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: Another Question about Electric Potential

  1. Apr 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Three charges are placed at the corners of an equilateral triangle with sides of length 2.0 m. Determine the electric potential at the midpoint of each side of the triangle. The three charges at the corners of the triangle at +4.0 X 10^-6 C, +2.0 X 10^-6 C, and -4.0 10^-6 C.

    2. Relevant equations

    3. The attempt at a solution
    I tried finding the electric potential of each of the charges to the midpoint, meaning that the r=1.0 m, and then adding the electric potential on either side of the midpoint together... but it's not working...
    For example:
    For the midpoint betweeen the charges +4.0 X 10^-6 C and +2.0 X 10^-6 C, i found the electric potential from the midpoint to the charge +4.0 X 10^-6 C, in which i got 36000 V and then i found the electric potential from the midpoint to the charge of +2.0 X 10^-6 C, and i got 18000 V. I added these two values together... but i didn't get the right answer. The answers for the three midpoints of the triangle are 1.0 X 10^4 V, 3.3 X 10^4 and 2.8 X 10^3 V. Can someone please help?
  2. jcsd
  3. Apr 10, 2007 #2
    Did you use 1m for all your distances? What distance did you use for the charge opposite the midpoint?
  4. Apr 11, 2007 #3
    oops! I didn't know before hand that i'm suppose to consider the charge across from the midpoint. Thank you for the hint :D
  5. Dec 15, 2011 #4
    I am stuck in the same question
    I tried to use V= kq/r but I don't know what is r for each point
    the triangle are 2m length so at one of the midpoint it is [itex]\sqrt{3}[/itex], 1, 1, away from each charge

    do you add v=kq/r1 + v=kq/r2 and so on?
    I got it!!
    Last edited: Dec 15, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook