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Consider a parallel plate capacitor with plates in the shape of square and in $X Y$-plane. The gap between the plates is filled with dielectric material. The dielectric constant $k$ varies with $X$-axis as $k(x)=\left[1+\left(\frac{x}{L}\right)^\alpha\right]$, where $\alpha$ is a constant. Let $C_d$ and $C_a$ are capacitance in the presence of dielectric and air, respectively. If the ratio $\frac{C_d}{C_a}=\frac{7}{6}$, then the value of $\alpha$ must be
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The correct answer is:
5
Area of square plate, $A=L^2$
Consider an elemental capacitor at distance $x$ from above of thickness $d x$. Capacitance of elemental capacitor,
$$
d C=\frac{k \varepsilon_0 d A}{d}=\frac{k \varepsilon_0(L \cdot d x)}{d}
$$
All such elemental capacitors will be in parallel and equivalent capacitance will be sum of aII, so
$$
\begin{aligned}
& C_d=\frac{\varepsilon_0 L}{d} \int_0^L 1+\left(\frac{x}{L}\right)^a \cdot d x \\
& C_d=\frac{\varepsilon_0 L}{d}\left[x+\frac{\left(\frac{x}{L}\right)^{1+\alpha}}{(1+\alpha) \cdot \frac{1}{L}}\right]_0^L \\
& C_d=\frac{\varepsilon_0 L}{d}\left[L+\frac{L}{1+\alpha}\right] \Rightarrow C_d=\frac{\varepsilon_0 L^2}{d}\left(\frac{2+\alpha}{1+\alpha}\right)
\end{aligned}
$$
It is given, $\frac{C_d}{C_a}=\frac{7}{6}$
where, $C_a=$ capacitance in absence of dielectric.
$$
\begin{array}{ccc}
\Rightarrow \quad \frac{\frac{\varepsilon_0 L^2}{d}\left(\frac{2+\alpha}{1+\alpha}\right)}{\frac{\varepsilon_0 L^2}{d}}=\frac{7}{6} & \Rightarrow \frac{2+\alpha}{1+\alpha}=\frac{7}{6} \\
\Rightarrow \quad 12+6 \alpha=7+7 \alpha & \Rightarrow \quad \alpha=5
\end{array}
$$
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