a 5 microfarad capactior and a 10 microfarad capacitor are connected in parallel across a 50 volt battery as shown. the battery is disconnected and the two capacitors are separated from one another without altering the charge on either one. oil with a dielectric constant of 4 is inserted into the gap of the 5 microfarad capacitor, then the capacitors are reconnected.
what is the final potential across the parallel combination assuming:
_|_ _|_ _|_ 50volts
_a _ _b_ _
a = 5micro farads
b = 10 microfarads
a) the positive side of the each capacitor are connected together?
b) one capacitor was flipped so that the positive side of one was connected to the negative side of the other?
parallel C = C_1 + C_2 = Q/V + Q/V
Q = CV where C is capacitance, V is electric potential, Q is charge
parallel plate capacitor with dielectric
C = kappa[(epsilon_0*A)/d] where kappa is dielectric constant, in this case = 4, epsilon_0 is constant = 8.85*10^-12, A is area of plate, d is distance between plates
The Attempt at a Solution
since size is not specified, it is assumed area A is constant
i initially thought i would sum the two capacitor in parallel to get 15 microfarads and then sub that in for C in the last eq, along with 4 for kappa, but then came across distance d which is not given. also that equation does not involve the voltage from the battery.
then i tried using the second equation for charge Q where i summed the Q's for each capacitor getting 2.5*10^-4 + 5*10^-4 = 6.5*10^-4, after getting Q net value i was unsure about how to include kappa = 4
this is where i am at, not sure about the equations nor whether my attempts were along the right track.