The Electric Flux on all sides of a cube

In summary: And I think it would be helpful to see what the electric field looks like in order to solve this equation.
  • #1
ecsx00
14
0

Homework Statement


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Homework Equations


[itex]\phi[/itex] = [itex]\vec{E}[/itex] * A * d
phi is the electric flux
E is the Electric field from vector form
A is the area and A = [itex]\pi[/itex][itex]r^{2}[/itex]
d is the distance on the axis (x, y, or z)


The Attempt at a Solution


s1 = s3 = 0 *Since they are along the y-axis and there is no y-vector
s2 = 1.14e-1
s4 = - 1.14e-1
s5 = 2.69e-1
s6 = -2.69e-1

I don't know why I am getting these wrong but you need to arrange them from s1-s6 on the answer when you submit them and it tells me to check my sign on s5. Changing it to negative doesn't help.
I'll recheck my calculations again and try a different method. I probably forgot to put " [itex]\pi[/itex]" into the calculations but I will check on that.

Am I doing this right? Or is the equation to solve this problem totally different and I am off.
Pointing out where I went wrong helps, please do not post the answer as I just need a hint or tip.
Thank you!
 
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  • #2
Area for a square is the length times width not [itex]\pi r^2[/itex] .
You need to consider that flux is dependent on angle and that not all the surfaces are perpendicular to the electric field.
Also what is the deal with the distance component in you're equation for flux?
 
  • #3
I'd add as a hint: sketch the field.
 
  • #4
Bhumble said:
Area for a square is the length times width not [itex]\pi r^2[/itex] .
You need to consider that flux is dependent on angle and that not all the surfaces are perpendicular to the electric field.
Also what is the deal with the distance component in you're equation for flux?

I really have no clue. This equation is from the book directly.
 
  • #5
[itex]\varphi[/itex] = [itex]\vec{E}[/itex][itex]L^{2}[/itex]cos(θ)
or
[itex]\varphi[/itex] = -[itex]\vec{E}[/itex][itex]L^{2}[/itex]cos(θ)
Where E is the electric field from the [itex]\hat{n}[/itex] vector.
θ is either 90° or 180°

Is that the equation I am looking for?
 
Last edited:
  • #6
Instead of trying to guess which equation to apply - why don't you start with what "flux" actually means? And seriously: sketch the field.
 
  • #7
Were you able to figure it out?
That equation looks right. Like Simon said though you should make sure to understand what flux means and you'll know when to apply the appropriate equations.
 
  • #8
Use $$\oint \vec{E} \cdot \vec{dA}$$ What is E and dA? What is the value of the electric field along S5?
 
  • #9
Thanks. I solved it after looking at a few examples that people had on this site and eventually figured out how they got to the solution. I wouldn't guess it.
 
  • #10
Yeah - guessing seldom works without understanding.
 

FAQ: The Electric Flux on all sides of a cube

1. What is the electric flux on all sides of a cube?

The electric flux on all sides of a cube is the measure of the total electric field passing through the surface of the cube. It is a scalar quantity and is represented by the symbol Φ.

2. How is the electric flux on all sides of a cube calculated?

The electric flux on all sides of a cube is calculated by taking the dot product of the electric field and the surface area vector of each side of the cube. The flux for each side is then summed up to get the total electric flux on all sides.

3. What is the unit of measurement for electric flux?

The unit of measurement for electric flux is volts per meter (V/m) in SI units. It can also be expressed in other units such as newton meters squared per coulomb (N·m²/C) or joules per coulomb (J/C).

4. How does the electric flux on all sides of a cube relate to the charge enclosed within the cube?

The electric flux on all sides of a cube is directly proportional to the charge enclosed within the cube. This means that as the charge enclosed increases, the electric flux also increases.

5. What is the significance of calculating the electric flux on all sides of a cube?

Calculating the electric flux on all sides of a cube helps in understanding the flow of electric field through a closed surface. It also helps in determining the charge enclosed within the surface and in solving various problems related to electric fields and charges.

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