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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2025
  • Hard
  • Numerical


Space between the plates of a parallel plate capacitor of plate area \(4 \mathrm{~cm}^2\) 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. \(\left(\right.\) Given \(\left.\varepsilon_0=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)\)

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Answer & Solution

Correct Answer

(C) 15

Step-by-step Solution

Detailed explanation


\(\begin{aligned} & \mathrm{C}_1=\frac{5 \times 4 \times 10^{-4} \times 8.85 \times 10^{-12}}{\frac{1.77}{2} \times 10^{-3}}=20 \mathrm{pF} \\ & \mathrm{C}_2=\frac{3 \times 4 \times 10^{-4} \times 8.85 \times 10^{-12}}{\frac{1.77}{2} \times 10^{-3}}=12 \mathrm{pF} \\ & \mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{12 \times 20}{12+20}=7.5 \mathrm{pF}\end{aligned}\)

Finally equivalent capacitance
\(\left(\mathrm{C}_{\mathrm{eq}}\right)_{\text {final }}=7.5+7.5=15 \mathrm{pF}\)
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