Question

When a slab of insulating material $4\text{ mm}$ thick is introduced between the plates of a parallel plate capacitor of separation $4\text{ mm}$, it is found that the distance between the plates has to be increased by $3.2\text{ mm}$ 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

Hint:

Dielectric constants ($K$) are always greater than 1. If your calculation results in $K < 1$, check if you've swapped the shift ($\Delta d$) and the thickness ($t$) in the formula.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Inserting a dielectric slab increases the capacitance. To restore the capacitance to its original value, the plate separation must be increased.
Step 2: Key Formula or Approach:
The equivalent "air-gap" reduction when a dielectric slab of thickness \( t \) and constant \( K \) is inserted is given by the formula for the "shift" or change in effective distance:
\[ \Delta d = t \left( 1 - \frac{1}{K} \right) \]
This \( \Delta d \) is exactly the amount by which the plate separation must be increased to restore original capacitance.
Step 3: Detailed Explanation:
Given:
Thickness of slab \( t = 4 \, \text{mm} \)
Required increase in separation \( \Delta d = 3.2 \, \text{mm} \)
Substituting into the formula:
\[ 3.2 = 4 \left( 1 - \frac{1}{K} \right) \]
Divide by 4:
\[ 0.8 = 1 - \frac{1}{K} \]
Rearrange to solve for \( \frac{1}{K} \):
\[ \frac{1}{K} = 1 - 0.8 = 0.2 \]
\[ K = \frac{1}{0.2} = 5 \]
Step 4: Final Answer:
The dielectric constant of the material is 5.