A parallel plate capacitor has plates of area | vedclass

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Question

(A) Since the dielectric constant $k(x)$ varies with distance $x$,we consider a thin elemental slice of thickness $dx$ at a distance $x$ from one plat

Vedclass · Accepted Answer

$\frac{AK \varepsilon_{0}}{d}\left(1+\frac{\alpha d}{2}\right)$

Vedclass · Answer

(A) Since the dielectric constant $k(x)$ varies with distance $x$,we consider a thin elemental slice of thickness $dx$ at a distance $x$ from one plate.The capacitance of this elemental slice is $dC = \frac{\varepsilon_0 k(x) A}{dx} = \frac{\varepsilon_0 K(1+\alpha x) A}{dx}$.Since these elemental capacitors are connected in series,the equivalent capacitance $C$ is given by:$\frac{1}{C} = \int_0^d \frac{1}{dC} = \int_0^d \frac{dx}{\varepsilon_0 K A (1+\alpha x)}$.Integrating this,we get:$\frac{1}{C} = \frac{1}{\varepsilon_0 K A} \left[ \frac{\ln(1+\alpha x)}{\alpha} \right]_0^d = \frac{1}{\alpha \varepsilon_0 K A} \ln(1+\alpha d)$.Using the Taylor expansion $\ln(1+u) \approx u - \frac{u^2}{2}$ for $u = \alpha d $\frac{1}{C} \approx \frac{1}{\alpha \varepsilon_0 K A} (\alpha d - \frac{(\alpha d)^2}{2}) = \frac{d}{\varepsilon_0 K A} (1 - \frac{\alpha d}{2})$.Taking the reciprocal to find $C$ and using the binomial approximation $(1-u)^{-1} \approx 1+u$:$C \approx \frac{\varepsilon_0 K A}{d} (1 - \frac{\alpha d}{2})^{-1} \approx \frac{\varepsilon_0 K A}{d} (1 + \frac{\alpha d}{2})$.

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