# Electric Fields in Parallel Plates

1. Jun 19, 2007

### 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 im 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?

2. Jun 19, 2007

### cepheid

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

3. Jun 19, 2007

### 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

4. Jun 19, 2007

### 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 thats why i was thinking its more of a logical question that i dont get

rather than a mathematical proof .. but maybe it is i dont know...any advice?

5. Jun 19, 2007

### 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?

6. Jun 20, 2007

### cepheid

Staff Emeritus
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.