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Deriving an expression for the energy required to separate charges

  1. Feb 27, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi guys, would just too make sure my derivation and insight to why is correct.
    Question: a) only

    upload_2017-2-27_18-8-0.png

    2. Relevant equations


    3. The attempt at a solution
    $$dU=-wd_{ext}$$
    $$dU=-F_{ext} \cdot dx$$
    Now as the ##F_{ext}## is in the same direction and the direction vector { have not figured how to put direction vector in} then the equation becomes ##dU=-Fdx## this assume the direction is along the x axis
    So now if I intergrate both sides ##\int_{U(a)}^{U\infty}dU=-\int_{a}^{\infty}Fdx##

    Subbing in a factoring columbs force law I get:

    $$U(\infty)-U(a)=-kq_1q_2[-\frac{1}{x}]_a^\infty$$

    Now U infity is zero because there is no force acting on it anymore and so I am left with##-U(a)## on the LHS on the RHS I am left with ##-kq_1q_2\frac{1}{a}##
    thus the two negative cancel and I am left with the electric poteinal energy. Is this correct?
     
  2. jcsd
  3. Feb 27, 2017 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If your answer has a positive sign, it is correct.


    Beyond the scope of your homework problem: (c) has a very interesting story. The naive way to improve the estimate leads to something that is not well-defined in mathematics, and you have to be careful how to do it properly to get the correct result.
     
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