Electric flux calculatio in case of a cube

[gracy]

Homework Statement

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

Homework Equations

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^.

 

[blue_leaf77]

I want to know how to proceed from 800a^2 i^.a^2 i^.

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 $\hat{i} \cdot \hat{i}=1$.

 

[gracy]

Thanks again!

800a2i^⋅a2i^=800a4(i^⋅i^)

so do we take i/j/k separately?

 

[blue_leaf77]

so do we take i/j/k separately?

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.

 

[gracy]

OK.One more doubt from this problem

flux of cube =flux through ABCD

WHY?
why can not it be flux through EFGH as well?

Ex=800 x^2 N/C ,Ey=Ez=0

Here it is not mentioned that E negative x is also zero.

 

[blue_leaf77]

I thought it was already known to you when you write

here flux of cube =flux through ABCD

.
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?

 

[gracy]

∮E⋅da=E(ABCD)a2−E(EFGH)a2

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

 

[gracy]

What is the electric field in the EFGH plane?

Not given.

 

[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?

 

[gracy]

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

is this right?

 

[blue_leaf77]

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

Yes, that's correct.

 

[gracy]

now what is xx equal to for the EFGH plane?

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

 

[blue_leaf77]

it's zero.So E=0 for EFGH plane

Right.

 

[gracy]

Thanks .