Question

When a slab of insulating material 4 mm thick is introduced between the plates of a parallel plate capacitor 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.

• A. 2
• B. 5
• C. 3
• D. 7

Answer 1

The Correct Option is B

This analysis examines the impact of inserting an insulating slab into a parallel plate capacitor.

1. Initial Capacitance (C):

The initial plate separation is denoted as d = 4 mm.

The original capacitance is calculated as C = ε₀A / d.

2. Capacitance with Insulating Slab:

Upon introducing a slab of thickness t = 4 mm and dielectric constant K, the capacitance (C') is given by:

C' = ε₀A / [d - t + (t/K)]

3. Restoration of Original Capacitance:

To restore the original capacitance, the plate separation is increased by 3.2 mm, resulting in a new separation d'.

d' = d + 3.2 mm = 4 mm + 3.2 mm = 7.2 mm.

The capacitance with this new separation (C'') is:

C'' = ε₀A / d'.

The condition for restoring the original capacitance is C'' = C.

This implies ε₀A / d' = ε₀A / d, leading to d' = d.

However, the objective is to determine the dielectric constant K that restores the original capacitance when the distance is increased to 7.2 mm after inserting the dielectric. Therefore, the equation is:

ε₀A / d = ε₀A / [d - t + t/K].

Given that the original capacitance is restored by increasing the distance to 7.2 mm, we have:

d = d' - (d - (d - t + t/K))

d = d' - (t - t/K)

Substituting the known values: 4 mm = 7.2 mm - (4 mm - 4 mm/K).

Simplifying the equation: 4 = 7.2 - 4 + 4/K

4 = 3.2 + 4/K

0.8 = 4/K

Solving for K: K = 4 / 0.8 = 5.

The dielectric constant of the material is thus 5.

The correct answer is:

Option 2: 5