- #1
stunner5000pt
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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!
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!