Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electrostatic Potential and Capacitance

Question:

When a slab of insulating material 4mm thick is introduced between the plates of a parallel plate capacity of separation 4 mm, it is found that the distance between the plates has to be increased by 3.2 mm to restore the capacity to its original value. The dielectric constant of the material is _________.

Fill in the blank with the correct answer from the options given below.

Options:

2

5

3

7

Correct Answer:

5

Explanation:

Initial Capacitance is $ C = \frac{\epsilon_0 A}{d}$

After introduction of slab and increasing distance by 2.3 mm the capacitance becomes

$ C' = \frac{\epsilon_0 A}{d-t+\frac{t}{k}} $

According to question 

$ C' = C$

$\Rightarrow \frac{\epsilon_0 A}{d}=\frac{\epsilon_0 A}{d'-t+\frac{t}{k}}$

$\Rightarrow 4 = 7.2 - 4 +\frac{4}{k} \Rightarrow k = \frac{4}{0.8} = 5$

The correct answer is Option (2) → 5