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Electric Field due to Two Point Charges

  1. Jan 23, 2008 #1
    1. The problem statement, all variables and given/known data

    Two point charges are placed on the x axis. The first charge, q1= 8 nC, is placed a distance 16m from the origin along the positive x axis; the second charge, q2= 6 nC , is placed a distance 9m from the origin along the negative x axis.

    Find the electric field at the origin, point O.

    Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

    Now, assume that charge q2 is negative; q2 = -6 nC. What is the net electric field at the origin, point O?

    Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

    2. Relevant equations

    E = [tex]\frac{kQ}{r^2}[/tex]

    3. The attempt at a solution

    I start by "placing" a test charge at the origin.

    Here's what I think I should do:

    E = [tex]\frac{kQ_1}{r^2}[/tex] + [tex]\frac{kQ_2}{r^2}[/tex]

    E_1 = [tex]\frac{(9x10^9)(8x10^-9)}{16^2}[/tex] = 0.261 N/C

    E_2 = [tex]\frac{(9x10^9)(6x10^-9)}{9^2}[/tex] = 0.667 N/C

    E = E_1 + E_2 = 0.261 + 0.667 = 0.928 N/C

    Am I right?
     
    Last edited: Jan 23, 2008
  2. jcsd
  3. Jan 23, 2008 #2
    The direction of the E-field is taken to be the direction of the force it would exert on a positive test charge so when you draw a diagram with a positive test charge how would the forces from the q1 and q2 be acting on it? In the same direction or opposite?
     
  4. Jan 23, 2008 #3
    Since all of the charges are positive, they would repel, and thus create a force pushing it to the left.

    Still not sure (electircity isnt my strong point)
     
  5. Jan 23, 2008 #4
    q1 is to the right of the test charge (placed at the origin) and q2 is to the left of it.. draw a force diagram to see
     
    Last edited: Jan 23, 2008
  6. Jan 23, 2008 #5
    ohhhh.... I thought the second charge was also placed on the positive x-axis

    so would it be 0.667 - 0.261 since the force is pushing the test charge to the left?

    and I just noticed that I posted it in the wrong section. Would someone be able to move it to intro physics?
     
    Last edited: Jan 23, 2008
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