Q. 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, the 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 them
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C = \ \frac{K\varepsilon_{0}(l \times b)}{d}$
$\Rightarrow 2 \times 10^{- 6} = \ \frac{2.5 \times 8.85 \times 10^{- 12}(l \times 400 \times 10^{- 3})}{0.15 \times 10^{- 3}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ l = 33.9\ m.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.