Capacitors

JEE Main level, numerical answer, medium

25 of 180

Question 25 of 180


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)

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Answer: 15


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}
Finally equivalent capacitance

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

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