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JenniferSmart1
 2 years ago
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JenniferSmart1
 2 years ago
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JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1361317472233:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1361317624407:dw

JenniferSmart1
 2 years ago
Best 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.

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0Oh and we're moving an electron from the negatively charged plate to the positively charged plate.

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I calculated the electric field already

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1what is the question?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I have to calculate the work done to move an electron from the negative plate to the positive plate.

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1361318208383:dw This is the force I need to overcome correct?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1361318290447:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0My class is about to start...

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0but the electric field is pointing in the opposite direction of where the electron is moving to

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1but 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...

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0oops! \[W=\triangle U=q\triangle V\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1I 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.

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0so shouldn't use the relationship I wrote earlier

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I'm given the potential difference between the plates

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I mean shouldn't *I use the relationship I wrote earlier

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1right, then that's all

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1coulombs*volts=jouls, yeah

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1Welcome "Jennifer/sophya"

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0LOL!!!!!! Not even close haha

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0since KE=U the kinetic energy would be the same correct?
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