Chapter 16, Problem 43P

To determine

To find: The charge enclose by the box.

Answer to Problem 43P

  $1.28\times {10}^{-8}\text{C}$

Explanation of Solution

Given:

  $\begin{array}{l}{\mathrm{\varphi }}_{\mathrm{E}}=1.45\times {10}^3{\text{Nm}}^{\text{2}}\text{/C}\\ \mathrm{a}=28\text{cm}\end{array}$

Formula used:

Gauss’s law states that the net electric flux through the closed surface is $\frac{1}{{\mathrm{\epsilon }}_0}$ charge and it can be expressed as follows:

  ${\mathrm{\varphi }}_{\mathrm{E}}=\frac{{\mathrm{q}}_{\mathrm{e}\mathrm{n}}}{{\mathrm{\epsilon }}_0}$

Here, ${\mathrm{\varphi }}_{\mathrm{E}}$ is the electric flux, ${\mathrm{q}}_{\mathrm{e}\mathrm{n}}$ is the charge enclosed in the closed region, and ${\mathrm{\epsilon }}_0$ is the permittivity of free space.

Calculation:

Rearrange the equation ${\mathrm{\varphi }}_{\mathrm{E}}=\frac{{\mathrm{q}}_{\mathrm{e}\mathrm{n}}}{{\mathrm{\epsilon }}_0}$ for ${\mathrm{q}}_{\mathrm{e}\mathrm{n}}$ as follows,

  ${\mathrm{q}}_{\mathrm{e}\mathrm{n}}={\mathrm{\varphi }}_{\mathrm{E}}{\mathrm{\epsilon }}_0$

Substitute $1.45\times {10}^3\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\mathrm{\varphi }}_{\mathrm{E}}$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$

  ${\mathrm{\epsilon }}_0$ .

  $\begin{array}{c}{\mathrm{q}}_{\mathrm{e}\mathrm{n}}=\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(1.45\times {10}^3\text{N}\cdot {\text{m}}^2/\text{C}\right)\\ =1.28\times {10}^{-8}\text{C}\end{array}$