Physics › Electrostatics › Capacitors With Dielectric
The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric…
The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:
$$
\mathrm{K}(\mathrm{x})=\mathrm{K}_{\mathrm{o}}+\lambda \mathrm{x}(\lambda=\mathrm{a} \text { constant })
$$
The capacitance $\mathrm{C}$, of the capacitor, would be related to its vacuum capacitance $\mathrm{C}_{\mathrm{o}}$ for the relation :
$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\mathrm{K}_{\mathrm{o}} \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$
$\left.\mathrm{C}=\frac{\lambda}{\mathrm{d} \cdot \ln \left(1+\mathrm{K}_{\mathrm{o}} \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$
$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\lambda \mathrm{d} / \mathrm{K}_{\mathrm{o}}\right.} \mathrm{C}_{\mathrm{o}}\right)$
$\left.\mathrm{C}=\frac{\lambda}{\mathrm{d} . \ln \left(1+\mathrm{K}_{\mathrm{o}} / \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$ Solution
The value of dielectric constant is given as,
$
\begin{aligned}
&\mathrm{K}=\mathrm{K}_0+\lambda \mathrm{x} \\
&\text { And, } \mathrm{V}=\int_0^{\mathrm{d}} \mathrm{Edr} \\
&\mathrm{V}=\int_0^{\mathrm{d}} \frac{\sigma}{\mathrm{K}} \mathrm{dx}
\end{aligned}
$
$
\begin{aligned}
&=\sigma \int_0^{\mathrm{d}} \frac{1}{\left(\mathrm{~K}_0+\lambda \mathrm{x}\right.} \mathrm{dx} \\
&=\frac{\sigma}{\lambda}\left[\ln \left(\mathrm{K}_0+\lambda \mathrm{d}-\ln \mathrm{K}_0\right]\right. \\
&=\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)
\end{aligned}
$
Now it is given that capacitance of vacuum $=\mathrm{C}_0$.
Thus, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$
$
\begin{aligned}
&=\frac{\sigma . \mathrm{s}}{\mathrm{v}}(\text { Let surface area of plates }=\mathrm{s}) \\
&=\frac{\sigma . \mathrm{s}}{\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)}
\end{aligned}
$
$=\mathrm{s} \lambda \cdot \frac{\mathrm{d}}{\mathrm{d}} \frac{1}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)}\left(\because\right.$ in vacuum $\left.\varepsilon_0=1\right)$
$\mathrm{c}=\frac{\lambda \mathrm{d}}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)} \cdot \mathrm{C}_0 \quad\left(\right.$ here, $\left.\mathrm{C}_0=\frac{\mathrm{s}}{\mathrm{d}}\right)$
Asked in: JEE Main 2014 (12 Apr Online)
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