Chapter 16, Problem 50P

(a)

To determine

The electric field as a function of r(a) for r less than the inner radius of the shell.

Answer to Problem 50P

The electric field is $\frac{Q}{4\pi {\epsilon }_0{r}^2}$ .

Explanation of Solution

Given information:

Consider figure 16.33

Formula Used:

Calculate the required electric field.

  ${\displaystyle \oint \overline{E}dA}=\frac{Q}{{\epsilon }_0}$

Here, Q is the charge, and ${\epsilon }_0$ is the relative permittivity.

Calculation:

Solve the above equation.

  $\begin{array}{c}{\displaystyle \oint \overline{E}dA}=\frac{Q}{{\epsilon }_0}\\ \overline{E}{\displaystyle \oint dA}=\frac{Q}{{\epsilon }_0}\\ E\left(4\pi r^2\right)=\frac{Q}{{\epsilon }_0}\\ E=\frac{Q}{4\pi {\epsilon }_0{r}^2}\end{array}$

Conclusion:

Thus, the electric field is $\frac{Q}{4\pi {\epsilon }_0{r}^2}$ .

(b)

To determine

The electric field inside the shell.

Answer to Problem 50P

The electric field is zero.

Explanation of Solution

Given information:

Consider figure 16.33

Formula Used:

Calculate the required electric field.

  ${\displaystyle \oint \overline{E}dA}=\frac{Q}{{\epsilon }_0}$

Here, Q is the charge, and ${\epsilon }_0$ is the relative permittivity.

Calculation:

Solve the above equation.

  $\begin{array}{c}{\displaystyle \oint \overline{E}dA}=\frac{Q_{net}}{{\epsilon }_0}\\ {\displaystyle \oint \overline{E}dA}=\frac{+Q-Q}{{\epsilon }_0}\\ {\displaystyle \oint \overline{E}dA}=\frac{0}{{\epsilon }_0}\\ \overline{E}=0\end{array}$

Conclusion:

Thus, the electric field is 0

(c)

To determine

The electric field beyond the shell.

Answer to Problem 50P

The electric field is $\frac{Q}{4\pi {\epsilon }_0{r}^2}$ .

Explanation of Solution

Given information:

Consider figure 16.33

Formula Used:

Calculate the required electric field.

  ${\displaystyle \oint \overline{E}dA}=\frac{Q}{{\epsilon }_0}$

Here, Q is the charge, and ${\epsilon }_0$ is the relative permittivity.

Calculation:

Solve the above equation.

  $\begin{array}{c}{\displaystyle \oint \overline{E}dA}=\frac{Q_{net}}{{\epsilon }_0}\\ {\displaystyle \oint \overline{E}dA}=\frac{+Q-Q+Q}{{\epsilon }_0}\\ E\left(4\pi r^2\right)=\frac{Q}{{\epsilon }_0}\\ E=\frac{Q}{4\pi {\epsilon }_0{r}^2}\end{array}$

Conclusion:

Thus, the electric field is $\frac{Q}{4\pi {\epsilon }_0{r}^2}$ .

(d)

To determine

Whether the field and the charge affect the shell.

Answer to Problem 50P

No, the shell does not affect the field but the charge affects the shell

Explanation of Solution

Given information:

Consider figure 16.33

Formula Used:

Calculate the required electric field.

  ${\displaystyle \oint \overline{E}dA}=\frac{Q}{{\epsilon }_0}$

Here, Q is the charge, and ${\epsilon }_0$ is the relative permittivity.

Calculation:

No, the shell does not affect the field but the charge affects the shell

Conclusion:

No, the shell does not affect the field but the charge affects the shell