The gap between the plates of a parallel plat | vedclass

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(D) Let the permittivity $\varepsilon(x)$ vary linearly with distance $x$ from one plate $(x=0)$ to the other $(x=d)$:$\varepsilon(x) = \varepsilon_1

Vedclass · Accepted Answer

$\varepsilon_0(\varepsilon_2 - \varepsilon_1)A / [d \ln(\varepsilon_2/\varepsilon_1)]$

Vedclass · Answer

(D) Let the permittivity $\varepsilon(x)$ vary linearly with distance $x$ from one plate $(x=0)$ to the other $(x=d)$:$\varepsilon(x) = \varepsilon_1 + \frac{\varepsilon_2 - \varepsilon_1}{d} x$Consider a thin elemental slab of thickness $dx$ at distance $x$. This acts as a capacitor with capacitance $dC = \frac{\varepsilon(x) A}{dx}$.Since these elemental capacitors are in series,the equivalent capacitance $C$ is given by $\frac{1}{C} = \int_0^d \frac{1}{dC} = \int_0^d \frac{dx}{\varepsilon(x) A}$.$\frac{1}{C} = \frac{1}{A} \int_0^d \frac{dx}{\varepsilon_1 + \frac{\varepsilon_2 - \varepsilon_1}{d} x}$Let $u = \varepsilon_1 + \frac{\varepsilon_2 - \varepsilon_1}{d} x$,then $du = \frac{\varepsilon_2 - \varepsilon_1}{d} dx$.$\frac{1}{C} = \frac{1}{A} \cdot \frac{d}{\varepsilon_2 - \varepsilon_1} \int_{\varepsilon_1}^{\varepsilon_2} \frac{du}{u} = \frac{d}{A(\varepsilon_2 - \varepsilon_1)} [\ln u]_{\varepsilon_1}^{\varepsilon_2} = \frac{d \ln(\varepsilon_2/\varepsilon_1)}{A(\varepsilon_2 - \varepsilon_1)}$.Therefore,$C = \frac{A(\varepsilon_2 - \varepsilon_1)}{d \ln(\varepsilon_2/\varepsilon_1)}$.

The gap between the plates of a parallel plate capacitor of area $A$ and distance between plates $d$ is filled with a dielectric whose permittivity varies linearly from $\varepsilon_1$ at one plate to $\varepsilon_2$ at the other. The capacitance of the capacitor is

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