JEE Main 2025 Apr 8 Shift 2, Physics Q23: Capacitance For Geometrical Figures With Dielectric

JEE Main2025Apr 8, Shift 2Physics

Space between the plates of a parallel plate capacitor of plate area $4$ cm $^{2}$ and separation of ( $d$ ) $1.77$ mm, is filled with uniform dielectric materials with dielectric constants ( $3$ and $5$ ) as shown in figure. Another capacitor of capacitance $7.5$ pF is connected in parallel with it. The effective capacitance of this combination is _____ pF.

(Given $ε_{0} = 8.85 \times 1 0^{- 12}$ F/m)

Your answer

Solution

With two dielectric slabs stacked one above the other between the plates, the arrangement is equivalent to two capacitors in series, each with plate separation $d /2$ and one of the dielectric constants.

Capacitance of top slab: $C_{1} = \frac{ε _{0} k _{1} A}{d /2} = \frac{8.85 \times 1 0 ^{- 12} \times 5 \times 4 \times 1 0 ^{- 4}}{1.77 \times 1 0 ^{- 3} /2} = 20$ pF.

Capacitance of bottom slab: $C_{2} = \frac{ε _{0} k _{2} A}{d /2} = \frac{8.85 \times 1 0 ^{- 12} \times 3 \times 4 \times 1 0 ^{- 4}}{1.77 \times 1 0 ^{- 3} /2} = 12$ pF.

Series combination: $C_{eq} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{20 \times 12}{32} = 7.5$ pF.

The parallel-connected external capacitor of $7.5$ pF adds directly: total $= 7.5 + 7.5 = 15$ pF.

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JEE Main 2025 Apr 8 Shift 2, Physics Q23: Capacitance For Geometrical Figures With Dielectric

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