Finding the electric flux through the right face, confused on integration!by mr_coffee Tags: confused, electric, face, flux, integration 

#1
Jul305, 05:52 PM

P: 1,629

I'm having troubles understanding whats 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 xaxis j = vector j in yaxis . means the dot product. Given: A nonuniform electric feild 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 feild 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. 



#2
Jul305, 06:24 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

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. 



#3
Jul405, 05:06 PM

P: 1,629

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? 



#4
Jul405, 05:18 PM

HW Helper
P: 2,280

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




#5
Jul405, 05:20 PM

Mentor
P: 40,889

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 [itex]\hat i [/itex] direction. The component of the field in that direction is [itex]3.0 x \hat i[/itex]; at x = 3 m, that component equals [itex]9.0 \hat i[/itex] (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.




#6
Jul405, 08:49 PM

P: 1,629

ohhh i think i finally get it... so because the y component of the electric feild 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. [itex] \zeta [(3.0x)(dA)\hat i \bullet \hat i][/itex] You said you took the dot product, if A is pointing to the right, and also the electric feild is point right, wouldn't that be cos(0) = 1? how did they get 0? [itex] \zeta [(3.0x)(dA) + 0][/itex] 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.




#7
Jul405, 08:52 PM

HW Helper
P: 2,280

It looks like you don't know this:
[tex] \vec{i} \cdot \vec{i} = \vec{j} \cdot \vec{j} = \vec{k} \cdot \vec{k} = 1 [/tex] [tex] \vec{i} \cdot \vec{j} = \vec{j} \cdot \vec{k} = \vec{i} \cdot \vec{k} = 0 [/tex] Ah and the integral is [tex] \int [/tex] 



#8
Jul405, 10:42 PM

P: 1,629

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



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