A container has a base of 50 cm × 5 cm and height 50 cm , as shown in the figure. It has two parallel electrically conducting walls each of area 50 cm × 50 cm . The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant 3 at a uniform rate of 250 cm 3 s − 1 . What is the value of the capacitance of the container after 10 seconds

Step-by-step solution

Height of the liquid column = volume base area ⇒ h = 250   cm 3   s - 1 × 10   s 50   cm × 5   cm = 10   cm Now capacitance of the upper part can be written as, C 1 = A 1 ε 0 d = 0 . 50 - 0 . 10 × 0 . 50 × 9 × 10 - 12 5 × 10 - 2 = 0 . 36 × 10 – 10   F Capacitance for the lower part can be written as, C 2 = K A 2 ε 0 d = 3 × 0 . 10 × 0 . 5 × 9 × 10 - 12 5 × 10 - 2 ⇒ C 2 = 0 . 27 × 10 &