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Finding the energy field due to three positive point charges on a 4th negative charge

  1. Aug 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Three equal positive point charge of magnitude Q=7.00uC are located at three corners of a square of edge length d=4.8cm. A negative charge of -21.00uC is placed on the fourth corner.
    A) At the position of the negative charge, what is the magnitude of the electric field due to the three positive charges?
    B) What is the magnitude of the attractive force on the negative charge?


    2. Relevant equations
    Ea=kQ/d^2 (also applies to Ec)
    Eb=sqrt (Ea^2 + Ec^2)
    e=KQ/r^2
    r=sqrt(2d)


    3. The attempt at a solution
    I haven't gotten to part B yet, but here is the work I have done for part A.
    Ea=(8.99x10^9)(7x10^-6)/.048^2=2.7x10^7
    Ec=2.7x10^7
    Eb=sqrt(Ea^2+Ec^2)=3.86x10^7
    The electric field at the negative point due to the corner charge: e=KQ/r^2
    e=(8.99x10^9)(21x10^-6)/0.3098^2=1967053.444
    Since e and Eb are parallel, to find the magnitude of the electric field you would need to add them together: e+Eb=4.1x10^7

    I have no idea why this is wrong. I also tried breaking it down into vectors, but that also got a wrong answer as well. If that was the right line of thinking I can post the work for that, but it felt off to me. I am including the diagram I drew up a well. I only have one try left so any guidance to help me see what I am doing wrong would be greatly appreciated.

    Untitled-1.png
     
  2. jcsd
  3. Aug 30, 2011 #2

    tiny-tim

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    Hi Marie! :smile:

    (try using the X2 icon just above the Reply box :wink:)
    No, read the question again, the charge is still 7 :redface:

    (and where does 0.3098 come from?)
     
  4. Aug 30, 2011 #3
    Re: Finding the energy field due to three positive point charges on a 4th negative ch

    Heh, I see the x2 button now. :) Thanks!

    0.3098 is r, where r=sqrt(2d).

    I had originally kept the charge the same for that part where e=(8.99x109)(7x10-6)/r2 but the answer was still wrong. It gave me 6.6x105 and when adding that to Eb, I got 3.9x107 N/C, which is still incorrect.
     
  5. Aug 30, 2011 #4

    tiny-tim

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    nooooooooooo! :cry:

    r=sqrt(2)d (or, easier, r2 = 2d2)​

    start again, and post it here first so we can check it! :smile:

    then get some sleep! :zzz:​
     
  6. Aug 30, 2011 #5
    Re: Finding the energy field due to three positive point charges on a 4th negative ch

    So r2=2d2=2*0.0482=0.004608
    e=KQ/r2=(8.99x109)*(7x10-6)/0.004608=1.4x107

    Then adding that to Eb would give us: E= 5.2x107 N/C
    Is that right? Which would bring us to part B, which would be:
    F=Eq=(5.2x107)(21x10-6)= 1092 N
     
  7. Aug 30, 2011 #6

    tiny-tim

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    Yes, that all looks ok :smile:, except you should have at least three https://www.physicsforums.com/library.php?do=view_item&itemid=523" in all your intermediate steps, or you might have a rounding error …

    you always need more sig figs in the intermediate steps than in the final result

    (and the newtons should presumably be to only 2 sig figs)

    do it again to check :wink:
     
    Last edited by a moderator: Apr 26, 2017
  8. Aug 30, 2011 #7
    Re: Finding the energy field due to three positive point charges on a 4th negative ch

    Thank you so much! This was correct and you really helped me to understand what I did wrong. When I get this problem right on the test I will thank you again! :)
     
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