Answer to Question #88915 in Electricity and Magnetism for Shivam Nishad

Question

Using Gauss's law, obtain expressions
for electric field at a point (i) outside, and
(ii) inside a spherical charge distribution.

Answer 1

According to the Gauss's law,

$$\oint \vec{E} \cdot d\vec{S} = \frac{q}{\varepsilon_0}$$

Assuming a uniform charge distribution inside the sphere and taking into account that due to the symmetry both E and dS vectors a parallel (or antiparrallel depending on the sign of the charge), we can derive:

1) outside the sphere (r > R, where R is the radius of the sphere):

$$E \cdot 4 \pi r^2 = \frac{q}{\varepsilon_0}$$

Hence,

$$E = \frac{q}{4 \pi \varepsilon_0 r^2 }$$

2) inside the sphere (r < R):

$$E \cdot 4 \pi r^2 = \frac{q_r}{\varepsilon_0}$$

where q_r is the net electric charge that is inside a spherical volume of radius r:

$$q_r = \rho V_r = \frac{q}{\frac{4}{3} \pi R^3} \frac{4}{3} \pi r^3 = q \frac{r^3}{R^3}$$

Hence,

$$E = \frac{q_r}{4 \pi \varepsilon_0 r^2} = \frac{q r^3}{4 \pi \varepsilon_0 r^2 R^3} = \frac{q r}{4 \pi \varepsilon_0 R^3}$$