NEET 2025 Physics Capacitance MCQ Question

To solve this problem, we start with the concept of capacitance and how dielectrics affect it. The capacitance $C$ of a parallel plate capacitor without any dielectric is given by:

$$C_0 = \frac{\varepsilon_0 A}{d}$$

where $\varepsilon_0$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the separation between the plates.

When dielectrics are introduced, the effective capacitance can be calculated by considering the individual contributions of the dielectrics. In this case, we have two dielectrics with thicknesses $\frac{3d}{8}$ and $\frac{d}{2}$, and we know their dielectric constants $K_1$ and $K_2$ respectively.

Step 1: Calculate the Effective Capacitance

The total thickness of the capacitor with the dielectrics is:

$$d_{total} = \frac{3d}{8} + \frac{d}{2} = \frac{3d}{8} + \frac{4d}{8} = \frac{7d}{8}$$

The remaining space between the dielectrics is:

$$d_{empty} = d - d_{total} = d - \frac{7d}{8} = \frac{d}{8}$$

Now we can treat the capacitor as three capacitors in series: one with $K_1$, one with $K_2$, and one with no dielectric.

1. Capacitance with dielectric $K_1$:

$$C_1 = \frac{K_1 \varepsilon_0 A}{\frac{3d}{8}} = \frac{8K_1 \varepsilon_0 A}{3d}$$

1. Capacitance with dielectric $K_2$:

$$C_2 = \frac{K_2 \varepsilon_0 A}{\frac{d}{2}} = \frac{2K_2 \varepsilon_0 A}{d}$$

1. Capacitance of the empty space:

$$C_3 = \frac{\varepsilon_0 A}{\frac{d}{8}} = \frac{8\varepsilon_0 A}{d}$$

Step 2: Calculate the Total Capacitance

The total capacitance $C$ of the three capacitors in series can be calculated using the formula for capacitors in series:

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

Substituting the values:

$$\frac{1}{C} = \frac{3d}{8K_1\varepsilon_0 A} + \frac{d}{2K_2\varepsilon_0 A} + \frac{d}{8\varepsilon_0 A}$$

To simplify, factor out $\frac{d}{\varepsilon_0 A}$:

$$\frac{1}{C} = \frac{d}{\varepsilon_0 A} \left( \frac{3}{8K_1} + \frac{1}{2K_2} + \frac{1}{8} \right)$$

Step 3: Setting Up the Equation

According to the problem, the new capacitance $C$ is double the original capacitance $C_0$:

$$C = 2C_0 = 2 \cdot \frac{\varepsilon_0 A}{d}$$

Equating the two expressions for $C$:

$$\frac{d}{\varepsilon_0 A} \left( \frac{3}{8K_1} + \frac{1}{2K_2} + \frac{1}{8} \right) = 2 \cdot \frac{\varepsilon_0 A}{d}$$

Canceling $\frac{d}{\varepsilon_0 A}$:

$$\frac{3}{8K_1} + \frac{1}{2K_2} + \frac{1}{8} = 2$$

Step 4: Solve for $K_1$ and $K_2$

Substituting $K_1 = 1.25 K_2$:

1. Replace $K_1$ in the equation:

$$\frac{3}{8(1.25 K_2)} + \frac{1}{2K_2}$$