A container has a base of 50cm × 5cm and height 50cm , as shown in...

Height of the liquid column $= \dfrac{\text{volume}}{\text{base area}}$

$⇒ h = \dfrac{\left(250{cm}^{3}{s}^{- 1}\right) \times \left(10s\right)}{\left(50cm\right) \times \left(5cm\right)} = 10cm$

Now capacitance of the upper part can be written as,

${C}_{1} = \dfrac{{A}_{1}{ε}_{0}}{d} = \dfrac{\left[\left(0 . 50 - 0 . 10\right) \times 0 . 50\right] \times \left(9 \times {10}^{- 12}\right)}{\left(5 \times {10}^{- 2}\right)}$

$= 0 . 36 \times {10}^{– 10}F$

Capacitance for the lower part can be written as,

${C}_{2} = \dfrac{K{A}_{2}{ε}_{0}}{d} = \dfrac{3 \times \left(0 . 10 \times 0 . 5\right) \times \left(9 \times {10}^{- 12}\right)}{5 \times {10}^{- 2}}$

$⇒ {C}_{2} = 0 . 27 \times {10}^{– 10}F$

Both part of the capacitor can be considered as connected in parallel,

${C}_{eff} = {C}_{1} + {C}_{2}$

$= 63pF$