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Electric Field Strength magnitude

  1. Nov 19, 2011 #1
    Hi,

    Four point charges of equal magnitude are held at the corners of a square. The length of each side of the square is 2a and the sign of the charges are the following: Two positive charges on the top two corners of the square and two negative charges on the bottom two corners of the square.

    The point P is at the center of the square.

    a) Deduce that the magnitude of the electric field strength at Point P due to one of the point charges is equal to (kQ)/2a2:

    I managed to do this one by proving that r2, that is, the distance between the center of the charge and the point P is equal to 2a2 by doing Pythagoras

    The next two parts of the question however, ask me to draw an arrow showing the resultant force and then determine, in terms of Q, a and k the magnitude of the electric field at point P. I simply can't do it however because it seems to me that there will be no resultant electric field at P! There are two positive charges and two negative charges, all at the same distance from point P and all have the same magnitude!

    Can anyone help me with this (but don't give out the answer, I want to to try and work it out :smile:)

    Thanks,
    Peter
     
  2. jcsd
  3. Nov 19, 2011 #2

    Simon Bridge

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    But different directions - don't forget the sign of the charges.
     
  4. Nov 19, 2011 #3
    But if they have opposite signs they have opposite directions and that's why I thought they would cancel. Or should I assume point P is a positive test charge or something?
     
  5. Nov 19, 2011 #4
    Electric field strength means the force that a unit + charge would experience so you are very much on the right track.
    The unit of field strength is N/C
    draw a diagram showing the + charges and the - charges
    Draw the arrows showing the forces on a + charge at the centre of the square.
    Can you see the resultant?
     
  6. Nov 19, 2011 #5

    Simon Bridge

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    If they were all the same sign, then opposite corners would cancel out.


    Draw the situation for a positive charge at P.
    The charge wants to move towards -ve charges and away from +ve charges.
    See?

    [edit] I was beaten to it:
    +1 the tech person: always draw the picture.
     
  7. Nov 19, 2011 #6
    very gracious of you to say that Simon !
    Cheers
     
  8. Nov 19, 2011 #7
    Thanks a lot for the contribution guys! I got it now. From what I drew I believe there would be a resultant downward electric field. Thanks!
     
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