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

Let's analyze the effect of introducing an insulating slab into a parallel plate capacitor.

1. Original Capacitance (C):

Let the original separation between the plates be d = 4 mm.

C = ε₀A / d

2. Capacitance with Insulating Slab:

When a slab of thickness t = 4 mm and dielectric constant K is introduced, the capacitance (C') becomes:

C' = ε₀A / [d - t + (t/K)]

3. Restoring Original Capacitance:

To restore the original capacitance, the distance between the plates is increased by 3.2 mm. Let the new separation be d'.

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

The new capacitance (C'') is:

C'' = ε₀A / d'

We are given that C'' = C:

ε₀A / d' = ε₀A / d

d' = d

However, we are told to find the K value where the original capacitance is restored after inserting the dielectric, and increasing the distance to 7.2 mm. So:

ε₀A / d = ε₀A / [d - t + t/K], and we are told that to restore the original capacitance the distance needs to be increased to 7.2 mm. therefore:

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

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

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

4mm = 7.2mm - (4mm-4mm/K)

4 = 7.2 - 4 + 4/K

4 = 3.2 + 4/K

0.8 = 4/K

K = 4/0.8 = 5

Therefore, the dielectric constant of the material is 5.

The correct answer is:

Option 2: 5