1. Not finding help here? Sign up for a free 30min 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!

Strength and Direction of Electric Field

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    What are the strength and direction of the electric field at the position indicated by the dot in the figure? Answer in component form.
    [​IMG]

    2. Relevant equations

    E=kQ/r^2

    3. The attempt at a solution

    E(+5) = (8.99E9)(5E-9)/(0.02^2) = (1.1E5i, 0j)
    E(-5) = (8.99E9)(5E-9)/(0.02^2) = (0i, 2.8E4j)
    r(+10) = SQR(0.02^2+0.04^2) = 0.0447
    E(+10) = (8.99E9)(10E-9)/(0.0447^2) = 4.5E4
    arctan(2/4)=26.57DEG
    = (4.5E4sin(26.57)i, 4.5cos(26.57)j)
    = (2.0E4i, -4.0E4j)​

    E(+5) + E(-5) + E(10) = (1.3E5i, -1.2E4j)

    Mastering Physics is telling me that this answer is wrong and I want to know where I screwed up.
     
  2. jcsd
  3. Feb 5, 2013 #2

    TSny

    User Avatar
    Homework Helper

    Your work looks good to me. I don't see any mistake.
     
  4. Feb 5, 2013 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The problem may well be too much round-off error.

    Keep at least two extra digits for all intermediate results.

    Round-off your final result to the correct sig. dig. if whoever/whatever is grading your work is anal about such things.
     
  5. Feb 6, 2013 #4
    Turns out it was the comma in my vector notation. Once I removed the comma, the computer accepted it. Looks like I'm a bit rusty on that stuff. Thank you both for your help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?