Recent content by Donnyboy

hmmm the circuit is kinda correct... here is what i gathered ...... ...__________... ..|...|...|... ..V...|...|... ..|...|...|... ..|...C_1...C_2.. ..|...|...|... ..|...|...|... ..|_____|____|... ......

i'll be baffled, you are right O.O no wonder i went wrong in my ways

err helmholtz, they are in parallel lol

well the charge Q_1= C_1*V and Q_2=C_2*V applies to which the capacitors are still connected to the power supply. after it disconnects and one of them is inserted with a dielectric the capacitance changes Furthermore the Electric Field is the same when it is connected (charging) in parallel...

Homework Statement 2 parallel plate capacitors of distance separation d between the plates are connected in parallel of capacitance C_1 and C_2 respectively and charged with a voltage V after that the Voltage is disconnected and a dielectric of κ is inserted into C_2 I) Express the charge on...