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

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 cm3 s1. What is the value of the capacitance of the container after 10 seconds?

[Given: Permittivity of free space ε0=9×1012 C2 N1 m2, the effects of the non-conducting walls on the capacitance are negligible]

  1. 27 pF
  2. 63 pF
  3. 81 pF
  4. 135 pF

Solution

Height of the liquid column =volumebase area

h=250 cm3 s-1×10 s50 cm×5 cm=10 cm

Now capacitance of the upper part can be written as,

C1=A1ε0d=0.50-0.10×0.50×9×10-125×10-2

=0.36×1010 F

Capacitance for the lower part can be written as,

C2=KA2ε0d=3×0.10×0.5×9×10-125×10-2

C2=0.27×1010 F

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

Ceff=C1+C2

=63 pF

Asked in: JEE Advanced 2023 (Paper 1)

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