Chapter 25, Problem 24PQ

Three particles and three Gaussian surfaces are shown in Figure P25.24. All the surfaces are three-dimensional. Use the net electric flux through each surface indicated on the figure to find the charge of each particle.

FIGURE P25.24

To determine

The charge on each particle.

Answer to Problem 24PQ

The charge on each particle are $\underset{\_}{0\text{C}}$, $\underset{\_}{-2.66\times {10}^{-10}\text{C}}$ and $\underset{\_}{7.97\times {10}^{-10}\text{C}}$.

Explanation of Solution

Write the expression for the total flux enclosed by the surface A.

${\Phi }_{E,\text{A}}=\frac{Q_1+Q_2}{{\epsilon }_0}$ (I)

Here, ${\Phi }_{E,\text{A}}$ is the total flux enclosed by the surface A, $Q_1$ is the charge on the first particle, $Q_2$ is the charge on the second particle and ${\epsilon }_0$ is a constant.

Write the expression for the total flux enclosed by the surface B.

${\Phi }_{E,\text{B}}=\frac{Q_2+Q_3}{{\epsilon }_0}$ (II)

Here, ${\Phi }_{E,\text{B}}$ is the total flux enclosed by the surface B and $Q_3$ is the charge on the third particle.

Write the expression for the total flux enclosed by the surface C.

${\Phi }_{E,\text{C}}=\frac{Q_1+Q_3}{{\epsilon }_0}$ (III)

Here, ${\Phi }_{E,\text{C}}$ is the total flux enclosed by the surface C and $Q_3$ is the charge on the third particle.

Rearrange equation (I) to find $Q_1$.

$Q_1={\epsilon }_0{\Phi }_{E,\text{A}}-Q_2$

Substitute ${\epsilon }_0{\Phi }_{E,\text{A}}-Q_2$ for $Q_1$ in equation (III).

${\Phi }_{E,\text{C}}=\frac{{\epsilon }_0{\Phi }_{E,\text{A}}-Q_2+Q_3}{{\epsilon }_0}$

Rearrange the above equation with charges on one side.

${\epsilon }_0{\Phi }_{E,\text{C}}-{\epsilon }_0{\Phi }_{E,\text{A}}=-Q_2+Q_3$ (IV)

Rearrange equation (II).

${\epsilon }_0{\Phi }_{E,\text{B}}=Q_2+Q_3$ (V)

Add equation (IV) and (V).

$2Q_3={\epsilon }_0{\Phi }_{E,\text{C}}-{\epsilon }_0{\Phi }_{E,\text{A}}+{\epsilon }_0{\Phi }_{E,\text{B}}$

Rearrange the above equation to find the charge $Q_3$.

$Q_3=\frac{{\epsilon }_0}{2}\left({\Phi }_{E,\text{C}}-{\Phi }_{E,\text{A}}+{\Phi }_{E,\text{B}}\right)$ (VI)

Rearrange equation (II) to find $Q_2$.

$Q_2={\epsilon }_0{\Phi }_{E,\text{B}}-Q_3$ (VII)

Rearrange equation (III) to find $Q_1$.

$Q_1={\epsilon }_0{\Phi }_{E,\text{C}}-Q_3$ (VIII)

Conclusion:

Substitute $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$, $90.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\Phi }_{E,\text{C}}$, $-30.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\Phi }_{E,\text{A}}$ and $60.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\Phi }_{E,\text{B}}$ in equation (VI).

$\begin{array}{c}{Q}_3=\frac{8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2}{2}\left(90.0\text{N}\cdot {\text{m}}^2/\text{C}-\left(-30.0\text{N}\cdot {\text{m}}^2/\text{C}\right)+60.0\text{N}\cdot {\text{m}}^2/\text{C}\right)\\ =7.97\times {10}^{-10}\text{C}\end{array}$

Substitute $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$, $60.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\Phi }_{E,\text{B}}$ and $7.97\times {10}^{-10}\text{C}$ for $Q_3$ in equation (VII).

$\begin{array}{c}{Q}_2=\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(60.0\text{N}\cdot {\text{m}}^2/\text{C}\right)-7.97\times {10}^{-10}\text{C}\\ \text{=}-2.66\times {10}^{-10}\text{C}\end{array}$

Substitute $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$, $90.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\Phi }_{E,\text{C}}$ and $7.97\times {10}^{-10}\text{C}$ for $Q_3$ in equation (VIII).

$\begin{array}{c}{Q}_1=\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(90.0\text{N}\cdot {\text{m}}^2/\text{C}\right)-7.97\times {10}^{-10}\text{C}\\ \text{=}0\text{C}\end{array}$

Therefore, the charge on each particle are $\underset{\_}{0\text{C}}$, $\underset{\_}{-2.66\times {10}^{-10}\text{C}}$ and $\underset{\_}{7.97\times {10}^{-10}\text{C}}$.