1. Not finding help here? Sign up for a free 30min 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!

What is the magnitude if the total force exerted by the four charges

  1. Jun 30, 2013 #1
    Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude if the total force exerted by the four charges on a charge Q located a distance b along a line perpendicular to the plane of the square and equidistant from the four charges?


    The result of my attempt was:

    F= 4kqQb/((d^2/2)+b^2)^3/2

    but I dont know if is
    F= kqQb/(b^2+l^2/2)^(3/2)
     
  2. jcsd
  3. Jun 30, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    The above is almost right. Look again at the term d^2/2. Is it correct?
     
  4. Jun 30, 2013 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Perhaps you could explain your reasoning.

    What is the l as in l^2/2?

    It appears to me that F= 4kqQb/((d^2/2)+b^2)^3/2 is incorrect. Check the d^2/2 term in the denominator. What is the distance of a charge from the centre of the square?

    AM
     
  5. Jun 30, 2013 #4
    This is what i was thinking:
    F because of one charge=kqQ/(b^2+l^2/2) * cos(theta)

    =kqQ/(b^2+l^2/2) *b/sqrt(b^2+l^2/2)

    =kqQb/(b^2+l^2/2)^(3/2) (along the perpendicular )

    as in plane parallel to the square net F=0
     
  6. Jun 30, 2013 #5
    explanation

    This is what i was thinking:

    F because of one charge=kqQ/(b^2+l^2/2) * cos(theta)

    =kqQ/(b^2+l^2/2) *b/sqrt(b^2+l^2/2)

    =kqQb/(b^2+l^2/2)^(3/2) (along the perpendicular )

    as in plane parallel to the square net F=0
     
  7. Jun 30, 2013 #6
    explanation

    this is what i was thinking
    F because of one charge=kqQ/(b^2+l^2/2) * cos(theta)

    =kqQ/(b^2+l^2/2) *b/sqrt(b^2+l^2/2)

    =kqQb/(b^2+l^2/2)^(3/2) (along the perpendicular )

    as in plane parallel to the square net F=0
     
  8. Jun 30, 2013 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That looks right to me, but I see I'm outvoted by AM and rude man :confused:. Maybe I'm missing something.
     
  9. Jun 30, 2013 #8

    rude man

    User Avatar
    Homework Helper
    Gold Member

    How about d^2/2 → (d/2)^2 = d^2/4 ?
     
  10. Jun 30, 2013 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, why? As I read the OP, the point Q is distance b from the centre of the square.
     
  11. Jun 30, 2013 #10
    I tend to agree with haruspex :-)
     
  12. Jun 30, 2013 #11

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Curses, haruspex and barryj, you are right. :blushing:

    @Andrew Mason:
    Locate the four charges at (d/2,d/2,0), (-d/2,d/2,0), -d/2,-d/2,0) and (d/2,-d/2,0) and the observation point at (0,0,b).

    Then the distance from any charge is

    sqrt[(d/2-0)^2 + (d/2-0)^2 + (0-b)^2] = sqrt{d^2/2 + b^2}.

    My excuse: spatial relations was never my strong suit!
     
  13. Jul 1, 2013 #12

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    You are right. I was using the distance as √2d/2 and it got me a bit confused. Squared it is d^2/2. So I agree the OP's first answer was right. Sorry for any confusion!

    AM
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted