## anonymous 3 years ago Question

1. anonymous

|dw:1361317472233:dw|

2. anonymous

|dw:1361317624407:dw|

3. 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.

4. anonymous

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

5. anonymous

@TuringTest

6. anonymous

I calculated the electric field already

7. TuringTest

what is the question?

8. anonymous

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

9. anonymous

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

10. anonymous

|dw:1361318290447:dw|

11. anonymous

My class is about to start...

12. TuringTest

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

13. anonymous

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

14. 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...

15. anonymous

Oh I see...

16. anonymous

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

17. anonymous

@phi

18. anonymous

oops! $W=-\triangle U=-q\triangle V$

19. 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.

20. anonymous

so shouldn't use the relationship I wrote earlier

21. anonymous

I'm given the potential difference between the plates

22. anonymous

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

23. anonymous

sorry typo

24. 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

25. TuringTest

right, then that's all

26. anonymous

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

27. TuringTest

coulombs*volts=jouls, yeah

28. anonymous

Thanks Max!

29. TuringTest

Welcome "Jennifer/sophya"

30. anonymous

LOL!!!!!! Not even close haha

31. anonymous

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