Potential difference

[dartmed]

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A potential difference of 10V is present between the plates of a capacitor. How much work must be done to move (6x10^18) electrons from the positive plate to the negative plate?

Options:
5J
10J
20J
30J

I got 5 J because I did PE = 1/2 q V = 1/2 (1 C) (10V) = 5 J

Note: 1 C = 6x10^18 electrons

However, the answer is 10J and I don't understand the explanation. Can someone why my calculation is wrong? Thanks!

 

[karafox]

|W| = |-change in PE| = qV = 1.6e-19*6e18*10

remember that an electric field exists between parallel plate capacitors.
work done by an electric field: |W| = change in PE = qV

 

[dartmed]

|W| = |-change in PE| = qV = 1.6e-19*6e18*10

remember that an electric field exists between parallel plate capacitors.
work done by an electric field: |W| = change in PE = qV

Thanks! I just found a post as to why we shoudn't use 1/2QV when finding work.