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 ar

Question

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? [Given: Permittivity of free space ε 0 = 9 × 10 − 12 C 2 N − 1 m − 2 , the effects of the non-conducting walls on the capacitance are negligible]

TestHub · Accepted Answer

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 – 10 F Both part of the capacitor can be considered as connected in parallel, C e f f = C 1 + C 2 = 63 pF

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