Question:

A container has a base of 50 cm Γ— 5 cm and height 50 cm, as shown in the figure. It has two parallel electrically conducting walls each of area 50 cm Γ— 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 cm3sβˆ’1. What is the value of the capacitance of the container after 10 seconds? [Given: Permittivity of free space πœ–0 = 9 Γ— 10βˆ’12 C2Nβˆ’1mβˆ’2, the effects of the non-conducting walls on the capacitance are negligible]

Updated On: Jul 11, 2024
  • 27 pF

  • 63 pF

  • 81 pF

  • 135 pF

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The Correct Option is B

Solution and Explanation

\(h = \frac {250*10}{50*5} = 10 cm \) 

\[C1 = \frac{(0.40 * 0.50) * 9 * 10}{5 * 10}\]

= 0.36 * 10-10  

\(C2 = \frac {3*0.10*0.5*9*10-12}{5*10-2}\)

C2 = 0.27 * 10-10 F

C = C1 + C2

= 63pF

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Concepts Used:

Electrostatic Potential

The electrostatic potential is also known as the electric field potential, electric potential, or potential drop is defined as β€œThe amount of work that is done in order to move a unit charge from a reference point to a specific point inside the field without producing an acceleration.”

SI Unit of Electrostatic Potential:

SI unit of electrostatic potential - volt

Other units - statvolt

Symbol of electrostatic potential - V or Ο†

Dimensional formula - ML2T3I-1

Electric Potential Formula:

The electric potential energy of the system is given by the following formula:

U = 1/(4πΡº) Γ— [q1q2/d]

Where q1 and q2 are the two charges that are separated by the distance d.