JEE Physics 2023 Question with Solution

Given: Initial capacitance is $15 \, \text{pF}$. The plate separation becomes twice, and a dielectric of constant $K = 3.5$ is inserted.

Find: The required numerical value asked in the problem.

For a parallel plate capacitor,

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

With dielectric constant $K$ and new separation $2d$, the new capacitance becomes

$$C' = K \frac{\epsilon_0 A}{2d} = \frac{K}{2} \cdot \frac{\epsilon_0 A}{d} = \frac{K}{2}C$$

Substituting $K = 3.5$ and $C = 15 \, \text{pF}$,

$$C' = \frac{3.5}{2} \times 15 = 26.25 \, \text{pF}$$

From the extracted solution, the problem further interprets this as $C' = \frac{x}{4} \, \text{pF}$. Therefore,

$$\frac{x}{4} = 26.25$$

So,

$$x = 26.25 \times 4 = 105$$

Therefore, the required numerical answer is $105$.

Given: Initial capacitance $C = 15 \, \text{pF}$, dielectric constant $K = 3.5$, and new separation is $2d$.

Find: The required numerical value.

Use the direct change rule for a parallel plate capacitor:

$$C' = K \frac{C}{n}$$

where $n$ is the factor by which separation increases. Here $n = 2$.

Thus,

$$C' = 3.5 \times \frac{15}{2} = 26.25 \, \text{pF}$$

Using the extracted relation

$$C' = \frac{x}{4} \, \text{pF}$$

we get

$$x = 4 \times 26.25 = 105$$

Hence, the required answer is $105$.