Question

Capacitance of a capacitor made by a thin metal foil is $2\mu F$. If the foil is folded with paper of thickness $0.15mm$, dielectric constant of paper is $2.5$ and width of paper is $400mm$, the length of foil will be

• A. $0.34mm$
• B. $1.33m$
• C. $13.4m$
• D. $33.9m$

Answer 1

The correct option is D $33.9m$

If length of the foil is $l$ then
$C=\frac{K{ϵ}_0(l\times b)}{d}$
$\Rightarrow 2\times {10}^{-6}=\frac{2.5\times {8.85}^{-12}(l\times 400\times {10}^{-3})}{0.15\times {10}^{-13}}$
$l=33.9m$
$l=33.9m$.