Electric Field due to Rectangular Sheet

[teleport]

Homework Statement

Find the electric field due to a rectangular plate with charge Q, length L, width W, at a distance s<< L perpendicular to the plate. The point at that location is exactly above (with respect to the plate) the center of mass of the rectangular plate

Homework Equations

Electric Field due to a rod: E = k q/[r*sqrt(r^2 + (L/2)^2)]

q is the charge of the rod.
r is the distance from the c.m. of the rod to the required location
L is the length of the rod

The Attempt at a Solution

Divide the plate into infinite amount of rods of length L and width dx, and charge dq. Use above equation to represent the electric field due to one of these rods and then integrate over the surface.

r^2 = x^2 + s^2. Unfortunately, I can't state that r rounds to x, because x could even be smaller than s.

The components of the electric field due to the plate that are parallel to it cancel out. So only the components up survive.

dq = dxQ/W. Now I would have to that integral which looks hard to do. My belief is that there must be an easier way to solve this. Is there?

Thank you.

 

[teleport]

Is it that perhaps I wrote all the previous to not receive the most trivial comment?

 

[teleport]

I certainly didn't think that no one in this forum was capable of giving at least a simple helpful hint.

 

[teleport]

OK. My book doesn't have a formula for electric field produced by a rectangular sheet with uniform charge. So, I tried to derive it with no success. Can anyone please tell me this expression? I would really appreciate it. Thanks.

 

[berkeman]

I don't think I'd start with rods. Start with point charges and do the integration. Figure out the point charges from the area charge density and integrate each little dA.

 

[teleport]

the double integral that I get has (x^2 + y^2 + s^2)^(3/2) in the denominator, and nothing convinient in the numerator. Is this right or I'm messing around with something?

 

[berkeman]

Does the simplification of s << L help you at all?

 

[teleport]

For the moment I don't see how it helps.

 

[siddharth]

the double integral that I get has (x^2 + y^2 + s^2)^(3/2) in the denominator, and nothing convinient in the numerator. Is this right or I'm messing around with something?

Looks right to me, but the integration seems tricky. If you need help with that, the online calculator at www.quickmath.com may help. Once you evaluate it, try using the simplification berkeman suggested.

 

[teleport]

hmm it seems they are having some kind of problem with the website. Broken link or something. Will try again later. Man, I need to get one of those maples...

 

[teleport]

done (look in the attachment). Integral done was

dydx/(x^2+y^2+s^2)^(3/2) with x limits of [-C/2, C/2] and y limits [-K/2, K/2] and the result is beyond words. Is this what you were talking about? Hmm.

 

[teleport]

Hmm, suddenly no more help?

 

[berkeman]

Hmm, suddenly no more help?

I didn't check the math, but if you set up the integration correctly and the engine worked correctly, it seems like you should have the correct answer. I did some googling yesterday for electric field rectangular charge sheet, and got some interesting hits, but nothing that I thought would be helpful for you. Can you check your answer with a classmate to see what they got?

 

[teleport]

I don't think there is anyone elese in my class trying to get that derivation. On the other hand, if that answer is correct, then it is really scary. How is it that such a nice simple figure like a rectangle could have such a complicated expression for the electric field. If that answer is indeed correct, then my book didn't derive it because the math is too complex, and not because it is supposed to be easy. I am sure that there must be a simpler equivalent answer.

 

[dubbya89]

I believe that since s<<L, you can treat the rectangle as an infinite plate of charge, or at least an infinite line of charge of variable width. That might simplify things.