Work done by moving electrons through electric potential?

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SUMMARY

The work done in charging a parallel-plate capacitor to an electric potential of 100 V by moving 4x1019 electrons is not simply calculated using W = q * V due to the non-constant voltage during charge movement. The correct approach involves integrating the voltage, leading to the formula W = (1/2) * Q * V, resulting in a total work of 320 J. This discrepancy arises because the voltage builds from 0 V to 100 V, necessitating the use of average voltage in calculations.

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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.
 
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nghpham said:

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.
 
I see the error. Thank you much.
 
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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