1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric field due to a charged square

  1. Jun 10, 2016 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    E=k*q/r^2

    3. 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.
     

    Attached Files:

    Last edited: Jun 10, 2016
  2. jcsd
  3. Jun 10, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  4. Jun 10, 2016 #3
    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?
     
  5. Jun 10, 2016 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Your method is correct.

    Can you write out your final expression for the answer?
     
  6. Jun 10, 2016 #5
    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
     
  7. Jun 10, 2016 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I don't think the factor of 4 is correct.
     
  8. Jun 10, 2016 #7
    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?
     
  9. Jun 10, 2016 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  10. Jun 10, 2016 #9

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  11. Jun 10, 2016 #10
    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 :/
     
  12. Jun 10, 2016 #11

    TSny

    User Avatar
    Homework Helper
    Gold Member

    OK. Good work!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Electric field due to a charged square
Loading...