Question 346: Electrostats & Capacitance

Question

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 	imes 10^{-12} $ F/m)

Accepted Answer

$$\begin{aligned}
& \mathrm{C}_1=\frac{5 	imes 4 	imes 10^{-4} 	imes 8.85 	imes 10^{-12}}{\frac{1.77}{2} 	imes 10^{-3}}=20 \mathrm{pF} \
& \mathrm{C}_2=\frac{3 	imes 4 	imes 10^{-4} 	imes 8.85 	imes 10^{-12}}{\frac{1.77}{2} 	imes 10^{-3}}=12 \mathrm{pF} \
& \mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{12 	imes 20}{12+20}=7.5 \mathrm{pF}
\end{aligned}$$
Finally equivalent capacitance
$\left(\mathrm{C}_{	ext {eq }}ight)_{	ext {final }}=7.5+7.5=15 \mathrm{pF}$

Electrostats & Capacitance

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