Question

A thin-walled metal spherical shell has radius$\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{cm}$and charge $\mathbf{2}\mathbf{.}\mathbf{88}\mathit{x}{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{C}$. Find Efor a point (a) inside the shell, (b) just outside it, and (c)from the center.

Short answer

a) The magnitude of the electric fieldinside the shell is zero

b) The magnitude of the electric field just outside the shell is $2.88x{10}^4\mathrm{N}/\mathrm{C}$.

c) The magnitude of the electric field 3.00 m from the center is$200\mathrm{N}/\mathrm{C}$

Step-by-step solution

Listing the given quantities

spherical shell has a radius of 25.0 cm and charge $2.00x{10}^{-7}\mathrm{C}$

Understanding the concept of electric field

Applying the concept of electric field for spherical shells we get the magnitude of the electric field.

Formula:

$\mathrm{E}=\frac{\left|\mathrm{q}\right|}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\mathrm{r}}^2}$

(a) Calculation of the magnitude of the electric field inside the shell

For r < R, E = 0

The magnitude of the electric field inside the shell is zero

(b) Calculation of the magnitude of the electric field just outside the shell

For r slightly greater than R,

$$\begin{gathered} \mathrm{E}=\frac{\left|\mathrm{q}\right|}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\mathrm{r}}^2} \\ =\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2} \\ =\frac{\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(2.00\times {10}^{-7}\mathrm{C}\right)}{{\left(0.250\mathrm{m}\right)}^2} \\ =2.88\times {10}^4\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field just outside the shell is $2.88\times {10}^4\mathrm{N}/\mathrm{C}$.

(c) Calculation of the magnitude of the electric field 3.00 m from the shell

For r > R

$$\begin{gathered} \mathrm{E}=\frac{\left|\mathrm{q}\right|}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\mathrm{r}}^2} \\ ={\mathrm{E}}_{\mathrm{R}}{\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^2 \\ =\left(2.88\times {10}^4\mathrm{N}/\mathrm{C}\right){\left(\frac{0.250\mathrm{m}}{3.00\mathrm{m}}\right)}^2 \\ =200\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field 3.00 m from the center is $200\mathrm{N}/\mathrm{C}$