Chapter 16, Problem 46P

(a)

To determine

The net flux through the cube.

Answer to Problem 46P

The net flux through the cube is zero.

Explanation of Solution

Given information:

Electric field intensity, $E=6.50\times {10}^3\text{ N/C}$

Formula Used:

Calculate the required amount of flux.

  $\varphi =EA$

Here, $\varphi$ is the flux, E is the electric field intensity and A is the area.

Calculation:

The electric flux is zero as the electric flux through the left face is equal to the right face.

Substitute the values in the above equation.

  $\varphi =0$

Conclusion:

Thus, net flux through the cube is 0

(b)

To determine

The flux through each of its 6 faces.

Answer to Problem 46P

The flux through each of its 6 faces is calculated.

Explanation of Solution

Given information:

Electric field intensity, $E=6.50\times {10}^3\text{ N/C}$

Formula Used:

Calculate the required amount of flux.

  $\varphi =EA$

Here, $\varphi$ is the flux, E is the electric field intensity and A is the area.

Calculation:

The electric flux is zero as the electric flux through the upper, lower, back and front faces.

Substitute the values in the above equation for left side.

  $\begin{array}{c}\varphi l_{}=EA\\ =\left(-E\right){\left(L\right)}^2\\ ={\left(8.50\text{ cm}\right)}^2\left(-7.5\times {10}^3\right)\\ =-6.37{\text{ Nm}}^2\text{/C}\end{array}$

Substitute the values in the above equation for right side.

  $\begin{array}{c}\varphi r_{}=EA\\ =\left(-E\right){\left(L\right)}^2\\ ={\left(8.50\text{ cm}\right)}^2\left(-7.5\times {10}^3\right)\\ =-6.37{\text{ Nm}}^2\text{/C}\end{array}$

Conclusion:

Thus, flux through each of its 6 faces is calculated.