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Determining Direction of Electric Field!

  1. Dec 12, 2009 #1
    Determining Direction of Electric Field! :)

    1. The problem statement, all variables and given/known data

    Calculate the electric field at one corner of a square 50 cm on a side if the other three corners are occupied by 250 x 10^-7 C charges.

    2. Relevant equations

    E = (kQ)/(r^2)

    Where:

    E = Electric Field Intensity in (N/C)
    k = Electrostatics Constant (9 X 10^9 Nm^2/C^2)
    Q = Charge in (C)
    r = separation in (m)

    3. The attempt at a solution

    Firstly, I calculated the E from one corner to each of the three charges:

    E1 = [(9 x 10^9 Nm^2/C^2)(250 x 10^7 C)]/(.50m)^2
    = 9 x 10^5 N/C (Not Sure of Direction)

    E2 = [(9 x 10^9 Nm^2/C^2)(250 x 10^7 C)]/(.50m)^2
    = 9 x 10^5 N/C (Not Sure of Direction)

    E3 = [(9 x 10^9 Nm^2/C^2)(250 x 10^7 C)]/(.71m)^2
    = 4.5 x 10^5 N/C (Not Sure of Direction)
    (Note the change in r, as this is diagonal from the corner)

    I then added the E.s up:

    Etotal = E1 + E2 + E3
    = (9 x 10^5 N/C) + (9 x 10^5 N/C) + (4.5 x 10^5 N/C)

    Because E1 + E2 are not on the same plane, I would have to use Pythagoras to add the two, and then add it to E3. So,

    Etotal = E1 + E2 + E3
    = (9 x 10^5 N/C) + (9 x 10^5 N/C) + (4.5 x 10^5 N/C)
    = (1.3 x 10^6 N/C) + (4.5 x 10^5 N/C)
    = 1.8 X 10^6 N/C

    Okay, wait. When I typed it out here and did all the calculations, I got the right answer! That's weird, considering every other time I did it it was wrong.

    Anyway, I still have to determine the direction of the Electric Field.
    In my answer key, it says: [AWAY FROM CENTER], which I do not understand. Is it because I am dealing with all positive charges, so they all repel each other?

    I am simply terrible when it comes to determining direction, so please help!
     
  2. jcsd
  3. Dec 12, 2009 #2
    Re: Determining Direction of Electric Field! :)

    Also, if someone could help me out as to the direction of each E. Field where I note (Not Sure of Direction), that would be great as well!
     
  4. Dec 12, 2009 #3

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Re: Determining Direction of Electric Field! :)

    Hi,

    If you do not have a figure given, or drawn yourself, describing this problem, I urge you to draw one now. Like in many physics problems, you stand little chance of understanding what's going on without a figure. Do not try to follow this discussion any further without first having a figure, showing the 3 charges arranged at the corners of a square, in front of you.

    The direction of each field is away from the charge you are using to calculate it. This means each E is in a different direction, and in particular E3 is diagonal, away from the charge in the opposite corner.

    Note, since the E's act in different directions you should be adding them as vectors, not simply adding their magnitudes as you are indicating here.

    Actually it is because E1 and E2 are at a right angle to one another that you are allowed to use Pythagoras here. In fact, all three E's are in the same plane, the plane of the square.

    Okay because the vector result E1+E2 happens to be in the same (diagonal) direction as E3.
    They mean the field's direction, at the corner, is away from the center of the square. I.e., diagonally outward.
    There are only two rules to keep in mind to figure out direction:
    1. The field of a positive point charge is directed away from that charge.
    2. The field of a negative point charge is directed towards that charge.​
    In this problem the 3 charges are all positive, so each field E1, E2, E3 is directed away from the respective charge.

    Hope that helps. Again, you do need to have a figure in front of you to understand this discussion.
     
  5. Dec 13, 2009 #4
    Re: Determining Direction of Electric Field! :)

    Thank you! It was very thorough, and easy to understand! :)
     
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