Question 698: Electrostats & Capacitance

Question

A parallel plate capacitor with air between the plates has a capacitance of $$15 \mathrm{~pF}$$. the separation between the plates is $$d$$. The space between the plates is now filled with two dielectrics constant $$k_1=3$$ and thickness $$d / 3$$ while the other one has dielectric constant $$k_2=6$$ and thickness $$2 d / 3$$. Capacitance of the capacitor is now

Accepted Answer

$$\frac{\varepsilon_0 A}{d}=15 \mathrm{pF}$$ (given)
$$C_{	ext {eq }}=\frac{C_1 C_2}{C_1+C_2}=\frac{\frac{k_1 \varepsilon_0 A}{d / 3} 	imes \frac{k_2 \varepsilon_0 A}{2 d / 3}}{\frac{k_1 \varepsilon_0 A}{d / 3}+\frac{k_2 \varepsilon_0 A}{2 d / 3}}$$
$$\begin{aligned}
= & \left(\frac{\varepsilon_0 A}{d}ight)\left(\frac{3 k_1 k_2}{2 k_1+k_2}ight)=15 	imes \frac{3 	imes 3 	imes 6}{2 	imes 3+6}=\frac{135}{2} \
\Rightarrow \quad & C_{	ext {eq }}=67.5 ~p F
\end{aligned}$$

Answer

18 pF

Answer

90 pF

Answer

67.5 pF

Answer

81 pF

Electrostats & Capacitance

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