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Electric Field question (two of them!)

  1. Sep 2, 2007 #1
    1. The problem statement, all variables and given/known data

    Four point charges have the same magnitude of 6.80 x 10^-12 C and are fixed to the corners of a square that is 1.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

    The magnitude of each of the charges in the figure is 7.50 10-12 C. The lengths of the sides of the rectangle are 3.00 cm and 4.50 cm. Find the magnitude of the electric field at the center of the rectangle in figures a and b.

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

    3. The attempt at a solution
    Ok so i have been ALL over the forums here and looking at other examples and still nto getting it. I am on my final response...i am normally very smart so i don't get why i am sucking in this section.

    I looked at it and said that the two positive forces opposite each other would negate each other leaving only the negative charge and the one positive charge across from it.

    With that I said E = k*6.8e-12/ (1/2 of 1 * sqrt(2) since it is a half of a diagonal in a square)

    i got .1224 for each charge...added the two together and it was wrong. I also tried an instance where i did it with all 4...so it was 4 * k * q / (.5*sqrt(2)) that was wrong too...

  2. jcsd
  3. Sep 3, 2007 #2


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    Did you convert from cm to m and still get the wrong answer? Your work looks right except you didn't convert to m... so I think it should be 1224 N/C for one charge... and using both charges the answer is 2448 N/C.
  4. Sep 3, 2007 #3
    Ah ha! yes apparently the one thing I didn't do was check for stupid mistakes on this one :)

    Any chance you can give me some guidance on the 2nd part? Thanks!
  5. Sep 3, 2007 #4


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    Hmmm... I'm confused by the 2nd question... Are there 4 equal charges at the four corners of the rectangle? Or is it 3 positive charges and 1 negative charge like the first part?
  6. Sep 3, 2007 #5
    It is two parts:

    this thread has the pictures.

    I understand it is E = kq / r^2...

    a) i pythagorean theoremed the hypoteneuse... so (1.5^2 + 2.25^2)^.5 = 2.704 cm
    2.704 / 100 = .02704 m

    8.99e9 7.5e-12 / (.02704^2) = 92.31 N/C x 2 (for the other side as well) => 184.4 N/C (already got this one correct)

    is it just 92.31 x 4 (already tried and that is wrong, but i knew it wasn't the right use of the principals anyways)?

    The other thread doesnt' really talk about part B (the one with the 2 negative and 2 positive charges).

    The only other thing i can think of is that you calculate the charge like you do in part A for each side...and then you use those two * cos (30) to get the resultant E vector?

    so 369 * cos(30) = 319.4 N/C

  7. Sep 4, 2007 #6


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    For part b)... the horizontal components all cancel... take the vertical component of the field due to one charge... then multiply by 4... the angle isn't 30...
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