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Question:

Three identical capacitors C1, C2, and C3 have a capacitance of 1.0µF each and they are uncharged initially. They are connected in a circuit as shown in the figure and C1 is then filled completely with a dielectric material of relative permittivity εr. The cell electromotive force (emf) Vo=8V. First, the switch S1 is closed while the switch S2 is kept open. When the capacitor C3 is fully charged, S1 is opened and S2 is closed simultaneously. When all the capacitors reach equilibrium, the charge on C3 is found to be 5µC. The value of 4 × εr is:

Solution:

The capacitor C3 charges to 8µC as the switch S1 is closed.
Ref. image 1
But when switch S1 is opened and S2 is closed, then capacitor C3 is charged to 5µC, thus the net charge 8µC - 5µC = 3µC resides on C1 and C2, as they are connected in series.
Ref. image 2
So, Apply the Kirchhoff loop,
q1/C1 + q2/C1εr = q3/C3
q/C(1 + 1/εr) = 5
3(1 + 1/εr) = 5
εr = 1.5