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In a parallel plate air capacitor of plate separation $d$, a dielectric slab of thickness $t$ is introduced between the plates $(t < d)$. The capacitance becomes one-third of the original value. The dielectric constant of the slab will be
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Verified Answer
The correct answer is:
$\frac{t}{2 d+t}$
Equivalent Capacitance of the dielectric-air system is given by
$\frac{1}{C^{\prime}}=\left(\frac{d-t}{\varepsilon_0 A}\right)+\left(\frac{t}{K \varepsilon_0 A}\right)=\frac{d-t\left(1-\frac{1}{K}\right)}{\varepsilon_0 A}$
If new capacitance is given by one third of the old value:
$\begin{aligned}
& C=\frac{\varepsilon_0 A}{d-t\left(1-\frac{1}{K}\right)}=\frac{\varepsilon_0 A}{3 d} \\
& \Rightarrow d-t+\frac{t}{K}=3 d \\
& \Rightarrow K=\frac{t}{2 d+t}
\end{aligned}$
$\frac{1}{C^{\prime}}=\left(\frac{d-t}{\varepsilon_0 A}\right)+\left(\frac{t}{K \varepsilon_0 A}\right)=\frac{d-t\left(1-\frac{1}{K}\right)}{\varepsilon_0 A}$
If new capacitance is given by one third of the old value:
$\begin{aligned}
& C=\frac{\varepsilon_0 A}{d-t\left(1-\frac{1}{K}\right)}=\frac{\varepsilon_0 A}{3 d} \\
& \Rightarrow d-t+\frac{t}{K}=3 d \\
& \Rightarrow K=\frac{t}{2 d+t}
\end{aligned}$
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