- anonymous

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- jamiebookeater

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- anonymous

|dw:1361317472233:dw|

- anonymous

|dw:1361317624407:dw|

- anonymous

\[W=-\triangle U=-\int F\cdot dl=-\int_{x_1}^{x_2} Fdx=-kq_1q_2\int_{x_1}^{x_2}\frac{dr}{r^2}=-\frac{-kq_1q_2}{r}\]
I figured that \[q=\sigma A\]
I'm told that the charge densities are equal and opposite on the two large plates.

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## More answers

- anonymous

Oh and we're moving an electron from the negatively charged plate to the positively charged plate.

- anonymous

- anonymous

I calculated the electric field already

- TuringTest

what is the question?

- anonymous

I have to calculate the work done to move an electron from the negative plate to the positive plate.

- anonymous

|dw:1361318208383:dw|
This is the force I need to overcome correct?

- anonymous

|dw:1361318290447:dw|

- anonymous

My class is about to start...

- TuringTest

there is no resistance force to overcome, just the inertia of the electron moving due to the E field

- anonymous

but the electric field is pointing in the opposite direction of where the electron is moving to

- TuringTest

but the E-field points in the direction that a *positive* test charge would move.
After all, an electron wants to go from negative to positive, right? Like repels like and so forth...

- anonymous

Oh I see...

- anonymous

In that case:
\[W=-\triangle U=-q\triangle\]
q= is the charge of an electron correct?

- anonymous

- anonymous

oops!
\[W=-\triangle U=-q\triangle V\]

- TuringTest

I would say the elementary charge is \(q=-e\)\[W=-\int(-e)\vec E\cdot d\vec\ell=-e\int Edx\]Note that three negatives multiply to give a negative result.
One negative from the charge, one from the fact that a positive charge moving along the E field lines goes down in energy, and one from the fact that E and the direction of motion are antiparallel.

- anonymous

so shouldn't use the relationship I wrote earlier

- anonymous

I'm given the potential difference between the plates

- anonymous

I mean shouldn't *I use the relationship I wrote earlier

- anonymous

sorry typo

- anonymous

I mean if I'm given delta V and I know e ....I should be able to just multiply those to get delta U

- TuringTest

right, then that's all

- anonymous

oh and http://en.wikipedia.org/wiki/Joule
are \[C\cdot V=Joules \] correct?

- TuringTest

coulombs*volts=jouls, yeah

- anonymous

Thanks Max!

- TuringTest

Welcome "Jennifer/sophya"

- anonymous

LOL!!!!!! Not even close haha

- anonymous

since KE=U the kinetic energy would be the same correct?

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