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:
When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted into a capacitor, the capacitance increases. To "restore" it to the original value, the air gap must be increased. This shift ($d'$) is related to $t$ and $K$.

Step 2: Identifying the Formula and Values:
The formula for the shift in plate separation required to restore original capacitance is: \[ \Delta d = t \left(1 - \frac{1}{K}\right) \] Given:
• $t = 4\text{ mm}$ (thickness of the slab)
• $\Delta d = 3.2\text{ mm}$ (increase in distance)

Step 3: Calculation:
Substitute the values into the formula: \[ 3.2 = 4 \left(1 - \frac{1}{K}\right) \] Divide by 4: \[ 0.8 = 1 - \frac{1}{K} \] \[ \frac{1}{K} = 1 - 0.8 = 0.2 \] \[ K = \frac{1}{0.2} = 5 \]

Step 4: Final Answer:
The dielectric constant of the material is 5.