JEE Physics 2025 Question with Solution

Given: Plate area $A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2$, plate separation $d = 1.77 \, \text{mm} = 1.77 \times 10^{-3} \, \text{m}$, dielectric constants $k_1 = 5$ and $k_2 = 3$, and an additional capacitor $C_3 = 7.5 \, \text{pF}$.

Find: The effective capacitance of the combination.

The two dielectric-filled parts behave as two capacitors in series. For each part, thickness is $d/2$.

Using the parallel plate capacitor formula,

$$C = \frac{k\varepsilon_0 A}{d}$$

for the first dielectric,

$$C_1 = \frac{k_1 \varepsilon_0 A}{d/2}$$

Substituting the values,

$$C_1 = \frac{5 \times 8.85 \times 10^{-12} \times 4 \times 10^{-4}}{1.77 \times 10^{-3}/2} = 20 \times 10^{-12} \, \text{F} = 20 \, \text{pF}$$

For the second dielectric,

$$C_2 = \frac{k_2 \varepsilon_0 A}{d/2}$$

Substituting the values,

$$C_2 = \frac{3 \times 8.85 \times 10^{-12} \times 4 \times 10^{-4}}{1.77 \times 10^{-3}/2} = 12 \times 10^{-12} \, \text{F} = 12 \, \text{pF}$$

Now combine these two capacitances in series:

$$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}$$ $$\frac{1}{C_s} = \frac{1}{20} + \frac{1}{12} = \frac{2}{15}$$

Hence,

$$C_s = \frac{15}{2} \, \text{pF} = 7.5 \, \text{pF}$$

This series combination is in parallel with another capacitor of $7.5 \, \text{pF}$, so

$$C_{\text{eff}} = C_s + C_3 = 7.5 + 7.5 = 15 \, \text{pF}$$

Therefore, the effective capacitance is $15 \, \text{pF}$.

Given: $A = 4 \times 10^{-4} \, \text{m}^2$, $d = 1.77 \times 10^{-3} \, \text{m}$, $k_1 = 5$, $k_2 = 3$, $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}$, and $C_3 = 7.5 \, \text{pF}$.

Find: Effective capacitance of the complete arrangement.

First calculate half the separation:

$$d/2 = \frac{1.77 \times 10^{-3}}{2} = 8.85 \times 10^{-4} \, \text{m}$$

Now for the upper dielectric region,

$$C_1 = \frac{\varepsilon_0 A}{d/2} \cdot k_1$$ $$C_1 = \frac{(8.85 \times 10^{-12})(4 \times 10^{-4})}{8.85 \times 10^{-4}} \cdot 5$$ $$C_1 = 4 \times 10^{-12} \cdot 5 = 20 \times 10^{-12} \, \text{F} = 20 \, \text{pF}$$

For the lower dielectric region,

$$C_2 = \frac{\varepsilon_0 A}{d/2} \cdot k_2$$ $$C_2 = \frac{(8.85 \times 10^{-12})(4 \times 10^{-4})}{8.85 \times 10^{-4}} \cdot 3$$ $$C_2 = 4 \times 10^{-12} \cdot 3 = 12 \times 10^{-12} \, \text{F} = 12 \, \text{pF}$$

Since these two parts are in series,

$$\frac{1}{C_{\text{total}}} = \frac{1}{20} + \frac{1}{12}$$ $$\frac{1}{C_{\text{total}}} = \frac{3}{60} + \frac{5}{60} = \frac{8}{60} = \frac{2}{15}$$

Therefore,

$$C_{\text{total}} = \frac{15}{2} \, \text{pF} = 7.5 \, \text{pF}$$

Now add the parallel capacitor:

$$C_{\text{effective}} = C_{\text{total}} + C_3 = 7.5 + 7.5 = 15 \, \text{pF}$$

Therefore, the final answer is $15 \, \text{pF}$.