Capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$, dielectric constant of paper is $2.5$ and width of paper is $400\,mm$, then length of foil will be.....$m$
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(d)If length of the foil is $l$ then $C = \frac{{k{\varepsilon _0}(l \times b)}}{d}$
$==>$ $2 \times {10^{ - 6}} = \frac{{2.5 \times 8.85 \times {{10}^{ - 12}}(l \times 400 \times {{10}^{ - 3}})}}{{0.15 \times {{10}^{ - 3}}}}$
$==>$ $ l = 33.9 \,m$
$==>$ $2 \times {10^{ - 6}} = \frac{{2.5 \times 8.85 \times {{10}^{ - 12}}(l \times 400 \times {{10}^{ - 3}})}}{{0.15 \times {{10}^{ - 3}}}}$
$==>$ $ l = 33.9 \,m$
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