring that the electric field outside depends only on the total charge on the shell itself.

## What happens to the electric field near the surface of a conducting shell?

The electric field near the surface of a conducting shell is perpendicular to the surface and has a magnitude given by $$E = \frac{\sigma}{\epsilon_0}$$, where $$\sigma$$ is the surface charge density and $$\epsilon_0$$ is the permittivity of free space. This is a result of the boundary conditions for electric fields at the surface of a conductor.

## How is the charge distributed on a conducting shell?

The charge on a conducting shell distributes itself uniformly over the outer surface of the shell. This uniform distribution occurs because the charges repel each other and move as far apart as possible, resulting in a uniform charge density on the outer surface in the case of a spherical shell.

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