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Asked in JEE Main 8th Jan 2nd Shift 2020 · Capacitance of a conductor and a capacitor
Take a strip of width dx at a distance x from the narrow edge; its gap is d+α x and its area is a dx. Strips side by side are in parallel, so their capacitances add.
C=∫₀^a(ε₀a dx)/(d+α x)=(ε₀a)/α ln (1+(α a)/d).
For α a d, ln(1+u)≈ u-(u²)/2 with u=(α a)/d.
C≈(ε₀a)/α((α a)/d-(α²a²)/(2d²))=(ε₀a²)/d(1-(α a)/(2d)).
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