a) E = 0
b) ![3.38\cdot 10^6 N/C](https://tex.z-dn.net/?f=3.38%5Ccdot%2010%5E6%20N%2FC)
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}](https://tex.z-dn.net/?f=%5Cint%20EdS%3D%5Cfrac%7Bq%7D%7B%5Cepsilon_0%7D)
where
E is the electric field
q is the charge contained by the Gaussian surface
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](https://tex.z-dn.net/?f=r_1%20%3D%2010%20cm%5C%5Cr_2%20%3D%2015%20cm)
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}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BQ%7D%7B4%5Cpi%20%5Cepsilon_0%20r%5E2%7D)
where in this problem:
is the charge on the shell
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](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B15%5Ccdot%2010%5E%7B-6%7D%7D%7B4%5Cpi%20%288.85%5Ccdot%2010%5E%7B-12%7D%29%280.20%29%5E2%7D%3D3.38%5Ccdot%2010%5E6%20N%2FC)