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Homework Help: 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


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    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


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    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.
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