Work done by moving electrons through electric potential?

[nghpham]

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.

 

[cepheid]

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.

 

[nghpham]

I see the error. Thank you much.

 

[truesearch]

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