Work In Capacitor

[RetailBoy]

I searched this question and someone else had a problem with it too this is the question from TPRH workbook (it did have a passage with it):

A potential difference of 10 V is present between the plates of a capacitor. How much work must be done to move 6.25 × 10^18 electrons from the positive plate to the negative plate?
A. 5 J
B. 10 J
C. 20 J
D. 40 J


Answer (highlight):
B 10 J

They used the equation W=QV. I always thought it was W=1/2QV. Someone on another thread said this confidently:

If they are considering work done to move electrons in between *towards the negative plate of a capacitor, W = QV

If there is no mention of moving electrons or charge moving across the capacitor, then they probably mean the potential energy built up in the capacitor itself (1/2)QV^2, which can represent work or potential energy

so if moving charge use W=QV across capacitor. Other wise use W=1/2QV.

Sound right?

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[chiddler]

W = qΔΦ = qV

Where ΔΦ = change in potential = kq/r(final) - kq/r(initial)

i'm not sure where you got W=1/2 qV i've never seen this equation before.

 

[UrshumMurshum]

Volt has units of Joules/Coulomb.

Work has units of Joules

W = V * Q

where Q is measured in coulombs.

 

[pm1]

W = qΔΦ = qV

Where ΔΦ = change in potential = kq/r(final) - kq/r(initial)

i'm not sure where you got W=1/2 qV i've never seen this equation before.

I think he is referring to the equation of potential energy (energy stored by capacitors)
U = 1/2QV but I don't know why this one wouldn't work..

oh, may be because potential energy would have become kinetic energy half way through the plates, but when the electron gets to the negative plate then there is not kinetic energy so you have to multiply potential energy (U=QV/2) by two getting you to W=QV.

Would that be the case?

 

[chiddler]

Oh! I usually see it in 1/2 CV^2 form.

but C = Q/V

1/2 QV...that works.

er. i have an idea why this is improper, but i'm not certain of its accuracy. sorry i cannot help more.

 

[RetailBoy]

W=1/2CV^2 and W=1/2QV are both valid equations.

Here is the explanation:
First, note that 6 .25 × 10^18 electrons carry a charge of q = −1 C( This is stated at the end of the passage.) If W is the work done against the electric field, then, by definition, W = q∆φ. Since we are moving from the positive plate to the negative plate, the potential decreases, so ∆φ. = −10 V. Therefore, W = q ∆φ. = (−1 C)( −10 V) = 10 J

 

[RetailBoy]

Later in the book they talk about work again the field again in a capacitor:

W against field=PE=1/2CV^2

 

[RetailBoy]

OK I think this thread explains it 2nd post:
http://forums.studentdoctor.net/showthread.php?t=729690

 

[EnginrTheFuture]

OK I think this thread explains it 2nd post:
http://forums.studentdoctor.net/showthread.php?t=729690

It's similar to a sping

PE = 1/2 kx^2

Why is that 1/2 there? Because the force at first is F= -kx and x is zero to start.

By the end of spring compression the work you did was the AVERAGE force (0+kx)/2 times the distance traveled.

A very similar thing applies here. The force exerted on the first charge is much greater than on the last charge when theres no one on the capacitor left to push em.

So when can you assume a similar condition to a battery not a capacitor? Well if they don't give you the capacitance you have your answer They are stating that the voltage across the capacitor does not change for that one or relatively few charge's travel... whereas if you released all of them on the plate (charge/discharge of a capacitor follows PE=1/2 CV^2) the voltage does change signifiicantly so the average voltage is more appropriate.