0
$\begingroup$

I was recently given a question that went something like this:

A spherical conductor is suspended within a hollow, conducting shell. The shell carries $0$ total charge, while the sphere carries a total charge $Q$. Graph the magnitude of the electric field along a radius extending from the centre of the charged sphere.


My (seemingly incorrect) logic went something like this. The electric field within the conducting sphere is $0$ (the electric field within any conductor at electrostatic equilibrium is necessarily $0$).

$[1] $ Then, between the outer surface of the sphere and the inner surface of the shell, the electric field will behave similarly to that of a point charge.

Next, the electric field returns to zero within the conducting shell.

$[2]$ Finally, outside of the shell, it again behaves akin to that of a point charge.

Graphically, something like this:

enter image description here


I'm trying to wrap my head around precisely where I went wrong. There are a few things that aren't quite clear to me.

At $[1]$, what will be the behavior of the electric field near the surface of the sphere? Will it simply tend to infinity? Further, there will be an induced charge polarization in the conducting shell. Therefore, there will be a charge $Q$ on the outer surface of the sphere, and a charge $-Q$ on the inner surface of the shell. Will this lead to "capacitor-like" behavior (using Gauss' Law, I think not, but I could well be wrong)?

At $[2]$, there is a charge of $Q$ on the outer surface of the conducting shell. It seems to me, then, that the behavior here will be identical to that of the behavior of the electric field between the sphere and the shell, except it will continue along the radial axis to infinity. Moreover, we can find this using gauss law (the charge enclosed has not changed). Is my intuition right?

$\endgroup$
1
$\begingroup$
  1. The electric field near the surface:

    The sheet charge density on the spherical surface $$ \sigma = \frac{Q}{4\pi r^2}. $$ This renders an electric field $$ E = \frac{\sigma}{\epsilon_o} = \frac{Q}{4\pi \epsilon_o r^2}$$ The field from considering the surface charge density is exactly the same as that using Gauss's law.

  2. Induced charge dipolar layer:

    Yes. You may consider the dipolar layer in between the two surfaces of the shell. The electric filed induced by the dipolar layer cancels the electric filed from the charge in sphere. These two fields are cancelled resulting zero field inside the shell.

  3. Field outside the shell

    Thus, you can apply Gauss's law in the radius outside the shell. The field intensity. $$ E(r) = \frac{Q}{4\pi \epsilon_o r^2}. $$ Therefore, the field intensity will continue from the space between sphere surface and shell.

In your drawing the electric field is not continuing.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.