Air filled capacitor #1 has area of plates A and gap d. It is charged to the charge Q. Capacitor #2 has area of plates A and gap d, but it is filled with dielectric of dielectric constant K=4. Capacitor #2 is also charged to the charge 2Q. At some moment capacitors are corrected, positive plate to positive plate. When equilibrium is reached, what is the charge of capacitor #2?
Q/5
3Q/5
0
Q/2
5Q
3Q/12
4Q
Q
2Q
5Q/12
12Q/5
capacitance of an air filled parallel plate capacitor = .
capacitance of parallel plate capacitor filled with dielectric of dielectric constant K = .
initial charge on first capacitor = Q
initial charge on 2nd capacitor = 2Q
When the capacitors are connected with postive plate to positive plate, flow of charge takes place until the potential becomes equal.
The common potential is given by Vcommon = sum of charges / sum of capacitance
Therefore, final charge on 2nd capacitor is
[answer]
[Correct option is last option]
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