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JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
dw:1361317472233:dw
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
dw:1361317624407:dw
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
\[W=\triangle U=\int F\cdot dl=\int_{x_1}^{x_2} Fdx=kq_1q_2\int_{x_1}^{x_2}\frac{dr}{r^2}=\frac{kq_1q_2}{r}\] I figured that \[q=\sigma A\] I'm told that the charge densities are equal and opposite on the two large plates.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Oh and we're moving an electron from the negatively charged plate to the positively charged plate.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@TuringTest
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I calculated the electric field already
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
what is the question?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I have to calculate the work done to move an electron from the negative plate to the positive plate.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
dw:1361318208383:dw This is the force I need to overcome correct?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
dw:1361318290447:dw
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
My class is about to start...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
there is no resistance force to overcome, just the inertia of the electron moving due to the E field
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
but the electric field is pointing in the opposite direction of where the electron is moving to
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
but the Efield 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...
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Oh I see...
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
In that case: \[W=\triangle U=q\triangle\] q= is the charge of an electron correct?
 one year ago

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@phi
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
oops! \[W=\triangle U=q\triangle V\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
so shouldn't use the relationship I wrote earlier
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I'm given the potential difference between the plates
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I mean shouldn't *I use the relationship I wrote earlier
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
sorry typo
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I mean if I'm given delta V and I know e ....I should be able to just multiply those to get delta U
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
right, then that's all
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
oh and http://en.wikipedia.org/wiki/Joule are \[C\cdot V=Joules \] correct?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
coulombs*volts=jouls, yeah
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Thanks Max!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
Welcome "Jennifer/sophya"
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
LOL!!!!!! Not even close haha
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
since KE=U the kinetic energy would be the same correct?
 one year ago
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