Electric Fields in Parallel Plates

[salman213]

1. The magnitude of the electric field between two plates of a parallel plate capicitor is 4.7 x 10^4 N/C, if the charge on each plate increases by a a factor of 3, what happens to the electric field?

increase by a factor of 3 or 9?
decrease by a factor of 3 or 9?
not effected?



2. Electric field for plates = V/d



3. Thate the only equation i know and we did not learn any equation that reprensets charges on each plate so I am totally confused about how to relate the question...

maybe logically?
Umm since the charge on each plate increases by 3 the electric field will get stronger by the same factor of 3?

 

[cepheid]

I think you're on the right track, but you have to SHOW it. Hint: what's the definition of capacitance?

 

[salman213]

so i have to use formulas?

is therer a formula for te charge on a plate..? cause i only know

E = v/d and this ha snothing to do with increasing or decreasing distance or voltage, it has to do with charges on the plate,... :s

 

[salman213]

from what i found in my notes
i also have this equation

Ee = qV

but that q represents the charge on an electron between the two plates

so that's why i was thinking its more of a logical question that i don't get

rather than a mathematical proof .. but maybe it is i don't know...any advice?

 

[salman213]

ok i found Q= CV on the net

now i guess that means since E = V/d if E is increased then so does V and in turn so does Q ? cause its not being divided ... therefore it increases by a factor of 3 as well?

 

[cepheid]

Yeah basically. The distance between plates doesn't change, therefore the capacitance doesn't, and tripling the charge means tripling the voltage means tripling the electric field.