Two parallel plate air capacitance `C`, were connected in series to a battery with `emf` `xi`. Then one of the capacitors was filled up with uniform dielectric with permittivity `epsilon`. How many times did the electric field strength in that capacitor decrease ? What amount of charge flows throgh the battery ?

Explanation

From the symmetry of the problem, the voltage across each capacitor, `Delta varphi = xi//2` and charge on each capacitor `q = C xi//2` in the absence of dielectric.
Now when the dielectric is filled up in one of the capacitor, the equivalent capacitance of the system,
`C'_(0) = (C epsilon)/(1 + epsilon)`
and the potential difference across the capacitor, which is filled with dielectric,
`Delta varphi' = (q')/(epsilon C) = (C epsilon)/((1 + epsilon)) (xi)/(C epsilon) = (xi)/((1 + epsilon))`
But `varphi alpha E`
So, as `varphi` decreases `(1)/(2) (1 + epsilon)` times, the field strength also decreases by the same factor and flow of charge, `Delta q = q' - q`
`= (C epsilon)/((1 + epsilon)) xi - (C)/(2) xi = (1)/(2) C xi ((epsilon - 1))/((epsilon + 1))`