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Electostatic forces - hard question

  1. Feb 2, 2005 #1
    There are four charges, each with a magnitude of 2.0 mC. Two are positive and two are negative. The charges are fixed to the corners of a 0.30-m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.

    any idea how to go about this?

    Id assume it would be alternate positive/negative charges around the square.. but then what do i do?
  2. jcsd
  3. Feb 2, 2005 #2
    errr...0? I think regardless of how you set them up that'd happen, but I'll check mathmatically

    Well, it'd be zero in the middle of the square, or something like that
    Last edited: Feb 2, 2005
  4. Feb 2, 2005 #3
    hint: try to figure out where the charges are located on the square.
    There are only two possibilities, either both the positives will be on the same side, or they will be across from each other (along the diagonal ).

    keep in mind that the net force is directed at the center, like charges repel - opposites attract, and the force between two point charges acts along the line between them
  5. Feb 3, 2005 #4
    im pretty sure that it would be

    - +

    + -

    but i dont know where to go from here :/
  6. Feb 3, 2005 #5
    What is the equation for the force between two point charges?
    These forces obey the principal of superposition, so to get the net force on a point charge, you just need to find the force from each of the other three charges and add them (keep in mind their vectors so you have to add corresponding components )

    Make use of any symmetry in the problem (do any of the components cancel? )
    Last edited: Feb 3, 2005
  7. Feb 3, 2005 #6
    I got 0.366N.. is this correct?

    (F on each charge at x and y component is k2q/0.3^2 - sin45(k2q/0.424^2))
  8. Feb 3, 2005 #7
    I can't tell exactly what you did from your equations, but your answer is off by a factor of 10^6... did you use millicoulombs (10^-3 C )...

    in case you didnt know... mC = 10^-3 C which is pronounced millicoulombs

    [itex] \mu C[/itex] = 10^-6 C and is pronounced microcoulombs
  9. Feb 3, 2005 #8
    yes i figured that out

    if you copy [itex] \mu C[/itex] from a site, and paste it in here, the mu changes to an m.. (for some reason)
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