Question

In each figure below is shown a rectangular box with dimensions L × W × H, placed in various electric fields of uniform magnitude E. For each case, find an expression for the net electric flux ΦE out of the box in terms of E, L, W, and H. The field in each case is assumed to be perpendicular to the side faces of the box.

Answer 1

Concepts and reason

The concept of electric flux is required to solve the problem.

First, determine the area of each face using the length and width of the face. Then, use the relation of electric flux and the angle between the area vector and electric field to determine the electric flux through each face. Finally, determine the net electric flux by adding the electric flux through each face.

Fundamentals

The electric flux is defined as the number of electric field lines passing through a surface. The electric flux through a surface of area A is,

${\Phi _E} = EA\cos \theta$

Here, E is the electric field vector, A is the area of the surface, and $\theta$ is the angle between the electric field and area vector which points perpendicular to the surface.

(1)

The electric flux is given as,

${\Phi _E} = EA\cos \theta$

For the right face:

The area of the right face of the rectangular box is given as,

$A = WH$

Here, W is the width of the rectangular box and H is the height of the rectangular box.

The area vector for the right face is normal to the right face of the box and it is towards right. The electric field is also towards right. Hence, the angle between the area vector and electric field is ${0^{\rm{o}}}$ .

Substitute ${\Phi _1}$ for ${\Phi _E}$ , $WH$ for A, and ${0^{\rm{o}}}$ for $\theta$ in the equation ${\Phi _E} = EA\cos \theta$ .

$\begin{array}{c}\\{\Phi _1} = EWH\cos {0^{\rm{o}}}\\\\ = EWH\\\end{array}$

For the left face:

The area of the left face of the rectangular box is given as,

$A = WH$

Here, W is the width of the rectangular box and H is the height of the rectangular box.

The area vector for the left face is normal to the left face of the box and it is towards left. The electric field is towards right. Hence, the angle between the area vector and electric field is ${180^{\rm{o}}}$ .

Substitute ${\Phi _2}$ for ${\Phi _E}$ , $WH$ for A, and ${180^{\rm{o}}}$ for $\theta$ in the equation ${\Phi _E} = EA\cos \theta$ .

$\begin{array}{c}\\{\Phi _2} = EWH\cos {180^{\rm{o}}}\\\\ = - EWH\\\end{array}$

For the other faces:

For all the other faces of the cube the area vector is perpendicular to the electric field. Hence, the angle between the area vector and electric field is ${90^{\rm{o}}}$ .

Substitute ${\Phi _3}$ for ${\Phi _E}$ and ${90^{\rm{o}}}$ for $\theta$ in the equation ${\Phi _E} = EA\cos \theta$ .

$\begin{array}{c}\\{\Phi _3} = EA\cos {90^{\rm{o}}}\\\\ = 0\\\end{array}$

Here, ${\Phi _3}$ is the electric flux for the remaining faces of the box.

The net electric flux is given as,

${\Phi _{\rm{E}}} = {\Phi _1} + {\Phi _2} + {\Phi _3}$

Substitute $EWH$ for ${\Phi _1}$ , $- EWH$ for ${\Phi _2}$ , and 0 for ${\Phi _3}$ in the equation ${\Phi _{\rm{E}}} = {\Phi _1} + {\Phi _2} + {\Phi _3}$ .

$\begin{array}{c}\\{\Phi _{\rm{E}}} = EWH + \left( { - EWH} \right) + 0\\\\ = 0\\\end{array}$

(2)

The electric flux is given as,

${\Phi _E} = EA\cos \theta$

For the right face:

The area of the right face of the rectangular box is given as,

$A = WH$

Here, W is the width of the rectangular box and H is the height of the rectangular box.

The area vector for the right face is normal to the right face of the box and it is towards right. The electric field is towards left. Hence, the angle between the area vector and electric field is ${180^{\rm{o}}}$ .

Substitute ${\Phi _1}$ for ${\Phi _E}$ , $WH$ for A, and ${180^{\rm{o}}}$ for $\theta$ in the equation ${\Phi _E} = EA\cos \theta$ .

$\begin{array}{c}\\{\Phi _1} = EWH\cos {180^{\rm{o}}}\\\\ = - EWH\\\end{array}$

For the left face:

The area of the left face of the rectangular box is given as,

$A = WH$

Here, W is the width of the rectangular box and H is the height of the rectangular box.

The area vector for the left face is normal to the left face of the box and it is towards left. The electric field is also towards left. Hence, the angle between the area vector and electric field is ${0^{\rm{o}}}$ .

Substitute ${\Phi _2}$ for ${\Phi _E}$ , $WH$ for A, and ${0^{\rm{o}}}$ for $\theta$ in the equation ${\Phi _E} = EA\cos \theta$ .

$\begin{array}{c}\\{\Phi _2} = EWH\cos {0^{\rm{o}}}\\\\ = EWH\\\end{array}$

For the other faces:

For all the other faces of the cube the area vector is perpendicular to the electric field. Hence, the angle between the area vector and electric field is ${90^{\rm{o}}}$ .

Substitute ${\Phi _3}$ for ${\Phi _E}$ and ${90^{\rm{o}}}$ for $\theta$ in the equation ${\Phi _E} = EA\cos \theta$ .

$\begin{array}{c}\\{\Phi _3} = EA\cos {90^{\rm{o}}}\\\\ = 0\\\end{array}$

Here, ${\Phi _3}$ is the net electric flux for the remaining faces of the box.

The net electric flux is given as,

${\Phi _{\rm{E}}} = {\Phi _1} + {\Phi _2} + {\Phi _3}$

Substitute $- EWH$ for ${\Phi _1}$ , $EWH$ for ${\Phi _2}$ , and 0 for ${\Phi _3}$ in the equation ${\Phi _{\rm{E}}} = {\Phi _1} + {\Phi _2} + {\Phi _3}$ .

$\begin{array}{c}\\{\Phi _{\rm{E}}} = - EWH + EWH + 0\\\\ = 0\\\end{array}$

(3)

The electric flux is given as,

${\Phi _E} = EA\cos \theta$

For the right face:

The area of the right face of the rectangular box is given as,

$A = WH$

Here, W is the width of the rectangular box and H is the height of the rectangular box.

The area vector for the right face is normal to the right face of the box and it is towards right. The electric field is also towards right. Hence, the angle between the area vector and electric field is ${0^{\rm{o}}}$ .

Substitute ${\Phi _1}$ for ${\Phi _E}$ , $WH$ for A, and ${0^{\rm{o}}}$ for $\theta$ in the equation ${\Phi _E} = EA\cos \theta$ .

$\begin{array}{c}\\{\Phi _1} = EWH\cos {0^{\rm{o}}}\\\\ = EWH\\\end{array}$

For the left face:

The area of the left face of the rectangular box is given as,

$A = WH$

Here, W is the width of the rectangular box and H is the height of the rectangular box.

The area vector for the left face is normal to the left face of the box and it is towards left. The electric field is also towards left. Hence, the angle between the area vector and electric field is ${0^{\rm{o}}}$ .

Substitute ${\Phi _2}$ for ${\Phi _E}$ , $WH$ for A, and ${0^{\rm{o}}}$ for $\theta$ in the equation ${\Phi _E} = EA\cos \theta$ .

$\begin{array}{c}\\{\Phi _2} = EWH\cos {0^{\rm{o}}}\\\\ = EWH\\\end{array}$

For the other faces:

For all the other faces of the cube the area vector is perpendicular to the electric field. Hence, the angle between the area vector and electric field is ${90^{\rm{o}}}$ .

Substitute ${\Phi _3}$ for ${\Phi _E}$ and ${90^{\rm{o}}}$ for $\theta$ in the equation ${\Phi _E} = EA\cos \theta$ .

$\begin{array}{c}\\{\Phi _3} = EA\cos {90^{\rm{o}}}\\\\ = 0\\\end{array}$

Here, ${\Phi _3}$ is the net electric flux for the remaining faces of the box.

The net electric flux is given as,

${\Phi _{\rm{E}}} = {\Phi _1} + {\Phi _2} + {\Phi _3}$

Substitute $EWH$ for ${\Phi _1}$ , $EWH$ for ${\Phi _2}$ , and 0 for ${\Phi _3}$ in the equation ${\Phi _{\rm{E}}} = {\Phi _1} + {\Phi _2} + {\Phi _3}$ .

$\begin{array}{c}\\{\Phi _{\rm{E}}} = EWH + EWH + 0\\\\ = 2EWH\\\end{array}$

Ans: Part 1

The net electric flux out of the box is 0.