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

1. Initial Capacitance:

The capacitance of a parallel plate capacitor without any dielectric is given by:

$C_0 = \frac{\epsilon_0 A}{d}$

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

2. Capacitance with Dielectric Slab:

When a dielectric slab of thickness $t = 4\times10^{-3}$ m and dielectric constant $K$ is introduced between the plates (without changing the separation $d$), the new capacitance $C'$ becomes:

$C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$

3. Capacitance with Increased Separation and Dielectric Slab:

To restore the capacitance to the original value $C_0$, the separation between the plates is increased by $\Delta d = 3.5\times10^{-3}$ m. The new separation is $d' = d + \Delta d = d + 3.5\times10^{-3}$ m. The capacitance with the increased separation and dielectric slab is $C_{new}$:

$C_{new} = \frac{\epsilon_0 A}{d' - t + \frac{t}{K}} = \frac{\epsilon_0 A}{(d + 3.5\times10^{-3}) - 4\times10^{-3} + \frac{4\times10^{-3}}{K}}$

4. Equating New and Original Capacitance:

We are given that the capacity is restored to the original value, so $C_{new} = C_0$. 

Therefore, we can equate the expressions for $C_{new}$ and $C_0$:

$\frac{\epsilon_0 A}{(d + 3.5\times10^{-3}) - 4\times10^{-3} + \frac{4\times10^{-3}}{K}} = \frac{\epsilon_0 A}{d}$

5. Simplifying the Equation by Equating Denominators:

Since $\epsilon_0 A$ is present in the numerator of both sides and is non-zero, we can equate the denominators:

$(d + 3.5\times10^{-3}) - 4\times10^{-3} + \frac{4\times10^{-3}}{K} = d$

6. Solving for the Dielectric Constant $K$:

Subtract $d$ from both sides of the equation:

$3.5\times10^{-3} - 4\times10^{-3} + \frac{4\times10^{-3}}{K} = 0$

Combine the numerical terms:

$-0.5\times10^{-3} + \frac{4\times10^{-3}}{K} = 0$

Add $0.5\times10^{-3}$ to both sides:

$\frac{4\times10^{-3}}{K} = 0.5\times10^{-3}$

Divide both sides by $10^{-3}$:

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

Solve for $K$:

$K = \frac{4}{0.5} = \frac{4}{1/2} = 4 \times 2 = 8$

Final Answer: The dielectric constant of the material will be 8.

Answer 2

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.