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

In summary, according to the attached picture, the electric field is constant over the right side of the cube, so no integration is necessary. The only component of the field that contributes to the flux through a side is the component perpendicular to that side.
  • #1
mr_coffee
1,629
1
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.
 

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  • #2
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
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
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
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
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. [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 field 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! :bugeye: that zeta is suppose to be an integral sign, i can't find the integral on the latex guide.
 
  • #7
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]
 
Last edited:
  • #8
ahhh! thanks so much, I had no idea that property even existed. Damn luckly I'm not going to be a mechanical engineeer.
 

1. What is electric flux?

Electric flux is a measure of the amount of electric field passing through a given area. It is represented by the symbol Φ and is given in units of volts per meter (V/m).

2. What is the right face in the context of finding electric flux?

In the context of finding electric flux, the right face refers to a specific surface or area where the electric field is being measured. This could be a flat surface or a curved surface, but it is important to define the boundaries of the right face in order to accurately calculate the electric flux.

3. Why is integration necessary for finding electric flux?

Integration is necessary for finding electric flux because it allows us to take into account the varying strength of the electric field over the entire right face. By breaking the surface into smaller, infinitesimal areas and calculating the electric field at each point, we can then sum up these values to get the total electric flux through the right face.

4. Can you provide an example of finding electric flux through the right face using integration?

Yes, for example, if we have a flat surface with a constant electric field of 10 V/m passing through it, and the surface has an area of 2 square meters, we can calculate the electric flux by multiplying the electric field strength by the area (10 V/m x 2 m²), giving us a total electric flux of 20 V.

5. How does the orientation of the right face affect the calculation of electric flux?

The orientation of the right face can affect the calculation of electric flux because it determines the angle at which the electric field lines pass through the surface. For example, if the surface is perpendicular to the electric field lines, the electric flux will be at its maximum, whereas if the surface is parallel to the field lines, the electric flux will be zero.

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