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The plates of a parallel plate capacitor are charged upto 200 volts. A dielectirc slab of thickness $4 \mathrm{~mm}$ is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{~mm}$. The dielectric constant of the dielectric slab is
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Verified Answer
The correct answer is:
$5$
$$
\begin{aligned}
& \frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{d^{\prime}-t+\frac{t}{K}} \\
& \Rightarrow \quad d=d^{\prime}-t+\frac{t}{K} \Rightarrow d^{\prime}-d=t\left(1-\frac{1}{K}\right)
\end{aligned}
$$
Here, $d^{\prime}-d=3.2 \mathrm{~mm}, t=4 \mathrm{~mm}$
$$
\begin{aligned}
& \therefore \quad 3.2=4\left(1-\frac{1}{K}\right) 1-\frac{1}{K} \\
& \Rightarrow \quad \frac{3.2}{4}=1-\frac{1}{K} \Rightarrow 1-\frac{1}{K}=\frac{4}{5} \\
& \therefore \quad K=5 \\
&
\end{aligned}
$$
\begin{aligned}
& \frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{d^{\prime}-t+\frac{t}{K}} \\
& \Rightarrow \quad d=d^{\prime}-t+\frac{t}{K} \Rightarrow d^{\prime}-d=t\left(1-\frac{1}{K}\right)
\end{aligned}
$$
Here, $d^{\prime}-d=3.2 \mathrm{~mm}, t=4 \mathrm{~mm}$
$$
\begin{aligned}
& \therefore \quad 3.2=4\left(1-\frac{1}{K}\right) 1-\frac{1}{K} \\
& \Rightarrow \quad \frac{3.2}{4}=1-\frac{1}{K} \Rightarrow 1-\frac{1}{K}=\frac{4}{5} \\
& \therefore \quad K=5 \\
&
\end{aligned}
$$
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