JEE MAIN - Physics (2023 - 6th April Morning Shift - No. 27)
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 $$x < d$$.
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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 __________ $$\mu \mathrm{F}$$
Answer
3
Explanation
Given,
$$
C_1=2 \mu \mathrm{F} \text { and } K=4
$$
$\Rightarrow C_1=\frac{\varepsilon_0 A}{\left(\frac{\frac{d}{3}}{K}+\frac{2 d}{3}\right)}=\frac{\varepsilon_0 A}{\left(\frac{d}{3 \times 4}+\frac{2 d}{3}\right)}$
$$ \Rightarrow 2 \mu F=\frac{\varepsilon_0 A \times 12}{9 d}=\frac{4}{3} \frac{\varepsilon_0 A}{d} $$
$\Rightarrow 2 \mu F=\frac{4}{3} \frac{\varepsilon_0 A}{d} \quad\left(\right.$ for $\left.x=\frac{1}{3} d\right)$
$\Rightarrow \frac{\varepsilon_0 A}{d}=\frac{3}{2} \mu \mathrm{F}$
Now,
$$ \begin{aligned} & C_2=\frac{\varepsilon_0 A}{\left(\frac{\frac{2 d}{3}}{K}+\frac{d}{3}\right)} \quad\left(\text { for } x=\frac{2 d}{3}\right) \\\\ & \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}\right)} \\\\ & \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}\right)} \\\\ & \Rightarrow C_2=\frac{12 \varepsilon_0 A}{2 d+4 d}=\frac{2 \varepsilon_0 A}{d}=\frac{3}{2} \times 2 \\\\ & =3 \mu \mathrm{F} \end{aligned} $$
$\Rightarrow C_1=\frac{\varepsilon_0 A}{\left(\frac{\frac{d}{3}}{K}+\frac{2 d}{3}\right)}=\frac{\varepsilon_0 A}{\left(\frac{d}{3 \times 4}+\frac{2 d}{3}\right)}$
$$ \Rightarrow 2 \mu F=\frac{\varepsilon_0 A \times 12}{9 d}=\frac{4}{3} \frac{\varepsilon_0 A}{d} $$
$\Rightarrow 2 \mu F=\frac{4}{3} \frac{\varepsilon_0 A}{d} \quad\left(\right.$ for $\left.x=\frac{1}{3} d\right)$
$\Rightarrow \frac{\varepsilon_0 A}{d}=\frac{3}{2} \mu \mathrm{F}$
Now,
$$ \begin{aligned} & C_2=\frac{\varepsilon_0 A}{\left(\frac{\frac{2 d}{3}}{K}+\frac{d}{3}\right)} \quad\left(\text { for } x=\frac{2 d}{3}\right) \\\\ & \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}\right)} \\\\ & \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}\right)} \\\\ & \Rightarrow C_2=\frac{12 \varepsilon_0 A}{2 d+4 d}=\frac{2 \varepsilon_0 A}{d}=\frac{3}{2} \times 2 \\\\ & =3 \mu \mathrm{F} \end{aligned} $$
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