Induces a charge in the inner surface of the shell?PeroK said:
Have you studied that? Or, learned any techniques to find the electric field outside an asymmetric charge configuration?lys04 said:
I've learnt Gauss's law, if that's applicable.PeroK said:
Gauss's law is useful. What can you say about the potential on the surface of a conductor?lys04 said:
In a way: you will have to argue why the required symmetry (*) is present.lys04 said:
It's uniform?PeroK said:
Yeah I'm not sure about thatBvU said:
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.lys04 said:
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.
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.
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.
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.
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.