Electrostatics > The potential of an electric field. Conductors in a constant electric field > Problem 6.3.16

Statement

$6.3.16.$ There is a charge $Q$ in the cavity of a metal ball of radius $R$. Find the charge induced by this charge on the surface of the cavity. Why will the charge be distributed with a constant density on the surface of the ball? What is the surface charge density of a sphere if its total charge is zero? Find the electric field strength outside the ball at a distance $L$ from its center if its total charge is $q$. Does this field depend on the location of the cavity in the ball? from its shape?

Solution

The presence of a charge $Q$ in the cavity produces an electric field that penetrates the walls of the cavity, inducing a charge $Q'$. Since the ball is made of metal, the sphere is conducting, so charge on it resides on its surface, then the electric field inside of it is zero. Applying Gauss law, for a Gaussian sphere centered at the position of charge $Q$ and radius $r$, always inside the ball,
$\int \vec{E}\cdot d\vec{S} = \frac{Q+Q'}{\varepsilon_0}$
as $\vec{E} = \vec{0}$,
$Q' = -Q$
and the outer surface of ball is charged with $-Q' = +Q$.

Let's take a Gaussian sphere centered at the ball's center with radius $R$, and consider the differential form of Gauss law,
$\vec{E}\cdot d\vec{S} = \frac{dq}{\varepsilon_0}$
$\sigma = \frac{dq}{dS} = \varepsilon_0 E(R)$
but E(R) has a constant value independently on cavity position inside the ball. So, $\sigma$ is constant (**the distribution over the outer surface is uniform**).

Applying Gauss law for a Gaussian sphere of radius $R$ centered at the ball (total charge = $Q'-Q'$ = 0),
$\int\vec{E}\cdot d\vec{S} = \frac{-Q'+Q'+Q}{\varepsilon_0} = \frac{Q}{\varepsilon_0}$ (1)
$\sigma = \frac{dq}{dS} = \varepsilon_0 E(R)$ (2)
From (1),
$E(R) = \frac{Q}{4\pi\varepsilon_0 R^2}$ (3)
Putting (3) into (2),
$\sigma = \frac{Q}{4\pi R^2}$

Applying Gauss law for a sphere centered at ball and radius $r > R$ and evaluating at $r=L$,
$\int \vec{E}\cdot d\vec{S} = \frac{q+Q}{\varepsilon_0}$
$E(L) = \frac{q+Q}{4\pi\varepsilon_0 L^2}$
The value of E(L) **doesn't depend on the cavity's position nor size of sphere, but the enclosed charge**.

Answer

$Q' = -Q$
Charge distribution obver the outer surface is uniform
$\sigma = \frac{Q}{4\pi R^2}$
$E(L) = \frac{q+Q}{4\pi\varepsilon_0 L^2}$
NO, NO