Consider a parallel plate capacitor with plat | vedclass

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(B) Area of the square plate,$A = L^2$. Consider an elemental capacitor at distance $x$ from the top with thickness $dx$. The capacitance of this elem

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$5$

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(B) Area of the square plate,$A = L^2$. Consider an elemental capacitor at distance $x$ from the top with thickness $dx$. The capacitance of this elemental capacitor is given by: $dC = \frac{k \varepsilon_0 dA}{d} = \frac{k \varepsilon_0 (L \cdot dx)}{d}$ Since all such elemental capacitors are connected in parallel,the equivalent capacitance $C_d$ is the sum (integral) of all these elemental capacitances: $C_d = \int_0^L \frac{\varepsilon_0 L}{d} \left[1 + \left(\frac{x}{L}\right)^\alpha\right] dx$ $C_d = \frac{\varepsilon_0 L}{d} \left[ x + \frac{x^{\alpha+1}}{L^\alpha (\alpha+1)} \right]_0^L$ $C_d = \frac{\varepsilon_0 L}{d} \left[ L + \frac{L^{\alpha+1}}{L^\alpha (\alpha+1)} \right] = \frac{\varepsilon_0 L^2}{d} \left( 1 + \frac{1}{\alpha+1} \right) = \frac{\varepsilon_0 L^2}{d} \left( \frac{\alpha+2}{\alpha+1} \right)$ The capacitance in the absence of a dielectric is $C_a = \frac{\varepsilon_0 L^2}{d}$. Given the ratio $\frac{C_d}{C_a} = \frac{7}{6}$: $\frac{\frac{\varepsilon_0 L^2}{d} \left( \frac{\alpha+2}{\alpha+1} \right)}{\frac{\varepsilon_0 L^2}{d}} = \frac{7}{6}$ $\frac{\alpha+2}{\alpha+1} = \frac{7}{6}$ $6(\alpha+2) = 7(\alpha+1)$ $6\alpha + 12 = 7\alpha + 7$ $\alpha = 5$

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