# Homework Help: Electric Flux

1. Feb 7, 2008

### tony873004

A cube of side L=2.0 m is centered at the origin, with the coordinate axes perpendicular to its faces. Find the flux of the electric field $$\overrightarrow E = \left( {15{\rm{ N/C}}} \right){\rm{\hat i + }}\left( {{\rm{27 N/C}}} \right){\rm{\hat j + }}\left( {{\rm{39 N/C}}} \right){\rm{\hat k}}$$ through each face of the cube.

Let's just concentrate on the cube's left face. From class notes $$\Phi _{{\rm{left}}} = \left\langle {\overrightarrow E \cdot {\rm{\hat n}}} \right\rangle {\rm{ Area}}$$

$$\begin{array}{l} \Phi _{{\rm{left}}} = \left\langle {\overrightarrow E \cdot {\rm{\hat n}}} \right\rangle {\rm{ Area = }}\left( {15{\rm{ N/C}} \cdot 1} \right){\rm{ + }}\left( {{\rm{27 N/C}} \cdot 0} \right){\rm{ + }}\left( {{\rm{39 N/C}} \cdot 0} \right)\left( {2{\rm{m}}} \right)^2 {\rm{ = }}\left( {15{\rm{ N/C}}} \right){\rm{ + }}\left( 0 \right){\rm{ + }}\left( 0 \right)\left( {2{\rm{m}}} \right)^2 = \\ \\ \left( {15{\rm{ N/C}}} \right)4{\rm{m}}^2 = 60{\rm{Nm}}^{\rm{2}} /{\rm{C}} \\ \end{array}$$

But the units are wrong. Shouldn't they be C/m2 ?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 7, 2008

### rl.bhat

Shouldn't they be C/m2 ?
This quantity defines surface charge density, not the flux.

3. Feb 7, 2008

### Dick

Integrated flux is N*m^2/C, right. Gauss' Law says integrated flux*epsilon_0=Q. The units of epsilon_0 are C^2/(N*m^2). So Q comes out to be coulombs. Everything looks ok to me.

4. Feb 7, 2008

### tony873004

Thanks. I'm glad I got the units right then.
Can someone look over this and tell me if I did it right? Did I get the signs right? It's the first time I've done this type of problem, and the answer isn't in the back of the book.

Although it didn't ask for total flux, should this add to 0?

Thanks!
$$\begin{array}{l} {\rm{\hat n}}_{{\rm{right}}} {\rm{ = \hat i}} \\ {\rm{\hat n}}_{{\rm{left}}} {\rm{ = }} - {\rm{\hat i}} \\ {\rm{\hat n}}_{{\rm{top}}} {\rm{ = \hat k}} \\ {\rm{\hat n}}_{{\rm{bottom}}} {\rm{ = }} - {\rm{\hat k}} \\ {\rm{\hat n}}_{{\rm{back}}} {\rm{ = \hat j}} \\ {\rm{\hat n}}_{{\rm{front}}} {\rm{ = }} - {\rm{\hat j}} \\ \\ \end{array}$$

$$\begin{array}{l} \Phi _{{\rm{left}}} = \left\langle {\overrightarrow E \cdot {\rm{\hat n}}} \right\rangle {\rm{ Area = }}\left( {15{\rm{ N/C}} \cdot 1} \right){\rm{ + }}\left( {{\rm{27 N/C}} \cdot 0} \right){\rm{ + }}\left( {{\rm{39 N/C}} \cdot 0} \right)\left( {2{\rm{m}}} \right)^2 {\rm{ = }}\left( {15{\rm{ N/C}}} \right){\rm{ + }}\left( 0 \right){\rm{ + }}\left( 0 \right)\left( {2{\rm{m}}} \right)^2 \\ \,\,\,\,\,\,\,\,\,\, = \,\,\left( {15{\rm{ N/C}}} \right)4{\rm{m}}^2 = - 60\,{\rm{Nm}}^{\rm{2}} /{\rm{C}} \\ \\ \Phi _{{\rm{right}}} = \, - \Phi _{{\rm{left}}} = 60\,{\rm{Nm}}^{\rm{2}} {\rm{/C}} \\ \\ \Phi _{{\rm{top}}} = \left\langle {\overrightarrow E \cdot {\rm{\hat n}}} \right\rangle {\rm{ Area = }}\left( {15{\rm{ N/C}} \cdot 0} \right){\rm{ + }}\left( {{\rm{27 N/C}} \cdot 0} \right){\rm{ + }}\left( {{\rm{39 N/C}} \cdot 1} \right)\left( {2{\rm{m}}} \right)^2 {\rm{ = }}\left( {39{\rm{ N/C}}} \right){\rm{ + }}\left( 0 \right){\rm{ + }}\left( 0 \right)\left( {2{\rm{m}}} \right)^2 \\ \,\,\,\,\,\,\,\,\,\, = \,\,\left( {39{\rm{ N/C}}} \right)4{\rm{m}}^2 = 156\,{\rm{Nm}}^{\rm{2}} /{\rm{C}} \\ \\ \Phi _{{\rm{bottom}}} = - \Phi _{{\rm{top}}} = - 156\,{\rm{Nm}}^{\rm{2}} /{\rm{C}} \\ \\ \Phi _{{\rm{front}}} = \left\langle {\overrightarrow E \cdot {\rm{\hat n}}} \right\rangle {\rm{ Area = }}\left( {15{\rm{ N/C}} \cdot 0} \right){\rm{ + }}\left( {{\rm{27 N/C}} \cdot - {\rm{\hat j}}} \right){\rm{ + }}\left( {{\rm{39 N/C}} \cdot 0} \right)\left( {2{\rm{m}}} \right)^2 {\rm{ = }}\left( {27{\rm{ N/C}}} \right){\rm{ + }}\left( 0 \right){\rm{ + }}\left( 0 \right)\left( {2{\rm{m}}} \right)^2 \\ \,\,\,\,\,\,\,\,\,\,\,\, = \,\,\left( {27{\rm{ N/C}}} \right)4{\rm{m}}^2 = - 108\,{\rm{Nm}}^{\rm{2}} /{\rm{C}} \\ \\ \Phi _{{\rm{back}}} = - \Phi _{{\rm{front}}} = 108\,{\rm{Nm}}^{\rm{2}} /{\rm{C}} \\ \end{array}$$

5. Feb 7, 2008

### Dick

Sure it adds to zero. There is no charge in the cube. Otherwise the E field wouldn't be constant. Right?