Question 386: Electrostats & Capacitance

Question

A parallel plate capacitor with plate area $$\mathrm{A}$$ and plate separation $$\mathrm{d}$$ is filled with a dielectric material of dielectric constant $$K=4$$. The thickness of the dielectric material is $$x$$, where $$x < d$$.

Let $$\mathrm{C}_{1}$$ and $$\mathrm{C}_{2}$$ be the capacitance of the system for $$\chi=\frac{1}{3} d$$ and $$\mathcal{X}=\frac{2 d}{3}$$, respectively. If $$\mathrm{C}_{1}=2 \mu \mathrm{F}$$ the value of $$\mathrm{C}_{2}$$ is __________ $$\mu \mathrm{F}$$

Accepted Answer

Given,
$$
C_1=2 \mu \mathrm{F} 	ext { and } K=4
$$
$\Rightarrow C_1=\frac{\varepsilon_0 A}{\left(\frac{\frac{d}{3}}{K}+\frac{2 d}{3}ight)}=\frac{\varepsilon_0 A}{\left(\frac{d}{3 	imes 4}+\frac{2 d}{3}ight)}$
$$
\Rightarrow 2 \mu F=\frac{\varepsilon_0 A 	imes 12}{9 d}=\frac{4}{3} \frac{\varepsilon_0 A}{d}
$$
$\Rightarrow 2 \mu F=\frac{4}{3} \frac{\varepsilon_0 A}{d} \quad\left(ight.$ for $\left.x=\frac{1}{3} dight)$
$\Rightarrow \frac{\varepsilon_0 A}{d}=\frac{3}{2} \mu \mathrm{F}$
Now,
$$
\begin{aligned}
& C_2=\frac{\varepsilon_0 A}{\left(\frac{\frac{2 d}{3}}{K}+\frac{d}{3}ight)} \quad\left(	ext { for } x=\frac{2 d}{3}ight) \\
& \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}ight)} \\
& \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}ight)} \\
& \Rightarrow C_2=\frac{12 \varepsilon_0 A}{2 d+4 d}=\frac{2 \varepsilon_0 A}{d}=\frac{3}{2} 	imes 2 \\
& =3 \mu \mathrm{F}
\end{aligned}
$$

Electrostats & Capacitance

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