Question

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is the magnitude of the electric field at a distance (a) r = 12 cm, and (b) r = 20 cm from the center of the shell? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)

Answer 1

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$