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.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5
A dielectric slab of thickness \( t \) and dielectric constant \( K \) in a capacitor of plate separation \( d \) gives
\( \displaystyle C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} \)
So the slab acts like a reduction of the air gap by \( t - t/K \). To restore the capacitance, the separation must be increased by exactly this amount:
\( \displaystyle t\left(1 - \frac{1}{K}\right) = 3.2\,\mathrm{mm} \)
Putting \( t = 4\,\mathrm{mm} \):
\( \displaystyle 1 - \frac{1}{K} = \frac{3.2}{4} = 0.8 \;\Rightarrow\; \frac{1}{K} = 0.2 \;\Rightarrow\; K = 5 \)