Question

The cube in the figure has edge length 1.40 m and is oriented as shown in a region of uniform electric field. Find the electric flux through the right face for the following electric fields, given in newtons per coulomb.

(a) 2.00i (I need to find the answer in N*m^2/C)

(b)-6.00j(I need to find the answer in N*m^2/C)

(c)-2.00i+4.00k(I need to find the answer in N*m^2/C)

(d) What is the total flux through the cube for each of these fields?

for (a)	N · m2/C
for (b)	N · m2/C
for (c)	N · m2/C

Answer 1

Concepts and reason

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

First, determine the area of each face by squaring the edge length. Then, to determine the electric flux through the right face find the dot product between the area vector and the electric field vector. Finally, determine the net electric flux by adding the net 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,

${\Phi _E} = \vec E \cdot \vec A$

Here, E is the electric field vector and A is the area vector which points perpendicular to the surface.

The magnitude of area vector is the area of the surface.

A vector is represented as,

$\vec A = {A_x}\hat i + {A_y}\hat j + {A_z}\hat k$

Here, ${A_x}$ is the x-component of the vector, ${A_y}$ is the y component of the vector, ${A_z}$ is the z component of the vector, $\hat i$is the unit vector along the x direction, $\hat j$ is the unit vector along the y direction, and $\hat k$ is the unit vector along the z direction.

The dot product of any unit vector with any other unit vector is zero. That is,

$\begin{array}{c}\\\hat i \cdot \hat j = 0\\\\\hat j \cdot \hat k = 0\\\\\hat k \cdot \hat i = 0\\\end{array}$

The dot product of any unit vector with itself is one. That is,

$\begin{array}{c}\\\hat i \cdot \hat i = 1\\\\\hat j \cdot \hat j = 1\\\\\hat k \cdot \hat k = 1\\\end{array}$

(a)

The area of each face of the cube is given as,

$A = {l^2}$

Here, l is the edge length of the face.

Substitute 1.40 m for l in the above equation.

$\begin{array}{c}\\A = {\left( {1.40{\rm{ m}}} \right)^2}\\\\ = 1.96{\rm{ }}{{\rm{m}}^2}\\\end{array}$

The area vector for the right face is given as,

$\vec A = \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat j$

The electric field is given as,

$\vec E = \left( {2.00{\rm{ N/C}}} \right)\hat i$

Use the expression of electric flux.

Substitute $\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat j$ for $\vec A$ and $\left( {2.00{\rm{ N/C}}} \right)\hat i$ for $\vec E$ in the equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _E} = \left( {2.00{\rm{ N/C}}} \right)\hat i \cdot \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat j\\\\ = \left( {2.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\hat i \cdot \hat j} \right)\\\end{array}$

Substitute 0 for $\hat i \cdot \hat j$ in the above expression of electric flux.

$\begin{array}{c}\\{\Phi _E} = \left( {2.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( 0 \right)\\\\ = 0{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}\\\end{array}$

(b)

Use the expression of electric flux.

Substitute $\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat j$ for $\vec A$ and $\left( { - 6.00{\rm{ N/C}}} \right)\hat j$ for $\vec E$ in the equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _E} = \left( { - 6.00{\rm{ N/C}}} \right)\hat j \cdot \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat j\\\\ = - \left( {11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right)\left( {\hat j \cdot \hat j} \right)\\\end{array}$

Substitute 1 for $\hat j \cdot \hat j$ in the above expression of electric flux.

$\begin{array}{c}\\{\Phi _E} = - \left( {11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right)\left( 1 \right)\\\\ = - 11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}\\\end{array}$

(c)

Use the expression of electric flux.

Substitute $\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat j$ for $\vec A$ and $\left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}}$ for $\vec E$ in equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _E} = \left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}} \cdot \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat j\\\\ = \left( { - 2.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\hat i \cdot \hat j} \right) + \left( {4.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\hat k \cdot \hat j} \right)\\\end{array}$

Substitute 0 for $\hat i \cdot \hat j$ and $\hat k \cdot \hat j$ in the above expression of electric flux.

$\begin{array}{c}\\{\Phi _E} = \left( { - 2.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( 0 \right) + \left( {4.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( 0 \right)\\\\ = 0{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}\\\end{array}$

(d.1)

The electric field is given as,

$\vec E = \left( {2.00{\rm{ N/C}}} \right)\hat i$

For the front face:

The area vector pointing in the x direction is given as,

$\vec A = \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat i$

Substitute $\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat i$ for $\vec A$ and $\left( {2.00{\rm{ N/C}}} \right)\hat i$ for $\vec E$ in the equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _1} = \left( {2.00{\rm{ N/C}}} \right)\hat i \cdot \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat i\\\\ = \left( {2.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\hat i \cdot \hat i} \right)\\\end{array}$

Substitute 1 for $\hat i \cdot \hat i$ in the above expression of electric flux.

$\begin{array}{c}\\{\Phi _1} = \left( {2.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( 1 \right)\\\\ = 3.92{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}\\\end{array}$

For the back face:

The area vector pointing in the x direction is given as,

$\vec A = \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( { - \hat i} \right)$

Substitute $\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( { - \hat i} \right)$ for $\vec A$ and $\left( {2.00{\rm{ N/C}}} \right)\hat i$ for $\vec E$ in the equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _2} = \left( {2.00{\rm{ N/C}}} \right)\hat i \cdot \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( { - \hat i} \right)\\\\ = - \left( {2.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\hat i \cdot \hat i} \right)\\\end{array}$

Substitute 1 for $\hat i \cdot \hat i$ in the above expression of electric flux.

$\begin{array}{c}\\{\Phi _2} = - \left( {2.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( 1 \right)\\\\ = - 3.92{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}\\\end{array}$

For all the other faces of the cube the area vector is perpendicular to the electric field and the dot product of any unit vector with any other unit vector is zero. Hence, the electric flux for all the other faces is zero.

The net electric flux is given as,

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

Substitute $3.92{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}$ for ${\Phi _1}$ and$- 3.92{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}$ for${\Phi _2}$ in the equation${\Phi _{\rm{E}}} = {\Phi _1} + {\Phi _2}$.

$\begin{array}{c}\\{\Phi _{\rm{E}}} = 3.92{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}} + \left( { - 3.92{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right)\\\\ = 0{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}\\\end{array}$

(d.2)

The electric field is given as,

$\vec E = \left( { - 6.00{\rm{ N/C}}} \right)\hat j$

For the right face:

The area vector for the right face is given as,

$\vec A = \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat j$

Substitute $\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat j$ for $\vec A$ and $\left( { - 6.00{\rm{ N/C}}} \right)\hat j$ for $\vec E$ in the equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _1} = \left( { - 6.00{\rm{ N/C}}} \right)\hat j \cdot \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat j\\\\ = - \left( {11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right)\left( {\hat j \cdot \hat j} \right)\\\end{array}$

Substitute 1 for $\hat j \cdot \hat j$ in the above expression of electric flux.

$\begin{array}{c}\\{\Phi _1} = - \left( {11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right)\left( 1 \right)\\\\ = - 11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}\\\end{array}$

For the left face:

The area vector pointing in the x direction is given as,

$\vec A = \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( { - \hat j} \right)$

Substitute $\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( { - \hat j} \right)$ for $\vec A$ and $\left( { - 6.00{\rm{ N/C}}} \right)\hat j$ for $\vec E$ in the equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _2} = \left( { - 6.00{\rm{ N/C}}} \right)\hat j \cdot \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( { - \hat j} \right)\\\\ = \left( {11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right)\left( {\hat j \cdot \hat j} \right)\\\end{array}$

Substitute 1 for $\hat j \cdot \hat j$ in the above expression of electric flux.

$\begin{array}{c}\\{\Phi _2} = \left( {11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right)\left( 1 \right)\\\\ = 11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}\\\end{array}$

For all the other faces of the cube the area vector is perpendicular to the electric field and the dot product of any unit vector with any other unit vector is zero. Hence, the electric flux for all the other faces is zero.

The net electric flux is given as,

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

Substitute $11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}$ for ${\Phi _1}$ and$- 11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}$ for${\Phi _2}$ in the equation${\Phi _{\rm{E}}} = {\Phi _1} + {\Phi _2}$.

$\begin{array}{c}\\{\Phi _{\rm{E}}} = 11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}} + \left( { - 11.8\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right)\\\\ = 0{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}\\\end{array}$

(d.3)

The electric field is given as,

$E = \left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}}$

For the front face:

The area vector pointing in the x direction is given as,

$\vec A = \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat i$

Substitute $\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat i$ for $\vec A$, $\left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}}$ for $\vec E$, and 1 for $\hat i \cdot \hat i$ in the equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _1} = \left( {\left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}}} \right) \cdot \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat i\\\\ = \left( { - 2.00{\rm{ N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\hat i \cdot \hat i} \right)\\\\ = - 3.92{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\\\end{array}$

For the back face:

The area vector pointing in the x direction is given as,

$\vec A = \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( { - \hat i} \right)$

Substitute $- \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat i$ for $\vec A$, $\left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}}$ for $\vec E$, and 1 for $\hat i \cdot \hat i$ in the equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _2} = \left( {\left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}}} \right) \cdot \left( { - 1.96{\rm{ }}{{\rm{m}}^2}} \right)\hat i\\\\ = \left( { - 2.00{\rm{ N/C}}} \right)\left( { - 1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\hat i \cdot \hat i} \right)\\\\ = 3.92{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\\\end{array}$

For the upper face:

The area vector pointing in the z direction is given as,

$\vec A = \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\widehat k} \right)$

Substitute $\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\widehat k} \right)$ for $\vec A$, $\left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}}$ for $\vec E$, and 1 for $\widehat k \cdot \widehat k$ in the equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _3} = \left( {\left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}}} \right) \cdot \left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\widehat k\\\\ = \left( {{\rm{4}}{\rm{.00 N/C}}} \right)\left( {1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\widehat k \cdot \widehat k} \right)\\\\ = 7.84{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\\\end{array}$

For the lower face:

The area vector pointing in the z direction is given as,

$\vec A = \left( { - 1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\widehat k} \right)$

Substitute $\left( { - 1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\widehat k} \right)$ for $\vec A$, $\left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}}$ for $\vec E$, and 1 for $\widehat k \cdot \widehat k$ in the equation ${\Phi _E} = \vec E \cdot \vec A$.

$\begin{array}{c}\\{\Phi _4} = \left( {\left( { - 2.00\hat i + 4.00\hat k} \right){\rm{ N/C}}} \right) \cdot \left( { - 1.96{\rm{ }}{{\rm{m}}^2}} \right)\widehat k\\\\ = \left( {{\rm{4}}{\rm{.00 N/C}}} \right)\left( { - 1.96{\rm{ }}{{\rm{m}}^2}} \right)\left( {\widehat k \cdot \widehat k} \right)\\\\ = - 7.84{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\\\end{array}$

For all the other faces of the cube the area vector is perpendicular to the electric field and the dot product of any unit vector with any other unit vector is zero. Hence, the electric flux for all the other faces is zero. The flux due to left and right surfaces will be zero.

$\begin{array}{l}\\{\Phi _5} = 0\\\\{\Phi _6} = 0\\\end{array}$

The net electric flux is given as,

${\Phi _{\rm{E}}} = {\Phi _1} + {\Phi _2} + {\Phi _3} + {\Phi _4} + {\Phi _5} + {\Phi _6}$

Substitute $- 3.92{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}$ for ${\Phi _1}$ and$3.92{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}$ for${\Phi _2}$, 0 for ${\Phi _5}$ and ${\Phi _6}$,$7.84{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}$for ${\Phi _3}$, and -$7.84{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}$for ${\Phi _4}$ in the equation${\Phi _{\rm{E}}} = {\Phi _1} + {\Phi _2} + {\Phi _3} + {\Phi _4} + {\Phi _5} + {\Phi _6}$.

$\begin{array}{c}\\{\Phi _{\rm{E}}} = \left( { - 3.92{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right) + \left( {3.92{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right) + \left( {7.84{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) + \left( { - 7.84{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) + 0 + 0\\\\ = 0\\\end{array}$

Ans: Part a

The electric flux through the right face for the electric field $\left( {2.00} \right)\hat i$ is $0{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}$.