Q. Capacitance of a capacitor made by a thin metal foil is $2\mu F.$ If the foil is filled with paper of thickness $0.15mm,$ dielectric constant of paper is 2.5 and width of paper is $400mm,$ then length of foil will be
A) $0.34\ m$
B) $1.33\ m$
C) $13.4\ m$
D) $33.9\ m$
Explanation
(d) If length of the foil is $l$ then $C = \frac{k\varepsilon_{0}(l \times b)}{d}$
$\Rightarrow 2 \times 10^{- 6} = \frac{2.5 \times 8.85 \times 10^{- 12}(l \times 400 \times 10^{- 3})}{0.15 \times 10^{- 3}}$
$\Rightarrow l = 33.9\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.