In the given circuit, charge Q 2 on the 2 μ F capacitor changes as C is varied from 1 μF to 3 μ F . Q 2 as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale)

Step-by-step solution

∵ 1   μ F   &   2   μ F are in parallel. ∴   The equivalent capacitance of the series combination is, C eq is = 3 C C + 3 . So, the total charge supplied by the battery is, ⇒ Q = C e q E = 3 C E C + 3 . ∴ The potential difference across the parallel combination of 1   μ F and 2 μ F is, ∆ V = Q 3 = C E C + 3 . So charge on 2   μ F capacitor is, Q 2 = C 2 ∆ V = 2 C E C + 3 ⇒ Q 2 2 E = C C + 3 ⇒ Q 2 2 E = C