1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Electric Potential x-axis

  1. Jun 22, 2010 #1
    1. The problem statement, all variables and given/known data

    A -3.0 nC charge is on the x-axis at x = -9cm and a +4.0 nC charge is on the x-axis at x = 16 cm. At what point or points on the y-axis is the electric potential zero?


    2. Relevant equations

    V = kq/r

    3. The attempt at a solution

    I know the answer is +/- 12 cm but I have no clue as to how that number was achieved. Can anybody please help me out with this one?
     
  2. jcsd
  3. Jun 22, 2010 #2

    ehild

    User Avatar
    Homework Helper

    The potential at a point at distance r from a charge q is kq/r. If there are more charges, their potentials add up. You have two charges placed on the x axis. Take on a point on the y axis, and write out the distance from both charges. It is easier if you make a drawing. Use this distances as "r" in the formula for the potential. Add up both terms and take it equal to zero. Solve the equation for y.

    ehild
     
  4. Jun 22, 2010 #3
    So, I can take any arbitrary point on the y-axis, and it will eventually lead me to the right answer?
     
  5. Jun 22, 2010 #4

    ehild

    User Avatar
    Homework Helper

    Yes, if your procedure is correct. Let me see how you proceed.

    ehild
     
  6. Jun 22, 2010 #5
    So, I took an arbitrary point of (0, 5 cm), and this is what I got:

    r1 = 1.03 x 10^-1 m
    r2 = 1.68 x 10^-1 m

    Vi = kq1/r1 + kq2/r2
    = 2.62 x 10^2 V/m + 2.14 V/m
    = 4.76 x 10^2 V/m

    Now, I don't know where to go from here.
     
  7. Jun 22, 2010 #6

    ehild

    User Avatar
    Homework Helper

    Arbitrary means just y. Do not give any special value. Write the formula for the distance of the charge at point (-9,0) and (0,y), and do the same for the other charge. See attached picture.

    ehild
     

    Attached Files:

    Last edited: Jun 22, 2010
  8. Jun 22, 2010 #7
    I'm not sure I know how to write that formula...kq1/r1 + kq2/r2 = 0?
     
  9. Jun 22, 2010 #8

    ehild

    User Avatar
    Homework Helper

    First write r1. See the picture. You have a right triangle, don't you? Use Pythagorean Theorem to get r1.

    ehild
     
  10. Jun 22, 2010 #9
    So, kq1/(sqrt.9^2 + y^2) = 0?
     
  11. Jun 22, 2010 #10

    ehild

    User Avatar
    Homework Helper

    No, the sum of both potentials kq1/r1+kq2/r2=0!

    r1= sqrt(9^2+y^2)
    r2=sqrt(16^2+y^2).

    ehild
     
  12. Jun 22, 2010 #11
    So, how would I solve for r1 or r2 when there are two unknowns?
     
  13. Jun 22, 2010 #12

    ehild

    User Avatar
    Homework Helper

    What are those two unknowns?

    ehild
     
  14. Jun 22, 2010 #13
    y^2 and r1 for the first equation, and y^2 and r2 for the second.
     
  15. Jun 22, 2010 #14

    ehild

    User Avatar
    Homework Helper

    But you have three equations.

    ehild
     
  16. Jun 22, 2010 #15
    kq1/r1 + kq2/r2 = 0 being the third?
     
  17. Jun 22, 2010 #16
    [kq1/sqrt(9^2 + y^2)] + [kq2/sqrt(16^2 + y^2)] = 0?
     
  18. Jun 22, 2010 #17

    ehild

    User Avatar
    Homework Helper

    Plug in the data for q1 and q2 and solve for y.

    ehild
     
  19. Jun 22, 2010 #18
    The math isn't working for me. This is as far as I can get it:

    [kq1/sqrt(9^2 + y^2)] + [kq2/sqrt(16^2 + y^2) = 0
    (kq1)^2/(9^2 + y^2) = (kq2)^2/(16^2 + y^2)
    [(kq1)^2*(16^2 + y^2)]/(kq1)^2 = [(kq2)^2*(9^2 + y^2)]/(kq1)^2
    (16^2 + y^2)/(9^2 + y^2) = [(kq2)^2*(9^2 + y^2)]/[(kq1)^2*(9^2 + y^2)

    Is this the right equation? If so, can you help me with the mathematics?
     
  20. Jun 22, 2010 #19

    ehild

    User Avatar
    Homework Helper

    It is right up to here.

    Why don't you plug in the data for q1 and q2?

    ehild
     
  21. Jun 22, 2010 #20
    Ah, I finally got it! Thank you very much for your time and patience.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook