work to move an electron in a capacitor

[drbeck]

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i have come across several questions in the TPR workbook about the work required to move an electron from a negative plate to a positive plate. I thought it was change in potential energy meaning either 1/2 CV^2 or
1/2 QV

the question reads:

A potential difference of 10V is present b/w the plates of a capacitor. How much work must be done to move 6.25 X 10^18 elelctrons from the neg plate to the pos plate? (The passage states 6.25 X 10^18 electrons equals -1Coulomb.)

TPR uses Work= qV for the answer (5 J) and i'm not sure when to use that equation versus potential energy equation of a capacitor (1/2 QV).

Another question states: How much work is required to insert a dielectic when a battery is still connected? The answer uses 1/2CV^2 w/ dielectric - 1/2CV^2 w/o dielectric.

Can someone please clarify when to use which equation?

 

[HawkeyePostOp]

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)CV^2, which can represent work or potential energy

*
U = W = 1/2(QV) = 1/2(CV^2) = (Q^2/2C)

 

[niranjan162]

why is the answer 5 joules. Shouldnt it be 10joules. W=qv, q=-1columbs *10 v (10j/c)= -10joules, because ur doing negative work moving an electron from negative to positive.

 

[HawkeyePostOp]

I wanna see this solution

is it W = QV - (1/2)QV?

 

[Soccerdoc11]

The answer should definitely only include the potential difference across the plates (V) * The charge of the electron which you are saying is -1 coulomb according to the equation W(energy) = (Q)(V). I would like to see the explanation given also...

 

[drbeck]

sorry! i stated the wrong answer. it is W=qV and the answer is +10J. i stated the answer i got wrong.

 

[drbeck]

W= (-1C)(-10V) = +10J work against the field means -10V.