two parallel plates have equal and opposite charges of 0.01 coulombs. the electric field between the plates is 10^5 Newtons/coulomb and the separation between the plates is 2mm. what is the capacitance of the plates?
capacitance C = Q/V where Q is charge in coulombs, V is electric potential in volts
electric field E = V/d where d is distance in meters
Q = [epsilon_0*A/d]V where A is area and epsilon_0 is a constant = 8.85*10^-12
The Attempt at a Solution
i solved the electric field equation for V, so that V = dE and substituted that dE for V in the capacitance equation to get:
C = Q/dE = Q/(0.002*10^5) = Q/200
my questions are how do i factor in two, opposite charges or two equal charges for that matter, into the equation or am i interpreting the problem incorrectly?
is the problem asking me to find the capacitance in two different cases?
case1 --> two plates, with the same charge
case2---> two plates, opposite charges
how can i use the equation C=Q/200 and incorporate the charges of TWO plates? there is only one Q in the equation.
if that is the case, will what i attempted earlier work to find the capacitance, it seems it will?