When as labofinsulating material 4 mm thick is introduced between the plates of a parallel plate capacitor of separation 4mm, it is found that the distance between the plates has to be increased by 3.2mm to restore the capacity to its original value. The dielectric constant of the material is _________ .

The correct option is 5.

Let us understand the step-by-step physical principles and mathematical derivation that lead to this result.

1. Original Capacitance of the Parallel Plate Capacitor:
Initially, the plates of the capacitor are separated by air/vacuum with a distance $d$. The capacitance $C_0$ is given by the formula:
$C_0=\frac{{\epsilon }_{0}A}{d}$
where ${\epsilon }_0$ is the permittivity of free space and $A$ is the area of the capacitor plates. Here, the initial separation is $d=4\text{mm}$.

2. Capacitance with a Dielectric Slab:
When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced between the plates, and the separation is increased to a new distance $d^{\prime }$, the new capacitance $C$ is given by:
$C=\frac{{\epsilon }_{0}A}{d^{\prime }-t+\frac{t}{K}}$
where $t=4\text{mm}$ is the thickness of the slab.

3. Restoring the Original Capacitance:
To restore the capacitance to its original value ($C=C_0$), the denominators of both equations must be equal:
$d^{\prime }-t+\frac{t}{K}=d$
Rearranging the terms, we can find the shift in the plate distance ($d^{\prime }-d$):
$d^{\prime }-d=t(1-\frac{1}{K})$

4. Calculating the Dielectric Constant (K):
We are given that the separation between the plates has to be increased by $3.2\text{mm}$ to restore the original capacitance. Therefore, $d^{\prime }-d=3.2\text{mm}$ and the thickness of the slab is $t=4\text{mm}$. Substituting these values:
$3.2=4(1-\frac{1}{K})$
Divide both sides by 4:
$0.8=1-\frac{1}{K}$
Rearranging to solve for $K$:
$\frac{1}{K}=1-0.8$
$\frac{1}{K}=0.2$
$K=\frac{1}{0.2}=5$
Thus, the dielectric constant of the material is 5.