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More Gauss Law

  • #1
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An electron remains stationary in na electric field directed downward in teh Earth's gravitational field. If the electric field is due to charge on the two large conducting plates oppositely charged and separated by 2.3cm what is the surface charge density assumed to be uniform on the plates?

force on the elctric due to the plates is qE
and E = [itex] \sigma/epsilon_{0} [/itex]
force of gfravity is mg
so [tex] \sigma = \frac{mg\epsilon_{0}}{q} [/tex]
and this is the surface charge density on the plates, isn't it ?

Does the distnace (which does NOT appear on the formula and is totally irrelevant i think) matter?
Please do advise on this matter!
Thank you!
 

Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
Gold Member
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I say distance doesn't matter.

Maybe, just maybe, the distance was given to convince you that the infinite-plate approximation was justified.

Btw - the thread title advertises Gauss law but I don't see none. I was ready for some good ol' Gauss action here. kinda disapointed :P
 
Last edited:

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