Chapter 16, Problem 45P

(a)

To determine

The electric flux through the surface $A_1$ that enclose all three objects.

Answer to Problem 45P

The required direction is right.

Explanation of Solution

Given info:

The charge on object $O_1$ is $q_1=+1.0\mu \text{C}$ .

The charge on object $O_2$ is $q_2=-2.0\mu \text{C}$ .

The third object $O_3$ is electrically neutral that is $q_3=0\mu \text{C}$ .

Formula used:

Gauss law states that the net electric flux passing through closed surface is equal to the $\frac{1}{{\epsilon }_0}$ times the net charge enclosed within the surface.

The expression for the Gauss Law is,

  ${\varphi }_E=\frac{q_{\text{net}}}{{\epsilon }_0}$

The net charge enclosed within the surface $A_1$ is equal to the sum of the charges of objects $O_1$ , $O_2$ , and $O_3$ .

  $q_{net}=q_1+q_2+q_3$

Calculation:

Substituting the given values in formula $q_{net}=q_1+q_2+q_3$ , we get

  $\begin{array}{c}{q}_{net}=+1.0\mu \text{C}-2.0\mu \text{C}+0\mu \text{C}\\ =-1.0\mu \text{C}\end{array}$

Substituting the given values in formula ${\varphi }_E=\frac{q_{\text{net}}}{{\epsilon }_0}$ , we get

  $\begin{array}{c}{\varphi }_E=\frac{-1.0\mu \text{C}}{8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/Nm}}^{\text{2}}}\left(\frac{{10}^{-6}\text{C}}{1\mu \text{C}}\right)\\ =1.1\times {10}^5{\text{Nm}}^{\text{2}}\text{/C}\end{array}$

Conclusion:

Thus, the electric flux through the surface $A_1$ that enclose all three objects is $1.1\times {10}^5{\text{Nm}}^{\text{2}}\text{/C}$ .

(b)

To determine

The electric flux through the surface $A_2$ that encloses the third object only.

Answer to Problem 45P

The electric flux through the circle when its face is at $45°$ to the filed lines is $41.8{\text{Nm}}^{\text{2}}\text{/C}$ .

Explanation of Solution

Given info:

The charge on object $O_1$ is $q_1=+1.0\mu \text{C}$ .

The charge on object $O_2$ is $q_2=-2.0\mu \text{C}$ .

The third object $O_3$ is electrically neutral that is $q_3=0\mu \text{C}$ .

Formula used:

Gauss law states that the net electric flux passing through closed surface is equal to the $\frac{1}{{\epsilon }_0}$ times the net charge enclosed within the surface.

The expression for the Gauss Law is,

  ${\varphi }_E=\frac{q_{\text{net}}}{{\epsilon }_0}$

The net charge enclosed within the surface $A_1$ is equal to the charges of object $O_3$ .

  $q_{net}=q_3$

Calculation:

Substituting the given values in formula ${\varphi }_E=\frac{q_{\text{net}}}{{\epsilon }_0}$ , we get

  $\begin{array}{c}{\varphi }_E=\frac{0\mu \text{C}}{8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/Nm}}^{\text{2}}}\\ =0{\text{Nm}}^{\text{2}}\text{/C}\end{array}$

Conclusion:

Thus, the electric flux through the surface $A_2$ that encloses the third object only is $0{\text{Nm}}^{\text{2}}\text{/C}$ .