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

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

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

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

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

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

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

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

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

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

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

JenniferSmart1
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1right, then that's all

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

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

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

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

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