parallel plate capacitor capcitance question?

Question

I know that the capacitance is depended on area and distance etc etc. However, can someone explain to me the statement that as the area increases the, Elctrical field decreases.
C would increase because of the area, and V would decrease because of the E, so for this scenario would the charge be constant or increasing.

Can someone show in equation forms, please show how E would decrease.

Answer 1

If you hold the charge and the distance constant and double the area of the plates, the same charge is spread over twice the area, so the electric field density is reduced to half.

Answer 2

E = electric field, V = Voltage, d = distance of the gap and epsilon_0 is the vacuum permittivity.

->The E is independant of A.

->it is true that c is proportional to A.
you can easily derive the formula from Gauss law:

from Gauss law: EA = Q/epsilon_0
since E = V/d and Q = cV
u will come to this solution
A/d = c/epsilon_0
or c = epsilon_0 X (A/d)
the capacitance does not depend on the charge nor the charging voltage. Basically, it depends on the dimension of the two plates, A and d.

-> E = V/d, got nothing to do with A.

Sorry, i really don't understand your question.

Answer 3

The increase in area decreases the surface charge density, assuming the charge placed on the plate to be constant.This will reduce the E(E=surface ch density /epsilon). Hence the reduction in V and the increase in C.

Answer 4

C = eA/d
C reps capacitance and ar is A and d is the dist bet plates....... e is const if air is only presenst bet the plates..........this eqn is true for any PPC so u can c tat C is propotional to ar directly and thus inc with it......