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Energy Conservation in Multiple Point Charges

  1. Jan 19, 2006 #1
    Can somebody please check over my solution. I cannot figure out what's wrong. My final answer is not right. I think I may have an error with the signs of the point charges or may have a dumb calculation error. :grumpy:

    Thank you

    Four point charges, fixed in place, form a square with side length d.
    The particle with charge q is now released and given a quick push; as a result, it acquires speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m, what was its initial speed v?

    See picture here: http://ca.geocities.com/canbball/index.html [Broken]

    By using energy conservation I get:
    [tex]k_{i} = U_{f} -U_{i} [/tex]

    [tex] \frac{\ 1}{2} *mv^2 =k[( \frac{\ -3q^2}{\frac{\ d}{\sqrt{2}}} + \frac{\ 5q^2}{\frac{\ d}{\sqrt{2}}} + ( \frac{\ 2q^2}{\frac{\ d}{\sqrt{2}}}) - (( \frac{\ -3q^2}{d} + ( \frac{\ 2q^2}{d} + ( \frac{\ 5q^2}{d})] [/tex]

    after simplifying, I finally get:
    [tex] v= \sqrt {\frac{\ 3.31kq^2}{md}} [/tex]

    ETA: Sorry, I'm not sure how to fix the tex...
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 20, 2006 #2


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    Just scooping quickly, the distance(s) in U_i don't look right. The charge is not a distance d away from all three other charges.
  4. Jan 20, 2006 #3
    :blushing: You're right...one of the them is [tex] \sqrt{2} [/tex] away.

    Thanks Galileo!
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