The area of the plates of a parallel plate ca | vedclass

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Question

(A) Consider a small elemental strip of thickness $dy$ at a distance $y$ from one plate. This strip acts as a capacitor with capacitance $dC = \frac{K

Vedclass · Accepted Answer

$\pi \varepsilon_0 \lambda A / 2d$

Vedclass · Answer

(A) Consider a small elemental strip of thickness $dy$ at a distance $y$ from one plate. This strip acts as a capacitor with capacitance $dC = \frac{K \varepsilon_0 A}{dy}$.Since these elemental capacitors are in series,the equivalent capacitance $C$ is given by $\frac{1}{C} = \int_0^d \frac{dy}{K \varepsilon_0 A}$.Substituting $K = \lambda \sec(\pi y / 2d)$,we get $\frac{1}{C} = \frac{1}{\varepsilon_0 A \lambda} \int_0^d \cos(\pi y / 2d) dy$.Evaluating the integral: $\int_0^d \cos(\pi y / 2d) dy = [\frac{2d}{\pi} \sin(\pi y / 2d)]_0^d = \frac{2d}{\pi} \sin(\pi/2) = \frac{2d}{\pi}$.Thus,$\frac{1}{C} = \frac{1}{\varepsilon_0 A \lambda} \cdot \frac{2d}{\pi} = \frac{2d}{\pi \varepsilon_0 \lambda A}$.Therefore,$C = \frac{\pi \varepsilon_0 \lambda A}{2d}$.

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