Electric Flux question [Surface integral]

In summary: But no matter where you put the charge, the electric field is always perpendicular to the surface of the plate, so you have to take that into account when integrating.
  • #1
rohanprabhu
414
2

Homework Statement


Q] A charge 'Q' is kept over a non-conducting square plate of side 'l' at a height l/2 over the center of the plate. Find the electric flux through the square plate surface. Neglect any induction that may occur.

Homework Equations



[tex]
\phi = \int \overrightarrow{E}\cdot \overrightarrow{dS}
[/tex]

The Attempt at a Solution



Well.. it was pretty simple to do using Gauss law. Just take a cube as the surface and then the flux will be equal from all surfaces.. so the flux through one surface would be [itex]\frac{q}{6\varepsilon_o}[/itex]..

but i wanted to actually try a surface integral.. so I found out the Electric field as a function of the coordinates (x, y) on the square plate. Here, one of the corners is taken as the origin (0, 0):

[tex]
E(x, y) = \frac{2Q\sqrt{(l - x)^2 + (l - y)^2}}{\pi \varepsilon_o \left[4(l - x)^2 + 4(l - y)^2 + l^2\right]^\frac{3}{2}}
[/tex]

Also..

[tex]
\phi = \int \overrightarrow{E}\cdot \overrightarrow{dS} = \int EdS\cos{\theta}
[/tex]

and
[tex]
dS = d(xy) = xdy + ydx
[/tex]

so..

[tex]
\phi = \int \overrightarrow{E}\cdot \overrightarrow{dS} = \int E(xdy + ydx)
[/tex]

I have no idea how to go from here.

Any help will be appreciated. Thanks...
 
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  • #2
Well you have two integrals there at the end, the integral of E*x*dy and the integral of E*y*dx, and you have that big expression for E in terms of x and y(which I only assume you did correctly)

Go nuts
 
  • #3
blochwave said:
Well you have two integrals there at the end, the integral of E*x*dy and the integral of E*y*dx, and you have that big expression for E in terms of x and y(which I only assume you did correctly)

Go nuts

I get that.. but how do I solve E*x*dy? I mean.. while integrating the term with 'dy' do I take 'x' to be constant and vice versa for dx?
 
  • #4
Yes...

Except the whole affair is going to be a little fishier than that because you have to take into account the angle between E and dS at every point. You smartly wrote E dot dS then ignored that in writing E(xdy+ydx), which would imply that the electric field and dS are always parallel, which is only true at one point; the one directly beneath the charge
 

Related to Electric Flux question [Surface integral]

1. What is electric flux?

Electric flux is a measure of the flow of an electric field through a given surface. It is defined as the dot product of the electric field and the surface area vector.

2. How is electric flux calculated?

Electric flux can be calculated by taking the surface integral of the dot product of the electric field and the surface area vector over the given surface. This can be expressed mathematically as Φ = ∫E⋅dA, where Φ is the electric flux, E is the electric field, and dA is the differential surface area element.

3. What is the unit of electric flux?

The unit of electric flux is volt meters squared (V⋅m2) or newton meters squared per coulomb (N⋅m2/C).

4. What is the significance of electric flux?

Electric flux is an important concept in electromagnetism as it helps to understand the behavior of electric fields and their interactions with charged particles. It can also be used to calculate the electric field produced by a charge distribution.

5. How does electric flux change with distance?

Electric flux is inversely proportional to the square of the distance from the source of the electric field. This means that as the distance increases, the electric flux decreases. This relationship is known as the inverse square law.

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