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Homework Help: Electric field in a square

  1. Jan 23, 2014 #1
    1. The problem statement, all variables and given/known data

    A square has a charge at each corner. The square is 2cm x 2cm. In the upper left corner there is 1μc, in the upper right corner there is 2μc, the lower right corner there is 3μc, and the lower left corner there is 4μc.
    A. What is the magnitude and direction of the electric field at the 4μc charge?
    B. wha Tia the magnitude and direction of the force on the 4μc charge?

    2. Relevant equations

    E=kq/r^2. And F=EQ

    3. The attempt at a solution

    1. (9*10^9)(1*10^-6)/(.02)^2 = 2.25*10^7 N/C

    2. Same process with 2*10^-6 to get. 4.5*10^7 N/ C

    3. Same with 3*10^-6 to get 6.75*10^7 N/C

    E net = 2.25+4.5+6.75=13.5*10^7 N/C.

    But not sure how to get direction. Since they are all positive, would it be 45° SE?

    For force, F=(13.5*10^7)(1.6*10^-19)=21.6*10^-12N.

    How to get direction? Pythagorean to be used but how?
    Any help is appreciated.
  2. jcsd
  3. Jan 23, 2014 #2


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    Homework Helper

    The question hasn't explicitly stated that all charges are positive. This has to be assume to make it doable.


    Wrong. What's the distance between the upper right and lower left corners?


    It's not simple addition. Electric field strength is a vector. Do vector addition.

    Why SE? Shouldn't it be SW?

    Isn't the charge at the lower left corner 4uC?

    Draw a force diagram. How would you add vectors?
  4. Jan 23, 2014 #3
    So for 2. I should use Pythagorean - Sqrt(.02^2+.02^2)?
    Ok for vector addition, tip to tail for E.
    I believe SW would be right.

    But what do you mean about the charge in lower left? Am I not supposed to use E times Q the constant at 1.6*10^-19? Or am I wrong about Q?
  5. Jan 23, 2014 #4


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    Homework Helper

    SW is the general direction. It's not exactly SW because the magnitudes of the field strength in a Southward direction (due to the upper left corner) and that in a Westward direction (due to the lower right corner) are not identical. So you need to use trig to get the correct orientation of the resultant vector.

    It's easiest to work out the equal component vectors (Southward and Westward) of the field strength due to the top right hand corner charge first. Then add up the Southern and Westward components of the three field strengths individually. Finally, use Pythagorean theorem to find the magnitude of the resultant, and use trig to find the direction.

    Your (correct) units for field strength are N/C. Your charge is 4uC. How would you get the force (in Newton)?
  6. Jan 23, 2014 #5
    Q the constant is in C, so would leave me with N.

    This helps me a lot, Thanks so much!
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