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Electric field due to a charged square

  • Thread starter WrongMan
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  • #1
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Homework Statement


there is a square on the XoY plane, centered at the origin (just outlines of the square) it has a charge Q (Q>0) and side 2L, i must evaluate the electric field along the z axis. see attached image

Homework Equations


E=k*q/r^2

The Attempt at a Solution


So first i divided the square into 4 lines, and noticed that due to the simmetry there is only a field in the z direction, and all the lines of the square contribute the same to that field.

So:
dE= k(Q/8L)1* dx/(x2+y2+z2)2*(z/(x2+y2+z2))½)3

where subscrip1 is Q/8L because the full Q is for the whole square, so we divide it by 4 to get the charge on the line and divide again by 2L to get the charge density.
subscript 2 is the distance of the charge to the line.
subscript 3 is cosine of E with Ez so we get only the z component
Now the integral, i get an not-so-straightforward integral, so if there is a simplification i could do here please advise.
so i get
E= k(Q/8L)z*∫dx/((x2+y2+z2)3/2))
and i integrate either over -L to L or from 0 to L and multiply it by 2.
and i get

E=kQz*1/((y2+z2)*(L2+y2+z2)½)

This result makes sense to me physically, as field is 0 when z is 0 (makes sense since field would no longer have a z component) and also goes to 0 when z goes to infinity.

So the total field is 4*E. Am i correct? any simplification suggestions for the integral

EDIT: oh and y should be L since its fixed.
 

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Answers and Replies

  • #2
TSny
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E= k(Q/8L)z*∫dx/)(x2+y2+z2)*(x2+y2+z2)3/2))
Did you intend to leave out the part that is marked in blue?
and i integrate either over -L to L or from 0 to L and multiply it by 2.
and i get

E=kQz*1/(y2+z2)*(L2+y2+z2))½)
Note that your parentheses don't match in this expression.
Did you already multiply by 4 to take into account the 4 sides? If so, I believe you are getting the correct answer (with y = L).
any simplification suggestions for the integral
I don't see anything that would simplify what you have done.
 
  • #3
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Did you intend to leave out the part that is marked in blue?

Note that your parentheses don't match in this expression.
Did you already multiply by 4 to take into account the 4 sides? If so, I believe you are getting the correct answer (with y = L).

I don't see anything that would simplify what you have done.
Im sorry about the confusion with the expressions ive been trying to find a way to write them properly here (sugestions??).
About your first quote
yes i did, i confused myself when writing the expressions here.

about your second quote, everything after "1" is supposed to be in the denominator (bellow "1"), again, confusion when writing the expressions here.
and i multiplied by 4 bellow.

im fixing the expressions.
So it's correct?
 
  • #4
TSny
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So it's correct?
Your method is correct.

Can you write out your final expression for the answer?
 
  • #5
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Your method is correct.

Can you write out your final expression for the answer?
E= 4*(kQz*1/((L2+z2)*(L2+L2+z2)½)) in the z direction

EDIT: i replaced y for L since when i evaluated the single line y didnt vary, it was fixed at y=L and i evaluated the change as x went from -L to L
 
  • #6
TSny
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E= 4*(kQz*1/((L2+z2)*(L2+L2+z2)½)) in the z direction
I don't think the factor of 4 is correct.
 
  • #7
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I don't think the factor of 4 is correct. And of course you should replace y by its value in terms of L.
So i eavluated the field created by one line, assuming the opposite line would counter the field in the direction from one line to the oposite one, and wont all 4 lines contribute the same to the field?
 
  • #8
TSny
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So i eavluated the field created by one line, assuming the opposite line would counter the field in the direction from one line to the oposite one, and wont all 4 lines contribute the same to the field?
Yes, all four sides will contribute equally. But, before multiplying by 4 to take care of the 4 sides, I get a 4 in the denominator for the expression due to 1 side alone. So, the 4's cancel in getting the final answer.
 
  • #9
TSny
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One way to check the overall numerical factor in your answer is to look at the limit where L → 0 (i.e., L << z). The square should "act like a point charge" in this limit.
 
  • #10
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Yes, all four sides will contribute equally. But, before multiplying by 4 to take care of the 4 sides, I get a 4 in the denominator for the expression due to 1 side alone. So, the 4's cancel in getting the final answer.
OHHH yeah you are right.
in solving the problem, i re-did it saying the line had charge Q (to simplify a bit).
As you might notice the Q/8L dissapears from one expression to the next, Q/8L should simplify with the 2L from the integral becoming Q/4 and then when multiplied by 4 they'd cancel out. my bad, i get distracted :/
 
  • #11
TSny
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OK. Good work!
 

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