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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2025 · 8 April · Evening Shift · Physics · Capacitance

Space between the plates of a parallel plate capacitor of plate...

Space between the plates of a parallel plate capacitor of plate area 4 ~cm^2 and separation of (d) 1.77 mm, is filled
Space between the plates of a parallel plate capacitor of plate area 4 cm24 \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 ε0=8.85×10−12 F/m)\left.\varepsilon_0=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)

Correct answer

0

Step-by-step solution

C1=5×4×10−4×8.85×10−121.772×10−3=20pFC2=3×4×10−4×8.85×10−121.772×10−3=12pFCeq=C1C2C1+C2=12×2012+20=7.5pF\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}Finallyequivalentcapacitance(Ceq)final =7.5+7.5=15pFFinally equivalent capacitance\left(\mathrm{C}_{\mathrm{eq}}\right)_{\text{final }}=7.5+7.5=15 \mathrm{pF}
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