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Electric Field between two Charges

  1. Apr 12, 2016 #1
    1. The problem statement, all variables and given/known data
    A positive charge of 3 microCouloumbs is at the origin. A negative charge with the same magnitude is 8 m away along the x direction.
    a) What are the magnitude and direction of the electric field at the points between the two charges, 2, and 4, and 6m from the positive charge?
    b) What is the electric potential at each of these three points? (The reference point, where V=0 is infinitely far away).

    2. Relevant equations
    V=Ed

    Electric Field = Qk/r^2

    3. The attempt at a solution
    a) (3*10^-6)(9*10^9)/2^2 = 6750
    (3*10^-6)(9*10^9)/6^2 = 750
    7500 to the right
    (3*10^-6)(9*10^9)/4^2 = 3375 to the right


    (3*10^-6)(9*10^9)/6^2 = 750
    (3*10^-6)(9*10^9)/2^2 = 6750
    7500 to the right

    b) V=Ed
    Electric field at 2m
    7500 * 2 = 15000

    Electric field at 4 m
    3375 * 4 = 13500

    Electric field at 6 m
    7500 * 6 = 45000


    I'm confident in A but not in b. Please help
     
  2. jcsd
  3. Apr 12, 2016 #2

    gneill

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    Staff: Mentor

    For part b, you can see from part a that the electric field is not uniform (same magnitude and direction) as you move from the origin towards the second charge. So your equation V = E*d doesn't hold because E is not a constant value. You would have to integrate: ∫E(x)⋅dx.

    Instead, use the formula for the electric potential vs distance for a point charge.
     
  4. Apr 12, 2016 #3
    What is that formula?
     
  5. Apr 12, 2016 #4

    gneill

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    It should be in your text or course notes. It's a very common formula based on Coulomb's Law. You could also do a web search on "Electric potential due to a point charge".
     
  6. Apr 12, 2016 #5
    Electric Potential = qsk/r ?
     
  7. Apr 12, 2016 #6

    gneill

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    Hmm. Maybe... Can you explain what "qsk" represents?
     
  8. Apr 12, 2016 #7
    point source charge
     
  9. Apr 12, 2016 #8

    gneill

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    I'd have to say almost but not quite. Is "qsk" one variable? Or is Coulomb's constant "k" a separate value?
     
  10. Apr 12, 2016 #9
    oh yeah sorry qs times k which is 9*10^9 all divided by r
     
  11. Apr 12, 2016 #10

    gneill

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    Okay, presumably qs is meant to be qs. (Note that you can use the x2 and x2 buttons in the edit panel header to create sub- and superscripts).

    So, using that formula in much the same way as you used the one for the electric field, find the electric potential at the given locations. Note that electric potential is a scalar quantity, so you don't have to worry about vector addition this time.
     
  12. Apr 12, 2016 #11
    b) V = qsk/r
    (9*10^9)(3*10^-6)/2 = 13500
    (9*10^9)(3*10^-6)/6 = 4500
    = 18000

    is this how I would do find the electric potential at point 2?
     
  13. Apr 12, 2016 #12

    gneill

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    Very close. You need to take into account the signs of the charges. Isn't point two equidistant from both charges? Point one was at 2 m from the positive charge, point two was at 4 m, and point three at 6 m?
     
  14. Apr 12, 2016 #13
    Wouldn't the signs all be positive because a positive particle would move to the right away from the origin and towards the negative particle at 8 m
     
  15. Apr 12, 2016 #14

    gneill

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    Nope. As I said before, electric potential is a scalar property, not a vector. All that matters is the sign of the charge and the distance.
     
  16. Apr 12, 2016 #15
    I'm not sure how to continue. Having a positive charge near the origin would increase electric potential and if it was closer to the negative charge electric potential would decrease
     
  17. Apr 12, 2016 #16

    gneill

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    Yes. That's not a problem.

    Edit: If you were to plot the electric potential vs position between the two charges you'd get a curve that looks something like this:

    upload_2016-4-12_23-45-7.png
     
    Last edited: Apr 12, 2016
  18. Apr 12, 2016 #17
    Need more help please
     
  19. Apr 12, 2016 #18

    gneill

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    Apply the potential formula at each point using the signed charges and distances from each charge. Sum the contributions from each.
     
  20. Apr 13, 2016 #19
    b) V = qsk/r
    (9*10^9)(3*10^-6)/2 = 13500
    (9*10^9)(3*10^-6)/6 = 4500
    = 18000

    I still don't understand why this is incorrect. Everything is in the positive direction
     
  21. Apr 13, 2016 #20

    gneill

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    You aren't paying attention. You must incorporate the signs of the charges. Direction does not matter: voltage is a scalar quantity and has no direction associated with it. All that matters is the signed charge and the distance from it.
     
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