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!

Uniform charge Electric field

  1. Sep 3, 2008 #1
    A 10cm long thin glass rod uniformly charged to 12nC and a 10cm long thin plastic rod uniformly charged to -12nC are placed side by side, 3.9cm apart. What are the electric field strengths at 1cm, 2 cm , and 3cm from the glass rod along the line connecting the midpoints of the two rods?

    E=k Q/(r*[tex]\sqrt{r^2 +(L/2)^2}[/tex])

    E=k Q/(r*[tex]\sqrt{r^2 +(.05m)^2}[/tex])

    From glass rod
    E at 1cm = E= k 12nC/((.01m)*[tex]\sqrt{(.01m)^2 +(.05m)^2}[/tex])= 211805 N/C
    E at 2cm = 100275 N/C
    E at 3cm = 61739.5 N/C

    From plastic rod
    E at (3.9cm - 1cm) = E= k 12nC/((.029m)*[tex]\sqrt{(.029m)^2 +(.05m)^2}[/tex]) = 64429.9 N/C
    E at (3.9cm - 2cm) = 106270 N/C
    E at (3.9cm - 3cm) = 236204 N/C

    E net at p1= 211805 N/C + 64429.9 N/C = 276235 N/C
    E net at p2= 100275 N/C + 106270 N/C = 206545 N/C
    E net at p3 = 61739.5 N/C + 236204 N/C = 297944 N/C

    Did I do this right?

    Thanks for the help.
  2. jcsd
  3. Sep 3, 2008 #2
    I am really unsure if I did this problem correctly. Any advise would be greatly appreciated.

    Thank you.

  4. Sep 3, 2008 #3


    User Avatar
    Homework Helper

    Without looking up the charged rod relationship or the math involved, the method looks good.
  5. Sep 4, 2008 #4
    Anybody have any opinions.

    I really appreciate the help.
  6. Sep 5, 2008 #5
    I have to turn this in very soon. Did I do this problem correctly?

    Any help would be appreciated. Thanks.
  7. Sep 6, 2008 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes, it looks good. You might watch the significant figures, if your professor cares about that sort of thing.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?