Here's the question you clicked on:
JenniferSmart1
Question
|dw:1361317472233:dw|
|dw:1361317624407:dw|
\[W=-\triangle U=-\int F\cdot dl=-\int_{x_1}^{x_2} Fdx=-kq_1q_2\int_{x_1}^{x_2}\frac{dr}{r^2}=-\frac{-kq_1q_2}{r}\] I figured that \[q=\sigma A\] I'm told that the charge densities are equal and opposite on the two large plates.
Oh and we're moving an electron from the negatively charged plate to the positively charged plate.
I calculated the electric field already
what is the question?
I have to calculate the work done to move an electron from the negative plate to the positive plate.
|dw:1361318208383:dw| This is the force I need to overcome correct?
|dw:1361318290447:dw|
My class is about to start...
there is no resistance force to overcome, just the inertia of the electron moving due to the E field
but the electric field is pointing in the opposite direction of where the electron is moving to
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...
In that case: \[W=-\triangle U=-q\triangle\] q= is the charge of an electron correct?
oops! \[W=-\triangle U=-q\triangle V\]
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.
so shouldn't use the relationship I wrote earlier
I'm given the potential difference between the plates
I mean shouldn't *I use the relationship I wrote earlier
I mean if I'm given delta V and I know e ....I should be able to just multiply those to get delta U
right, then that's all
oh and http://en.wikipedia.org/wiki/Joule are \[C\cdot V=Joules \] correct?
coulombs*volts=jouls, yeah
Welcome "Jennifer/sophya"
LOL!!!!!! Not even close haha
since KE=U the kinetic energy would be the same correct?