Question

If a slab of insulating material (conceptual) 4 × 10^-3 m thick is introduced between the plates of a parallel plate capacitor, the separation between the plates has to be increased by 3.5 × 10^-3m to restore the capacity to original value. The dielectric constant of the material will be

• A. 6
• B. 8
• C. 10
• D. 12

Answer 1

The Correct Option is B

The capacitance of a parallel plate capacitor is given by:

$C = \frac{\epsilon_0 A}{d}$

where:

• $\epsilon_0$ is the permittivity of free space,
• $A$ is the area of the plates, and
• $d$ is the separation between the plates.

When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced, the capacitance becomes:

$C' = \frac{\epsilon_0 A}{d' - t + \frac{t}{K}}$ Where d' is the new separation between the plates.

A slab of insulating material with thickness $t = 4 \times 10^{-3}$ m is introduced. The separation between the plates is increased by $3.5 \times 10^{-3}$ m to restore the capacitance to its original value. Therefore, $d' = d + 3.5 \times 10^{-3}$ m and $C'=C$.

$\frac{\epsilon_0 A}{d} = \frac{\epsilon_0 A}{d + 3.5 \times 10^{-3} - 4 \times 10^{-3} + \frac{4 \times 10^{-3}}{K}}$

$d = d - 0.5 \times 10^{-3} + \frac{4 \times 10^{-3}}{K}$

$0 = -0.5 \times 10^{-3} + \frac{4 \times 10^{-3}}{K}$

$\frac{4 \times 10^{-3}}{K}=0.5 \times 10^{-3}$

$K = \frac{4}{0.5} = 8$

The correct answer is (B) 8.