Figure shows two identical capacitors, C1 and C2, each of 1 μF capacitance connected to a battery of 6 V. Initially switch 'S' is closed. After sometime 'S' is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?
When switch S is closed, potential difference across capacitor is 6V
Potential across the battery, V1 = V2 = 6V
Capacitance of the capacitor, C1 = C2 = 1 
Charge on each capacitor, 
When switch S is opened, the potential difference across C1 remains 6 V, while the charge on capacitor C2 remains 6 
. After insertion of dielectric between the plates of each capacitor, the new capacitance of each capacitor becomes
i)
Now, charge on capacitor C1 ,q1’ = C1’V1 = 
Charge on capacitor C2 remains the same, i.e., 6 
ii)
Potential difference across C1 remains the same.
Potential difference across C2 is, 
