JEE MAIN - Physics (2025 - 8th April Evening Shift - No. 23)
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Space between the plates of a parallel plate capacitor of plate area 4 cm2 and separation of 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 $ \epsilon_0 = 8.85 \times 10^{-12} $ F/m)
Explanation
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$$\begin{aligned} & \mathrm{C}_1=\frac{5 \times 4 \times 10^{-4} \times 8.85 \times 10^{-12}}{\frac{1.77}{2} \times 10^{-3}}=20 \mathrm{pF} \\ & \mathrm{C}_2=\frac{3 \times 4 \times 10^{-4} \times 8.85 \times 10^{-12}}{\frac{1.77}{2} \times 10^{-3}}=12 \mathrm{pF} \\ & \mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{12 \times 20}{12+20}=7.5 \mathrm{pF} \end{aligned}$$
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Finally equivalent capacitance
$\left(\mathrm{C}_{\text {eq }}\right)_{\text {final }}=7.5+7.5=15 \mathrm{pF}$
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