# Capacitor and Potential. EN - weird solution

1. Jul 26, 2008

### 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. The problem statement, all variables and given/known data

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?

2. Relevant equations

W= -qV

3. The attempt at a solution

Plug in above, answer is (apparently) 10J

Second Question 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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

2. Jul 27, 2008

### 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.

3. Jul 27, 2008

### hulkster1988

Sorry for the second equation, since Q=CV also, it can be substituted in to 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

4. Jul 27, 2008

### 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.