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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorBasic capacitor analysis with dielectricHard2 minQB

PhysicsHardsingle choice

A parallel plate capacitor has plates of area A separated by distance $d$ between them. It is filled with a dielectric which has a dielectric constant that varies as $K (x) = K_{0} (1 + α x)$ where $x$ is the distance measured from one of the plates. If $(α d) < < 1,$ the total capacitance of the system is best given by the expression:

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Answer:

Solution:

Capacitance of element, $d C = \frac{K_{0} (1 + α x) ε_{0} A}{d x}$
$\int \frac{1}{d C} = \int_{0}^{d} \frac{d x}{K_{0} ε_{0} A (1 + α x)}$
$\Rightarrow \frac{1}{C} = \frac{1}{K_{0} ε_{0} A α} \ln (1 + α d)$
Given $α d ≪ 1$
$\Rightarrow \frac{1}{C} = \frac{1}{K_{0} ε_{0} A α} (α d - \frac{α^{2} d^{2}}{2})$
$\Rightarrow \frac{1}{C} = \frac{d}{K_{0} ε_{0} A} (1 - \frac{α d}{2})$
$\Rightarrow C = \frac{K_{0} ε_{0} A}{d} (1 + \frac{α d}{2})$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Basic capacitor analysis with dielectric

⏱ 2mℹ️ Source: QB

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