Two identical capacitors `1` and `2` are connected in series to a batery as shown in figure. Capacitor `2` contains a dielectric slab of dieletric constant `k` as shown. `Q_(1)` and `Q_(2)` are the charges stored in the capacitors. Now the dielectirc slab us removed and the corresponding charges are `Q'_(1)` and `Q'_(2)`. Then

Options

A. `(Q'_(1))/(Q_(1)) = (K + 1)/(K)`
B. `(Q'_(2))/(Q_(2)) = (K + 1)/(2)`
C. `(Q'_(2))/(Q_(2)) = (K + 1)/(2K)`
D. `(Q'_(2))/(Q_(2)) = (K)/(2)`

Answer

C

Explanation

Let `C` be the capacity of each condenser without slab Fig.
`:.` When slab from capacitor 2 is removed, net capacity `= C//2`
`:. Q'_(1) = Q'_(2) = (CE)/(2)`
Before the slab is removed, `C_(1) = C. C_(2) = KC`
`C_(net) = (KC)/(K + 1)`
`Q_(1) = Q_(2) = (KCE)/(K + 1)`
`:. (Q'_(2))/(Q_(2)) = (CE (K + 1))/(2 (KCE))`
`= (K + 1)/(2K)`