A capacitor of 10 capacitance whose plates are separated by 10 mm through air and each plate has area 4 cm² is now filled equally with two dielectric media of K₁ = 2, K₂ = 3 respectively as shown in figure. If new force between the plates is 8 N. The supply voltage is ________ V. Numerical answer type. Work it out, then open the answer.

Answer: 80. The two media sit side by side (parallel), each over half the area: C' = (K₁ + K₂)/(2)C = 2.5 × 10 = 25. Force between plates: F = (Q²)/(2C'd) = (C'V²)/(2d). 8 = 25 × 10⁻⁶ × V²2 × 10⁻² ⇒ V² = 6400 V = 80 V (using the given 10).

A capacitor of 10 capacitance whose plates are separated by 10 mm… | JEE Main 2024 (6 Apr, Shift 2)

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Dielectrics

A capacitor of $10\ \mu\text{F}$ capacitance whose plates are separated by $10\text{ mm}$ through air and each plate has area $4\text{ cm}^2$ is now filled equally with two dielectric media of $K_1 = 2, K_2 = 3$ respectively as shown in figure. If new force between the plates is $8\text{ N}$ . The supply voltage is ________ V.

Numerical answer type. Work it out, then open the answer.

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Answer: 80

The two media sit side by side (parallel), each over half the area: $C' = \frac{K_1 + K_2}{2}C = 2.5 \times 10 = 25\ \mu\text{F}$ .
Force between plates: $F = \frac{Q^2}{2C'd} = \frac{C'V^2}{2d}$ .
$8 = \frac{25 \times 10^{-6} \times V^2}{2 \times 10^{-2}} \Rightarrow V^2 = 6400$
$V = 80\text{ V}$ (using the given $10\ \mu\text{F}$ ).

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