Finding the electric flux through the right face, confused on integration

[mr_coffee]

I'm having troubles understanding what's going on here, with the integration. Here is the integral through the right face of the cube.
I don't know how to insert all the fancey symbols, so here is my key:
S = integral symbol
Flux = electric flux symbol, omega or somthing, a circle with a cross down the middle.
i = vector i in x-axis
j = vector j in y-axis
. means the dot product.
Given: A nonuniform electric field given by E = 3.0xi + 3.0j pierces the gaussian cube. x = 3.0m.

Flux = S (E).(dA) = S (3.0xi + 4.0j).(dAi)

= S [(3.0x)(dA)i.i + (4.0)(dA)j.i] //whats goin on here? are they just distrubting the dA? Why are they allowed to sperate the vector i from dA?

= S (3.0x dA + 0) = 3.0 S x dA //why is i now 0? wouldn't it be cos(0) = 1? or how do u figure out where the electric field is pointing with the equation: 3.0xi + 4.0j.

= 3.0 S (3.0)dA = 9.0 S dA.

How do you insert symbolic symbols so my future posts won't looks this messy? Thanks. Picture is attached.

 

[HallsofIvy]

According to your attachment, the "right face" of the cube is the plane x= 3.0 and the (outward) unit normal is i so the dA= dydz i. Therefore
(3.0xi+ 4.0j). dA= 3.0x dydz= 3.0 x dA where dA= dydz.

i did not become "0" the dot product of two vectors is a scalar (number).
(3.0xi).(i)= 3.0x, of course.

 

[mr_coffee]

Thanks for the responce but I'm still confused... how do you go from, dA = dydz i.
then you said dA = 3.0x
dydz = 3.0 x dA...You didn't take the derivative of anything did you?
^is this the variable x or meaning multiplcation?

 

[Pyrrhus]

Halls, simply did the dot product, the result was 3x dA, then if you look at the picture x = 3, so 9*A, should be the solution.

 

[Doc Al]

The only component of the field that contributes to the flux through a side is the component perpendicular to that side. For the right side of the cube, that perpendicular direction is the \hat i direction. The component of the field in that direction is 3.0 x \hat i; at x = 3 m, that component equals 9.0 \hat i (in units of N/C). Since the field is constant over the area of the right side, no integration is needed, just flux = E times Area.

 

[mr_coffee]

ohhh i think i finally get it... so because the y component of the electric field doesn't matter (4.0j), you can just discard it and only worry about the 3.0xi. and because x = 3, you end up with 9.0i. So really is i just telling the direction of the vector? you can just discard it? I'm still confused on one issue though. \zeta [(3.0x)(dA)\hat i \bullet \hat i] You said you took the dot product, if A is pointing to the right, and also the electric field is point right, wouldn't that be cos(0) = 1? how did they get 0? \zeta [(3.0x)(dA) + 0] Sorry I'm really really rusty on vectors! that zeta is suppose to be an integral sign, i can't find the integral on the latex guide.

 

[Pyrrhus]

It looks like you don't know this:

\vec{i} \cdot \vec{i} = \vec{j} \cdot \vec{j} = \vec{k} \cdot \vec{k} = 1

\vec{i} \cdot \vec{j} = \vec{j} \cdot \vec{k} = \vec{i} \cdot \vec{k} = 0

Ah and the integral is

\int

 

[mr_coffee]

ahhh! thanks so much, I had no idea that property even existed. Damn luckly I'm not going to be a mechanical engineeer.