Figure shows two idential capacitors, `C_1 and C_2`, each of `1mu F` 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?

Explanation

Initially,when the switch S is closed in the arrangment shown, we have
`C_1 = C_2 =1 muF, V_1 =V_2 = 6 V and Q_1 = Q_2 = CV = 6muF`
When the switch S is left open, the capacitor `C_1` is still connected to the battery but capacitor `C_2` is now disconnected from the battery.
On inserting dielectric slabs of dielectric constant K = 3, so as to fill completely the space between the plates of two capacitors,their capacity changes to `C._1 = C._2 = KC_1 or KC_2 = 3 xx 1 muF = 3 muF`
(i) As capacitor `C_1` is still connected to the battery, its new voltage `V._1 = V_1 =6 V` and new charge `Q._1 = C._1 V._1 = 3muF xx 6 V = 18 muC`
(ii) As capacitor `C_2` is now disconnected from the battery, its new charge `Q._2= Q_2 = 6muC` and new potential difference across its plates
`V._2 = (Q._2)/(C._2) = (6muC)/(3 muF) = 2V`.