Potential Difference between two positive point charges

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

    Two equal positive charges Q are fixed on the x-axis, one at +a and the other at -a.
    (a) The electric field E at the Origin O
    (b) The electric potential V at the origin O

    2. Relevant equations

    E=-dV/dr --> V=kQ/r


    3. The attempt at a solution

    VNet = V0 + V1

    I got V0 = -kQ/r (because the vector r is negative)
    ..and V1 = kQ/r
    V = 0

    Is this correct?

    Thank you for your help!
     
  2. jcsd
  3. Dick

    Dick 25,851
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    No, the potential of a charge q located at x=a is kq/|x-a|, notice the absolute value.
     
  4. Hi Flannabhra!

    Let's answer a few questions:

    a.) What kind of quantity is the electric potential, V?
    b.) Describe to me the significance of 'r'.
     
  5. SammyS

    SammyS 8,432
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    That's not correct.

    First of all, what was your answer for part (a) and how did you get it.


    For you attempt at part (b):
    What does the variable, r, represent?

    The operation of division is not defined for a vector --- that is to say, division by a vector is not defined.

    Electric potential is a scalar quantity.​
     
  6. I got E = 0 for part (a) but i now feel that it is wrong based on your responses. The variable r represents the distance from the point charge to the Origin.
     
  7. SammyS

    SammyS 8,432
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    That is the correct answer for part (a), but do you know why it is?

    Distance is a non-negative quantity, as pointed out by Dick. So r can't change the sign of the potential.
     
  8. Well a logical explanation sufficed for me(the Electric Fields would be of equal magnitude in opposite directions), but I would like to understand the explanation of it based on formulas.

    On the matter of r, I was under the impression that you could take the direction that r is pointing from the reference point and use that as the direction for r. I.e. the equation for E would be:

    E = kQ/r3 * r

    - in which r is the vector r and the r in the denominator is raised to the third power so as to keep the equation true.
     
  9. SammyS

    SammyS 8,432
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    Yes. That answers the question regarding part (a) very nicely.
     
  10. I don't understand this.. What is x in this situation if r is the distance from the point Q to the origin?
     
  11. Hi!

    a) I believe that in this situation, V is scalar.
    b) 'r' is the distance from the point Q to the origin
     
  12. SammyS

    SammyS 8,432
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    For the charge at x = -a, r = ai , where i is the unit vector in the positive x direction.

    For the charge at x = a, r = -ai .

    In each case the distance is a, and the vector, r, points away from the charge.
     
  13. So since the Electric Potential is scalar, is V = 2kQ/a correct?

    Thank you so much for your help SammyS!!!
     
  14. SammyS

    SammyS 8,432
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    That's correct.
     
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