Chapter 16, Problem 32P

To determine

To find: The magnitude of charge on each point charge.

Answer to Problem 32P

  $2.65\times {10}^{-10}\text{C}$

Explanation of Solution

Given:

  $\begin{array}{l}\mathrm{E}=745\text{N}/\text{C}\\ \mathrm{r}=16\text{cm}\end{array}$

Formula used:

The net electric field at the mid-point between the two charges is shown in the following figure:

  

In the figure, ${\mathrm{E}}_1$ is the electric field due to positive charge, ${\mathrm{E}}_2$ is the electric field due to negative charge, q is the magnitude of the charge, and $8\text{cm}$ is the distance between charge and the center.

The magnitude of electric fields at center due to positive charge and negative charge are as follows:

  $\begin{array}{l}{\mathrm{E}}_1=\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}\\ {\mathrm{E}}_2=\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}\end{array}$

The net electric field due to two charges is as follows:

  ${\mathrm{E}}_{\mathrm{n}\mathrm{c}\mathrm{t}}={\mathrm{E}}_1+{\mathrm{E}}_2$

The given electric field at center is E.

Rewrite the above equation as follows:

  $\mathrm{E}={\mathrm{E}}_1+{\mathrm{E}}_2$

Substitute $\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}$ for ${\mathrm{E}}_1$ and $\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}$ for ${\mathrm{E}}_2$ in the equation and solve for $\mathrm{q}$

  $\begin{array}{l}\mathrm{E}=\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}+\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}\\ \mathrm{E}=2\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}\\ \mathrm{q}=\frac{\mathrm{E}{\mathrm{r}}^2}{2\mathrm{k}}\end{array}$

Calculation:

Calculate the magnitude of the charge as follows:

Substitute $9\times {10}^9\text{N}\cdot \text{m}/{\text{C}}^2$ for $\text{k},745\text{N}/\text{C}$ for $\mathrm{E},$ and $8\text{cm}$ for $\mathrm{r}$ in the equation $\mathrm{q}=\frac{\mathrm{E}{\mathrm{r}}^2}{2\mathrm{k}}$ and solve for $\mathrm{q}$ .

  $\begin{array}{l}\mathrm{q}=\frac{(745\text{N}/\text{C}){(8\text{cm})}^2}{2\left(9\times {10}^9\text{N}\cdot \text{m}/{\text{C}}^2\right)}\\ =\frac{(745\text{N}/\text{C}){\left(8\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}{2\left(9\times {10}^9\text{N}\cdot \text{m}/{\text{C}}^2\right)}\\ =2.65\times {10}^{-10}\text{C}\end{array}$

Conclusion:

Therefore, the magnitude of charge on each is $2.65\times {10}^{-10}\text{C}$ .