Capacitor and Potential. EN - weird solution

[hulkster1988]

There are these two homework questions that I think are asking the exact same thing, yet use different equations (and get different answers for them)

First Question 1. Homework Statement

A potential difference of 10V is present between the plates of a capacitor. How much work must be done to move 6.25x10^18 electrons from the negative plate to the positive plate?

Homework Equations

W= -qV

The Attempt at a Solution

Plug in above, answer is (apparently) 10J


Second Question 1. Homework Statement

4 uC of negative charge is transferred from one plate of a 8uF capacitor to the other plate. How much work was done by the electric field in this charging process?

Homework Equations

W= PE stored in capacitor
W= -PE
= 1/2 QV

The Attempt at a Solution

Plug it in, I get -1J

The thing is the above equation is for work stored in the capacitor, and the 1/2 is there because it is the average of V as it is charged from 0 to V.

So why does the first question use the formula for work that is for a charge, q, placed into a field? Shouldn't it to use the 1/2QV formula?

I'm definitely stumped on this one

 

[dk_ch]

I don't understand how u get PE in the second case as 1 J using equation PE = 1/2QV, it is possible if capacitance of the capacitor, C = 8pF.

In the charging process of a capacitor charge increases from 0 to Q so 1/2QV equation is valid in the second case . but in the first case charge (Q)transfer occurs against a potential V so workdone shld be QV only

One ambiguity lies in the second question where both "charge transfer" as well as "charging" have been stated. This misleads the concept.

 

[hulkster1988]

Sorry for the second equation, since Q=CV also, it can be substituted into make an equation of PE = Q^2 / 2C.

So are you saying that in the since case there is no V between the capacitors, and thus the 1/2 is used?

ALso, when it says electrons are "transferred", does that mean through the circuit or a direct spark through the plates?

Thanks for the help

 

[Defennder]

The factor of 1/2 comes from a derivation of work done to charge a capacitor from 0V using elementary calculus. When a capacitor is charged, the electrons are transferred in the circuit, not across the the dielectric between the plates. We're assuming a perfect dielectric with 0 current leakage between capacitor plates here.