Fig. shows two identical capacitors `C_(1) and C_(2)` each of `1 muF` capacitance, connected to a battery of 6V Initially,swich S is closed. After sometime, S is left open and dielectric slabs of dielectric constant K = 3 are instered to fill compelelty the space between the plates of two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after teh slabs are inserted ?

Explanation

In Fig, when switch S is closed, both `C_(1) and C_(2)` are charged to 6V potential,
i.e, `V_(1) = 6V and V_(2) = 6V`
when S is left open,
`C_(1)` is still connected to battery. On introducing slab of dielectric constant. K = 3.
`C'_(1) = KC_(1) = 3mu F = 3xx10^(-6) F`,
`V'_(1) = V_(1) = 6V`
`Q_(1) = C'_(1) V'_(1) = 3xx10^(-6) xx6 = 18xx10^(-6)C`.
However `C_(2)` is disconneced from battery now, when S is left open. Due to introduced of slab,
`C'_(2) = KC_(2) = 3xx10^(-6)F`
`Q'_(2) = Q_(2) = C_(2) V_(2) = 1xx10^(-6)xx6 = 6xx10^(-6) C`
`:. V_(2) = (Q_(2))/(C_(2)) = (6xx10^(6))/(3xx10^(-6)) = 2V`