Question about capacitor

[icydragon123]

Im confused about the difference between charged stored by a capacitor and its electrical potential energy difference. Specifically, the two equations:

E = (1/2)QV
PE = qV

For the question below, I mixed the two equations up.

A photon strikes the negative plate of a capacitor of capacitance C and holding charge Q, jettisoning a single electron. This electron leaves the negative plate with negligible speed, but it is immediately accelerated toward the positive plate. With what kinetic energy will it strike the positive plate? Take the mass of the electron to be me and its charge to be –e.


Answer: eQ/C

Solution: ΔPE = qV. V = Q/C for a capacitor, so qV = –eQ/C. Therefore, KEf = –ΔPE= eQ/C.

I got: ½Q2/C (because I mixed the equations up and set this equal to KE instead)

Please help. Thank you!

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[babyoreo]

First, let's get these equations straight.

The energy stored in a capacitor E = 1/2CV^2 = 1/2QV
The potential energy of a charge in an electric field PE = qV
The capacitance of a capacitor C = Q/V.

The electron is accelerated due to the electric field of the capacitor. Therefore, the KE of the electron at the positive plate (where it ended up) = the PE of the negative plate (where it first started out). Here, we use the general equation for the PE of a charge due to an electric field, PE = qV. Therefore, KE = qV. However, the answer choices do not use these terms.

First, the charge (q) of the electron is given as e. Therefore, KE = eV.

Second, the voltage (V) is determined from the capacitor. Rearranging the capacitance equation, V = Q/C. Substituting this into the above equation, KE = eQ/C.

Why is the 1/2QV equation not used?? We are not concerned about the energy stored in the capacitor; we are concerned about the energy stored in the electron (qV). Hope this helps! Where did you find this question?

 

[icydragon123]

Wow very good explanation. Thank you so much that cleared it up for me.

It was from TPR online course