91Ó°ÊÓ

• The spherical shell of radius ‘a’ has a charge $q_{a.}$
• The spherical shell of radius b > a and charge $q_b.$

At all points where there is an electric field, it is radially outward. For each part of the problem, use a Gaussian surface in the form of a sphere that is concentric with the metal shells of charge and passes through the point where the electric field is to be found. The field is uniform on the surface, so

Formula:

$$\begin{gathered} \mathrm{Ï•}=âˆ\circledR \stackrel{→}{E}.\stackrel{→}{dA} \\ =4{\mathrm{Ï€°ù}}^2\mathrm{E} \\ =\frac{{\mathrm{q}}_{\mathrm{enc}}}{{\mathrm{Î\mu }}_0} \end{gathered}$$

(Where r is the radius of the Gaussian surface)

From Gauss’s law, the charge enclosed in the spherical metal shell is zero. Therefore, for r < a, the charge enclosed is enc $q_{enc}=0.$

Hence electric field for the region inside the shell is zero,$E=0.$

For a < r < b, the charge enclosed by the Gaussian surface is , so the field strength is:

$E=\frac{q_a}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}$

Therefore, the magnitude of the electric field at $a<r<b$is $\frac{q_a}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}.$.

For r > b, the charge enclosed by the Gaussian surface is $q_{enc}=q_a+q_b$, so the field strength is:

$E=\frac{q_a+q_b}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}$

Therefore, the magnitude of the electric field at $r>b$is $\frac{q_a+q_b}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}$

Since E = 0 for r < a, the charge on the inner surface of the inner shell is always zero, and the charge on the outer surface of the inner shell is, therefore $q_{a.}$

Since E = 0 inside the metallic outer shell, the net charge enclosed in a Gaussian surface that lies in between the inner and outer surfaces of the outer shell is zero. Thus, the inner surface of the outer shell has a charge $-q_a$, leaving the charge on the outer surface of the outer shell to be $q_b+q_a$.

Thus, because of the charge distribution on the inner metal spherical shell, the charge on the outer surface of the outer shell is$q_b+q_{a.}$