# Electric flux calculatio in case of a cube

1. Sep 13, 2015

### gracy

1. The problem statement, all variables and given/known data
the electric field components in the figure below are Ex=800 x^2 N/C ,Ey=Ez=0 calculate the flux through the cube
a=assume 0.1 m

2. Relevant equations

3. The attempt at a solution
here flux of cube =flux through ABCD
=800a^2 i^.a^2 i^
=800a^4
=800 × (0.1)^4
=8×10^2 ×10^-4
=8 ×10^-2 Nm^2/C
But I don't understand this step =800a^4
I want to know how to proceed from 800a^2 i^.a^2 i^.

2. Sep 13, 2015

### blue_leaf77

That's just a dot product between the E field on the ABCD plane, $800a^2 \hat{i}$, and the area vector of this plane, $a^2 \hat{i}$,
$$800a^2 \hat{i} \cdot a^2 \hat{i} = 800a^4 (\hat{i} \cdot \hat{i}) = 800a^4$$
with $\hat{i} \cdot \hat{i}=1$.

3. Sep 13, 2015

### gracy

Thanks again!
so do we take i/j/k separately?

4. Sep 13, 2015

### blue_leaf77

We don't normally call them to be "separately", $\hat{i}$, $\hat{j}$, and $\hat{k}$ are unit vectors along the x, y, and z directions, respectively. In Cartesian coordinate, these three directions are perpendicular to each other, so, if you take a dot product between any two vectors along any two of these three directions, it must equal zero. Remember the dot product formula between vectors $\mathbf{A}$ and $\mathbf{B}$, $\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos \alpha$, with $\alpha$ the angle between $\mathbf{A}$ and $\mathbf{B}$. The angles between $\hat{i}$, $\hat{j}$, and $\hat{k}$ are always $90^0$ hence the dot product between them must be zero.

5. Sep 18, 2015

### gracy

OK.One more doubt from this problem
WHY?
why can not it be flux through EFGH as well?
Here it is not mentioned that E negative x is also zero.

6. Sep 18, 2015

### blue_leaf77

I thought it was already known to you when you write
.
If I were to write the more complete expression of the total flux going through the cube, it will be
$$\oint \mathbf{E} \cdot d\mathbf{a} = \int_1 \mathbf{E_1} \cdot d\mathbf{a}_1 + \int_2 \mathbf{E}_2 \cdot d\mathbf{a}_2 + \ldots + \int_6 \mathbf{E}_6 \cdot d\mathbf{a}_6$$
where the integrals on the RHS account for the six surfaces of the cube. However since the E field is along the x direction, only two of those six integrals will survive upon the dot product with the area vector. Let's suppose that the surfaces ABCD and EFGH are the first and second integrals above, respectively. Then it becomes
$$\oint \mathbf{E} \cdot d\mathbf{a} = E(ABCD) a^2 - E(EFGH) a^2$$
What is the electric field in the EFGH plane?

7. Sep 20, 2015

### gracy

Why is there subtraction?because of i^.-i^ i.e 1mltiplied by cos 180 degrees i.e -1?

8. Sep 20, 2015

### gracy

Not given.

9. Sep 20, 2015

### blue_leaf77

That's why you have to calculate it. The field has the form of $E_x = 800x^2$, now what is $x$ equal to for the EFGH plane?

10. Sep 20, 2015

### gracy

is this right?

11. Sep 20, 2015

### blue_leaf77

Yes, that's correct.

12. Sep 20, 2015

### gracy

it's zero.So E=0 for EFGH plane.And therefore
flux of cube =flux through ABCD.Right?

13. Sep 20, 2015

### blue_leaf77

Right.

14. Sep 20, 2015

Thanks .