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: Finding potential of two points in a constant electric field

  1. Jan 27, 2010 #1
    1. The problem statement, all variables and given/known data
    The figure below shows two points in an E-field: Point 1: (3,4) and Point 2: (12,9) both in m. The electric field is constant, with a magnitude of 80 V/m, and is directed parallel to the +X axis. The potential at point 1 is 1100 V. What is the potential at point 2?

    2. Relevant equations
    E = change in V / change in X
    A^2 + B^2 = C^2

    3. The attempt at a solution
    I got that the distance (tangent) between points 1 and 2 is 10.296 m. If I plug that into the equation to solve for V, I get 80=(V2 - 1100) / 10.296, or V2 = 823.68. Since Point 2 is further away from the origin of the E-field, the potential should be lower, so 1100 - 823.68 = 276.32. This is wrong. I tried adding them to see if that was right, (1923.68) but that was wrong too.

    My main question is does the E-field value change? It says it is constant, so I don't know why it would. Thanks for the help!

    The second part of the question is to calculate the work required to move a negative charge of Q = -608 microC from point 1 to point 2, which I would use the formula: the change in PE = the change in V times the charge (q). Is this right?
  2. jcsd
  3. Jan 27, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi skibum143! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Potential = potential energy per charge …

    potential energy = work done …

    what is the work done if you go a distance 9 parallel to the x-axis?

    what is the work done if you go a distance 5 parallel to the y-axis? :wink:
  4. Jan 27, 2010 #3
    Hi tiny-tim. Thanks so much for the response!
    The work going a distance 5 parallel to the y-axis is zero, because it's on the same equipotential line.
    So should I only be using 9 as my "r" instead of the tangent of 10.296?
  5. Jan 27, 2010 #4
    sorry, i meant use 9 as my change in x...
  6. Jan 27, 2010 #5
    I guess my other main question is, if they say the electric field is constant, at 80 V/m, is that the value at every point in the electric field? If not, how do you calculate the change in the electric field?
  7. Jan 27, 2010 #6
    AH!!!! I got it!!!!!!!! Thank you so much!!!! :)
  8. Jan 27, 2010 #7


    User Avatar
    Science Advisor
    Homework Helper

    :biggrin: Woohoo! :biggrin:

    But, slow down in future! :smile:

    you don't usually get a quick response here, so you may as well take your time! :wink:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook