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Finding charge so that the electric field is zero

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Two tiny objects with equal charges of 65.0 µC are placed at two corners of a square with sides of 0.335 m, as shown. How far above and to the left of the corner of the square labeled A would you place a third small object with the same charge so that the electric field is zero at A?
    above corner A= ?m
    to the left of corner A= ?m

    image http://www.webassign.net/grr/p16-27.gif


    2. Relevant equations

    Coulomb's Law: F = (k * q1 * q2 ) / r^2
    Electric Field: = k*q / r^2

    3. The attempt at a solution
    .287933617
    .56332741

    Force from just below A.

    d = 0.335 m
    k = 9*10^9
    q1 = 65 *10^-6 C

    E = kq1/d^2
    E = 9*10^9*65*10^-6/(0.335)^2
    E =5212752.26 N/q

    Force from charge on the same diagronal as A
    ==================================
    d = sqrt(0.335^2 + 0.335^2) A diagronal of a square = sqrt(side^2 + side^2)
    d = 0.4737 m

    E2 = k*q1/d
    E2 = 9*10^9 * 65 * 10^-6/(0.4737)^2
    E2 = 2607048.416 N/q

    Summing the Fields.
    Vertical
    Total Vertical = 5212752.26 + 2607048.416 *Sin(45o)
    the angle a diagonal makes with with vertical is 45o
    Total Vertical = 5212752.26 N/q + 1843461.614 N/q
    Total Vertical = 7056213.874 N/q

    Total Horizontal = 2607048.416 * cos(45o)
    Total Horizontal = 1843461.614 N/q

    I used the following equation for the rest, but im not getting the right answer, please if anyone can help me, this is due in an hour. Thank you so much ive been trying this for hours

    d^2 = kq1/E
     
  2. jcsd
  3. Jan 23, 2012 #2

    Delphi51

    User Avatar
    Homework Helper

    Welcome to PF, trip!
    Just some tips - you will want to do this yourself. No need to work with the numbers for the charge or the distance until the very end. In fact, the charges all cancel out in the calcs.

    Call the .335 m by the letter d. The diagonal distance is the square root of 2d².
    Then the horizontal component of E at A due to the charge q on the right is
    Ex = kq/(2d²). The vertical Ey is a very similar expression, just 3/2 times Ex, I think.
    Next, let x,y be the location of the third charge from A.
    For its Ex and Ey due to the 3rd charge (exactly equal to the ones already found) I got something like Ex = kqx/Z, where Z is (x² + y²) to the power 3/2 for that after expressing the cos A in terms of x and the hypotenuse. A similar expression for Ey. Now I have two equations in x and y to solve simultaneously. I solved one for Z, subbed into the other and immediately got x =3y. (Better check that!) Only a little more work to get the actual values for x and y as simple multiples of d. Much, much less work than you did!
     
    Last edited: Jan 23, 2012
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