Answer to Question #154635 in Physics for salma

Question

Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field (a) 10.0 cm and (b) 20.0 cm from the center of the charge distribution

 

Answer 1

Let us draw two imaginary spherical surfaces: inside and outside the shell (see figure). Also let $R = 14cm = 0.14m$ be the radius of the shell, $R_1 = 10cm = 0.1m$, and $R_2 = 20cm = 0.2m$ be the radii of the inner and outer surfaces respectively.

1. Writting down the Gauss's theorem for the inner surface, obtain:

$$\oint_{S_1}\mathbf{E_1}d\mathbf{S_1} = \dfrac{Q}{\varepsilon_0}$$

where $\mathbf{E_1}$ is the electric field on the inner surface $S_1$, $\varepsilon_0 = 8.85\times 10^{-12}F/m$ is electric constant, and $Q = 0$ is the charge enclosed by this surface. Since all charge is located on the shell, $Q = 0$, and, thus,

$$\mathbf{E_1} = 0$$

2. Writting down the Gauss's theorem for the inner surface, obtain:

$$\oint_{S_2}\mathbf{E_2}d\mathbf{S_2} = \dfrac{Q}{\varepsilon_0}$$

where $\mathbf{E_2}$ is the electric field on the outer surface $S_2$, and $Q = 32\times 10^{-6}C$ is the charge enclosed by this surface. The elecrtic field is constant and perpendicular to the $S_2$ at any point (due to spherical symmetry). Thus, the right hand side will be:

$$E_2\cdot 4\pi R^2_2 = \dfrac{Q}{\varepsilon_0}$$

Expressing $E_2$, obtain:

$$E_2 = \dfrac{Q}{\varepsilon_0\cdot 4\pi R^2_2}\\ E_2 = \dfrac{32\times 10^{-6}}{8.85\times 10^{-12}\cdot 4\pi \cdot 0.2^2} \approx 7.19\times 10^6N/C$$

Answer. a) , b) $7.19\times 10^6N/C$.