Three identical capacitors `C_(1),C_(2) and C_(3)` have a capacitance of `1.0muF` each and they are uncharged initially. They are connected in a circuit as shown in the figure and `C_(1)` is then filled completely with a dielectric material of relative permittivity `epsilon_(r)`. The cell electromotive force (emf) First the switch `S_(1)` is closed while the switch `S_(2)` is kept open. When the capacitor `C_(3)` is fully charged, `S_(1)` is opened and `S_(2)` is closed simultaneously. When all the capacitors reach equilibrium, the charge on `C_(3)` is found to be `5muC`. The value of `epsilon_(r)` is :

Answer

1.5 or 1.50

Explanation

Initial circuit
Initial charge on `C_3=CV_0=8 muC ` br>
Final circuit
By conservation of chanrge
`implies q_1+q_2=8 muc implies q_2=8 muC-Q_1=8 muC-5mu C=3 muC`
Also by KVl
`q^(2)/(in_rC)+q_2/(c)=q_1/c implies 3/C(1/(in_r)+1)=5/cimpliesin_r=3/2=1.5`