A sheet of paper $40\;mm$ wide and $1.5 * 10^{-2}\;mm$ thick between metal foil of the same width is used to make a $2.04 \mu F$ capacitor. If the dielectric constant of the paper is $2.5 $, what length of the paper is required?

Given,
Width $(b) = 40 mm = 40 \; * \; 10^{-3}\;m$
Thickness$(d) = 1.5 \; * \; 10^{-2} = 1.5 \; * \; 10^{-5}\; m$
capacitance $(C) = 2.04 \; * \; 10^{-6}\;F$
Dielectric Constant $(k) = 2.5$
Length of paper $(l) = ?$

Capacitance of the capacitor $(C) = $ $ \frac{\epsilon\;A}{d} = \frac{k\; \epsilon_0\;* \; l\;*\;b}{d}$

⇒ $l = $ $\frac{2\;*\;10^{-6} \; * \; 1.5\;*\;10^{-5}}{2.5 \; * \; 8.85 \; * \; 10^{-12} \; * \; 40 \; * \; 10^{-3}}$

∴ $l = 33.9 \; m$

Thus, the required length of the paper is $33.9\;m$

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