Flux of the electric field that crosses the faces of a cube

[Guillem_dlc]

Homework Statement

Calculate the flux of the electric field that crosses a cube with vertices at the points (coordinated in meters): $A(0,0,0)$, $B(4,0,0)$, $C(4,0,4)$, $D(0,0,4)$, $E(0,4,4)$, $F(0,4,0)$, $G(4,4,0)$, $H(4,4,4)$ located in a region of space where there is an electric field:
a) $\vec{E}=10^4\, \widehat{i}\, (\textrm{N}/\textrm{C})$
b) $\vec{E}=300x\, \widehat{i}\, (\textrm{N}/\textrm{C})$
c) $\vec{E}=60x^2\, \widehat{i}-1000y\, \widehat{j}+3000\, \widehat{k}\, (\textrm{N}/\textrm{C})$

Answers: a) $\phi=0$, b) $\phi=1,9\cdot 10^4\, \textrm{Nm}^2/\textrm{C}$, c) $\phi=-4,9\cdot 10^4\, \textrm{Nm}^2/\textrm{C}$.

Relevant Equations

$\phi = \vec{E}\cdot \vec{S}$

a) $$\phi_T=\phi_F-\phi_I=10^4\cdot 4\cdot 4-10^4\cdot 4\cdot 4=0\, \textrm{Nm}^2/\textrm{C}$$

b) $$\phi_F=\underbrace{300\cdot 4}_{\vec{E}}\cdot \underbrace{4\cdot 4}_{\textrm{area}}=19200\, \textrm{Nm}^2/\textrm{C}$$
$$\phi_0 = 300\cdot 0\cdot 4\cdot 4=0\, \textrm{Nm}^2/\textrm{C}$$
Then,
$$\phi_T=\phi_F-\phi_0=19200\, \textrm{Nm}^2/\textrm{C}$$

c) $x$ axis: $$E_x=6x^2\, \widehat{i}\rightarrow \phi_x=\phi_F-\phi_0=60\cdot 16\cdot 4\cdot 4-0=15360\, \textrm{Nm}^2/\textrm{C}$$
$y$ axis: $$E_y=1000y\rightarrow \phi_y=\phi_F-\phi_0=0-1000\cdot 4\cdot 4\cdot 4=-64000\, \textrm{Nm}^2/\textrm{C}$$
$z$ axis: $$E_x=3000\rightarrow \phi_z=\phi_F-\phi_0\rightarrow \phi_F=\phi_0\rightarrow \phi_z=0\, \textrm{Nm}^2/\textrm{C}$$

How should I do that in part c)? I would do the module of this to calculate the flux but it doesn't give the answer.

 

[TSny]

a) $$\phi_T=\phi_F-\phi_I=10^4\cdot 4\cdot 4-10^4\cdot 4\cdot 4=0\, \textrm{Nm}^2/\textrm{C}$$

I'm not sure what the subscripts $F$ and $I$ refer to. I think it would be better to write the total flux as the sum of the fluxes through each face. Thus,

$$\phi_T=\phi_F+\phi_I=10^4\cdot 4\cdot 4+(-10^4\cdot 4\cdot 4)=0\, \textrm{Nm}^2/\textrm{C}$$ This gives the same answer that you got. But conceptually, the total flux through all of the faces is obtained by adding the fluxes for each face. The flux through an individual face could be positive, negative, or zero.

c) $x$ axis: $$E_x=6x^2\, \widehat{i}\rightarrow \phi_x=\phi_F-\phi_0=60\cdot 16\cdot 4\cdot 4-0=15360\, \textrm{Nm}^2/\textrm{C}$$
$y$ axis: $$E_y=1000y\rightarrow \phi_y=\phi_F-\phi_0=0-1000\cdot 4\cdot 4\cdot 4=-64000\, \textrm{Nm}^2/\textrm{C}$$
$z$ axis: $$E_x=3000\rightarrow \phi_z=\phi_F-\phi_0\rightarrow \phi_F=\phi_0\rightarrow \phi_z=0\, \textrm{Nm}^2/\textrm{C}$$

How should I do that in part c)? I would do the module of this to calculate the flux but it doesn't give the answer.

Wouldn't $\phi_{\rm total} = \phi_x+\phi_y + \phi_z$?

 

[Guillem_dlc]

Wouldn't $\phi_{\rm total} = \phi_x+\phi_y + \phi_z$?

Would that be the total flow not? Add each component but with this "wouldn't" I don't know what you mean?

 

[TSny]

You have found the flux, $\phi_x,$ through the surface of the cube due to the $x$-component of the field $E_x$.

Likewise for $\phi_y$ and $\phi_z$ due to $E_y$ and $E_z$.

So how would you calculate the total flux through the surface of the cube from $\phi_x$, $\phi_y$, and $\phi_z$?

Earlier, you said

I would do the module of this to calculate the flux but it doesn't give the answer.

Can you show how you would "do the module"?

 

[Guillem_dlc]

Can you show how you would "do the module"?

Adding the result already comes.

I did this:
$$\phi =\sqrt{\phi_x^2+\phi_y^2+\phi_z^2}$$

 

[TSny]

Adding the result already comes.

I'm not sure what you mean here.

I did this:
$$\phi =\sqrt{\phi_x^2+\phi_y^2+\phi_z^2}$$

Flux is not a vector quantity. It is a scalar quantity.

In your calculation, $\phi_x$ represents the flux through the surface of the cube due to just the x-component of the electric field. But $\phi_x$ is not the x-component of some vector.

 

[Guillem_dlc]

I'm not sure what you mean here.Flux is not a vector quantity. It is a scalar quantity.

In your calculation, $\phi_x$ represents the flux through the surface of the cube due to just the x-component of the electric field. But $\phi_x$ is not the x-component of some vector.

I mean that $\phi=\phi_x+\phi_y+\phi_z$ gives the correct result.

Then the potential wouldn't be, right?

 

[TSny]

Then the potential wouldn't be, right?

I'm not sure what you are asking here. What potential are you referring to?

If you saying that electric potential is not a vector quantity, then you are right.