Work done by moving electrons through electric potential?

  • Thread starter nghpham
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Homework Statement


A parallel-plate capacitor is charged to an electric potential of 100 V by moving 4x10^19 electrons from one plate to the other. How much work was done?


Homework Equations


How much work was done?


The Attempt at a Solution


Work is then simply equals to -q*deltaV. Q= number of electrons times charge per electron.

W= +4E19 * 1.6E-19 * 100= 640 J

But the answer I was given was 320 J. I don't see a divisible of 2 anywhere that I can account for.

Please help. Thanks.
 

Answers and Replies

  • #2
cepheid
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Welcome to PF!

Homework Statement


A parallel-plate capacitor is charged to an electric potential of 100 V by moving 4x10^19 electrons from one plate to the other. How much work was done?


Homework Equations


How much work was done?


The Attempt at a Solution


Work is then simply equals to -q*deltaV. Q= number of electrons times charge per electron.

W= +4E19 * 1.6E-19 * 100= 640 J

But the answer I was given was 320 J. I don't see a divisible of 2 anywhere that I can account for.

Please help. Thanks.
The conceptual error you're making here is that V is not constant during the movement of the charges. In other words, the 100 V was not always there from the beginning, but rather it built up slowly from 0 V as the charge accumulated. So W = q*V won't work. Instead you need W = qdV = CVdV.

If you haven't done integrals before, then use the following (totally equivalent) method: look up the equation for the total energy stored in a capacitor.
 
  • #3
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I see the error. Thank you much.
 
  • #4
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The voltage built up from zero to 100V by transferring the charge. For a capacitor Q is proportional to V so to calculate the work done you need average voltage (V/2) x charge
 

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