JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_{1} = 2$ , $K_{2} = 3$ , and $K_{3} = 5$ (as shown in the figure). If new capacitance is $\frac{n}{3} C$ then the value of $n$ is ______.

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Solution

Let $C_{0} = \frac{ε _{0} A}{d}$ denote the base capacitance.

Bottom half — two side-by-side dielectrics in parallel, each of plate area $A /2$ and gap $d /2$ :

$C_{2} = \frac{K _{2} ε _{0} ( A /2 )}{d /2} = 3 C_{0}, C_{3} = \frac{K _{3} ε _{0} ( A /2 )}{d /2} = 5 C_{0}$

The corresponding upper-half slabs (each of area $A /2$ , gap $d /2$ , dielectric $K_{1}$ ) are: $C_{4} = C_{1}^{'} = 2 C_{0}$ each.

Recombine into two side-by-side columns, each a series combination of the top slab and one bottom slab:

Left column: $C_{1}^{'}$ in series with $C_{2}$ $= \frac{2 C _{0} \cdot 3 C _{0}}{5 C _{0}} = \frac{6 C _{0}}{5}$ .

Right column: $C_{1}^{'}$ in series with $C_{3}$ $= \frac{2 C _{0} \cdot 5 C _{0}}{7 C _{0}} = \frac{10 C _{0}}{7}$ .

The two columns are in parallel:

$C_{eq} = \frac{6 C _{0}}{5} + \frac{10 C _{0}}{7} = \frac{42 + 50}{35} C_{0} = \frac{92}{35} C_{0}$

Setting $\frac{92}{35} C_{0} = \frac{n}{3} C_{0}$ gives $n = \frac{92 \times 3}{35} = 7.886 \approx 8$ .

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