Consider a thin, spherical shell of radius 13.5 cm with a total charge of 31.3 µC distributed uniformly on its surface.
(a) Find the electric field 10.0 cm from the center of the charge distribution.
(b) Find the electric field 21.5 cm from the center of the charge distribution.
The field inside the shell should be zero. The field outside the shell is the same as the field created by a charged particle located at the center of the shell. Both of these results follow from the spherical symmetry in the situation.
We know that the field is directed radially outward and that it is uniform in magnitude over any sphere concentric with the shell. We will use spherical gaussian surfaces through the field points to find the magnitudes of the field at these points.
(a) A gaussian sphere with a radius less than the radius of the spherical shell encloses zero charge. Since we are asked for the magnitude of the electric field at 10.0 cm inside the shell, we have
→E = ( N/C)r̂.
(b) For a gaussian sphere of radius 21.5 cm, we apply the following equation.
ΦE = ∮ →E · d→A = qin / ε0
By symmetry, E is constant everywhere on the surface of the shell and is directed radially outward, so 4πr^2E = q/ε0. Because ke = 1/(4πε0), we have the following.
E = keq / r^2
= (8.99 × 10^9 N · m^2/C^2)( × 10^-6 C) / ( m)^2
In vector notation,
→E = ( × 10^6 N/C)r̂.