Chapter 16, Problem 48P

To determine

To find: The net charge on the ball.

Answer to Problem 48P

  $\mathrm{Q}=-3.74\times {10}^{-11}\text{C}$

Explanation of Solution

Given:

  $\begin{array}{l}{\mathrm{E}}_1=2.75\times {10}^2\text{N}/\text{C}\\ \mathrm{r}=3.50\text{cm}\end{array}$

Formula used:

The magnitude of electric flux through a closed surface is,

  ${\text{Φ}}_{\mathrm{E}}=\overrightarrow{\mathrm{E}}\cdot \overrightarrow{\mathrm{A}}$

The magnitude of electric flux through a closed surface is,

  $\begin{array}{cc}{\text{Φ}}_{\mathrm{E}}& =\overrightarrow{\mathrm{E}}\cdot \overrightarrow{\mathrm{A}}\\ & =|\overrightarrow{\mathrm{E}}\parallel \overrightarrow{\mathrm{A}}|\cos \mathrm{\theta }\end{array}$

For maximum flux, the angle must be equal to $0^{\circ }$ . The Gaussian surface due to the point charge is a sphere.

So the electric flux is,

  $\begin{array}{cc}{\text{Φ}}_{\mathrm{E}}& =\mathrm{E}\left(4\mathrm{\pi }{\mathrm{r}}^2\right)\cos 0^{\circ }\\ & =\mathrm{E}\left(4\mathrm{\pi }{\mathrm{r}}^2\right)\end{array}$

Gauss law states that the electric flux through a closed surface is proportional to $\frac{1}{{\mathrm{\epsilon }}_0}$ times the charge enclosed in it and is given by,

  ${\text{Φ}}_{\mathrm{E}}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$

Here, q is the charge and ${\mathrm{\epsilon }}_0$ is the permittivity.

Calculation:

Substitute $\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}\text{for}{\text{Φ}}_{\mathrm{E}}\text{in}{\text{Φ}}_{\mathrm{E}}=\mathrm{E}\left(4\mathrm{\pi }{\mathrm{r}}^2\right)$

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

Rearrange the equation for the charge.

  $\mathrm{q}=\mathrm{E}\left(4\mathrm{\pi }{\mathrm{\epsilon }}_0\right){\mathrm{r}}^2$

The charge on the metal sphere is,

  $\mathrm{q}=\mathrm{E}\left(4\mathrm{\pi }{\mathrm{\epsilon }}_0\right){\mathrm{r}}^2$

Substitute $2.75\times {10}^2\text{N}/\text{C for }{\mathrm{E}}_1\text{, }8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\text{for}{\mathrm{\epsilon }}_0,\text{and}3.50\text{cm}\text{for}\mathrm{r}$

  $\begin{array}{l}\mathrm{q}=\left(2.75\times {10}^2\text{N}/\text{C}\right)\left(4\mathrm{\pi }\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\right){\left(3.50\text{cm}\left(\frac{1\text{m}}{100}\right)\right)}^2\\ =3.746\times {10}^{-11}\text{C}\end{array}$

The electric field id directed away from the charge if positive and is directed towards the charge if the charge is negative.

Conclusion:

Here, the electric field is directed towards the metal ball. So, the charge is negative $-3.746\times {10}^{-11}\text{C}$ .