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Electricity and Magnetism

  1. Dec 9, 2004 #1
    I have a huge problem to work out..

    but first, i cant find the equations for

    (1) electrostatic potential of the center of a square (a charged particle at each vertice)


    (2) the magnitude of electric field at the center of the same square.

    These are both questions to two arrangements, one with positive and negative particles at opposite ends, and arrangement 2 with both negative on right vertices and both positive on left vertices..

  2. jcsd
  3. Dec 9, 2004 #2

    Doc Al

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    Staff: Mentor

    Figure it out for yourself. It's not so hard:
    (1) The potential at a distance r from a point charge is [itex]kq/r[/itex]. It's a scalar, so just add up the contributions from each charge.
    (2) The field at a distance r from a point charge is [itex]kq/r^2[/itex]. This is a vector directed radially outward (from a positive charge). Add up the contributions from each charge (but remember they are vectors).
  4. Dec 9, 2004 #3
    so let me get this straight...sorry if this isnt a hard question to you, but we are just learning this..

    for (1) in the situation where positives are at opposite vertices as well as negatives are at opposite vertices...the equation would be 4(9*10^9)(Q/(root2)s/2)? (with side s and charge +/-Q

    for (2) in the same situation..it would be (9*10^9)(-Q/(root2)s/2) + (9*10^9)(-Q/(root2)s/2) - (9*10^9)(Q/(root2)s/2) - (9*10^9)(Q/(root2)s/2)?
    Last edited: Dec 9, 2004
  5. Dec 9, 2004 #4


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    For (2) u can't add the contributions as if they were scalars.U have to add vectors.Make the correct picture in which draw every force.Project every force upon 2 perpendicular axes (parallel with the sides of the square) and add algebraically the projections which can be taken as scalars.
    Geometry in electrostatics is essential.U should be getting factors with sine and cosine of pi/4.
  6. Dec 9, 2004 #5
    i know this is aggravating for you, but how would u add up 4 vectors? Do you mean make 4 triangles with the square and make the forces on one axis and add/subtract?

    (and are the -Q's going toward P with the +Q's going away from P?)
    Last edited: Dec 9, 2004
  7. Dec 9, 2004 #6
    ok i think i know what you mean...but check me

    the top right force for instance (-Q) would be made the top vertice of a triangle. You would do cospi/4 (45 deg) x the hypotenuse force to get the x-axis force??
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