Three identical capacitors `C_(1),C_(2)` and `C_(3)` have a capacitance of `1.0 mu F` 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) `V_(0)=8V`. 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 `5 mu c`. The value of `epsilon_(r)` = ____________.

Answer

`1.50`

Explanation

When `S_(1)` is closed and `S_(2)` is open, then capacitor `C_(3) (1 muF)` gets charged up to a potential difference of 8 V. Hence using Q = CV, we can understand that charge stored by `C_(3)` is `8muC`.
Let C be the equivalent capacity for `C_(1) and C_(2)`, then
`(1)/(C )=(1)/(C_(1))+(1)/(C_(2))=(1)/(epsilon_(r ))=(1)/(1)=(1+epsilon_(r ))/(epsilon_(r )) rArr C=(epsilon_(r ))/(1+epsilon_(r ))`
When switch `S_(1)` is opened and `S_(2)` is closed, then final charge with `C_(3)` is `5muC`, hence remaining `3muC` must be stored in equivalent C.
C and `C_(3)` are in parallel, hence potential difference `(Q//C)` must be the same.
`(3)/(C )=(5)/(C_(3)) rArr (3)/(C )=(5)/(1)=5C=3`
`rArr 5 (epsilon_(r ))/(1+epsilon_(r ))=3`
`rArr 5epsilon_(r )=3+3epsilon_(r )`
`rArr epsilon_(r )=1.5`