Question

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

Answer #1

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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