JEE Physics 2026 Question with Solution

Given: Three parallel plate capacitors each have area $A$ and separation $d$. Two dielectrics with constants $k_1$ and $k_2$ are arranged differently, with $k_1 > k_2$.

Find: The correct order among $C_A$, $C_B$, and $C_C$.

From the solution, the basic capacitance relation is

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

Capacitance increases with larger dielectric constant and decreases when the effective arrangement behaves more like a series combination.

For configuration $A$, the dielectrics are arranged so that the lower dielectric $k_2$ contributes more in series combination. Therefore, the overall capacitance is the minimum, so $C_A$ is smallest.

For configuration $B$, the dielectrics are symmetrically arranged such that the effective dielectric contribution is maximum. Therefore, $C_B$ is the largest.

For configuration $C$, the dielectrics are arranged side by side, giving a parallel-type combination. Its equivalent capacitance lies between cases $A$ and $B$.

Thus,

$$C_B > C_C > C_A$$

Therefore, the correct option is A.

Given: Mixed dielectric capacitor arrangements.

Find: Which configuration gives minimum, intermediate, and maximum capacitance.

Use the quick comparison rule: arrangements behaving more like series give smaller capacitance, while arrangements behaving more like parallel give larger capacitance.

• Configuration $A$ has stronger series effect involving the lower dielectric $k_2$, so it gives the smallest capacitance.
• Configuration $B$ allows the higher dielectric $k_1$ to contribute most effectively, so it gives the largest capacitance.
• Configuration $C$ is side-by-side and behaves as a parallel combination, but remains between the other two as stated in the solution.

Hence,

$$C_B > C_C > C_A$$

So the correct option is A.