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^(/))/(Q_(1))=(K+1)/(K)`
B. `(Q^(/))/(Q_(2))=(K+1)/(2)`
C. `(Q^(/)2)/(Q_(2))=(K+1)/(2K)`
D. `(Q^(/)2)/(Q_(2))=(K)/(2)`

Answer

C

Explanation

`Q_(1) =Q_(2) =(CE)/(2)` , Before the slab is removed
`C_(1)=C and C_(2)=KC,C_("net")=((k)/(k+1))C`
`(Q_(2))/(Q_(2))=(k+1)/(2k)`.