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Capacitor and capacitance

  1. Jun 19, 2016 #1

    Hi everyone,

    1. The problem statement, all variables and given/known data


    It takes 18 J of energy to move a 0.30-mC charge from
    one plate of a capacitor to the other. How much
    charge is on each plate?

    2. Relevant equations

    -We/q = ΔV

    C = Q/V

    3. The attempt at a solution
    -18J/0.3.10-3 C * 15.10-6 F = Q

    Q = -0.9 C
    The problem is that in the book, the answer is 0.9 C but i found -0.9C. I had many problems with the formula

    -We/q = ΔV

    sometimes in the book they let the negative sign and sometimes not. If someone can explain me that..
     
  2. jcsd
  3. Jun 19, 2016 #2
    When you move a charge on a electric field, you do work against the electric field. So the work applied is equal to minus the work done by the electric field (assuming no changes in the kinetic energy).
     
  4. Jun 19, 2016 #3
    Thank you ....

    Can I say when the electric force is exerted in the same direction to the displacement that make a positive work ??
     
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