1
$\begingroup$

enter image description here

(The question has been taken just for reference)

What I know-

When a spherical conductor has a charge inside and outside, charges are induced in such a way that there’s no interaction between the fields(i.e. between the fields of inner charge and the field of outer charge)

My doubt-

If there’s no interaction between the two charges’ fields why will the induced charges on the outer surface (which was induced to cancel out electric field of the outer charge to prevent it from going inside) have any effect on inside charge?

Also why would the induced charges due to induced charges on inner surface have any effect on the outer charge?

How would we be able to find the magnitude? Because the distribution is not uniform.

$\endgroup$
2
  • $\begingroup$ > there’s no interaction between the fields(i.e. between the fields of inner charge and the field of outer charge) what does this mean? If we separate total field into its components, the components are non-zero everywhere. Why would they not interact? Only the total field vanishes inside the conductor, because there the components cancel out. Isn't this "interaction between the fields"? $\endgroup$ Jun 29, 2021 at 15:29
  • $\begingroup$ I meant the electric field due to inside charge + electric field due to charges induced on the inner surface is zero. And the electric field due to outside charge + electric field due to charges induced in the outer surface is zero. So no there won’t be any interaction. It’s called electrostatic shielding. $\endgroup$
    – Natru
    Jun 29, 2021 at 15:44

1 Answer 1

1
+50
$\begingroup$

enter image description here

Let consider first removing the conductor shell. The magnitude of force between these two charges is $$ F_{12} = \frac{q_1 q_2}{8\pi\epsilon R^2}. \tag{1} $$

Then put on the conductor shell. We first consider the induced charges from $q_1$. The induced charges on the inner surface (total amount $-q_1$) shielded the electric field from charge $q_1$, to ensure that there are no electric field inside the conductor. The total force exerted from these induced charges on $q_2$ is to cancel the electric force existing before the conductor is there.

There will be an induced charges (total amount $q_1$) on the outer surface. Since there are no field penetrated from the condutor, the induced charge on the outer surface will be even distributed. This induced chages in the outer surface (due to $q_1$) apply an electric force on the charge $q_2$, like a charge $q_1$ sits on the center: $$ F_{2, outer} = \frac{q_1q_2}{7\pi\epsilon R^2}, $$

enter image description here

Then we consider $q_2$. The outer surface will induce charge (total amount $-q_2$) to shield the electric field from penetrating into conductor. These induced charges on the outer surface exert an opposite $F_{12}$ to cancel the froce from $q_2$ to $q_1$ (Eq. 1).

enter image description here

There are induced charges in the inner surface (total amount $q_2$). they are evenly distributed, because the field from $q_2$ had been shielded. The even distributed inner charges render zero force to $q_1$.

The final situation will be the adding together of the above two figures.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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