Calculating Electric Flux Through Cube's Left Face

• tony873004
In summary, A cube of side L=2.0 m is centered at the origin, with the coordinate axes perpendicular to its faces. The flux of the electric field through each face of the cube can be calculated using the equation Phi = <E * n> * Area. By finding the unit vector normal to each face and using the given values for the electric field, the flux for each face can be calculated. The total flux for each face is as follows: Left face: -60 Nm^2/C, Right face: 60 Nm^2/C, Top face: 156 Nm^2/C, Bottom face: -156 Nm^2/C, Front face: -108 Nm^2/C,
tony873004
Gold Member
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 ?

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

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.

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

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

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

1. How do you calculate electric flux through a cube's left face?

To calculate electric flux through a cube's left face, you need to find the magnitude and direction of the electric field passing through the face, as well as the area of the face. Then, you can use the formula Φ = E * A * cos(θ) to calculate the electric flux, where Φ is the electric flux, E is the electric field, A is the area, and θ is the angle between the electric field and the normal vector to the face.

2. What is the significance of calculating electric flux through a cube's left face?

Calculating electric flux through a cube's left face allows us to quantify the amount of electric field passing through that particular face. This is important in understanding how electric fields behave and interact with different surfaces and materials.

3. What are the units of electric flux?

The units of electric flux are newtons per meter squared (N/m²) or volts per meter (V/m).

4. How does the angle between the electric field and the normal vector affect the electric flux through a cube's left face?

The angle between the electric field and the normal vector affects the electric flux through a cube's left face because it determines the component of the electric field that is perpendicular to the face. The greater the angle, the smaller the component of the electric field, and therefore, the smaller the electric flux through the face.

5. Can the electric flux through a cube's left face be negative?

Yes, the electric flux through a cube's left face can be negative. This occurs when the electric field is pointing in the opposite direction of the normal vector to the face. In this case, the angle between the electric field and the normal vector is 180 degrees, and the cosine of 180 degrees is -1, resulting in a negative value for the electric flux.

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