Question

The value of capacitor formed by a thin metallic foil is $2\mu F$ .the foil is attached to both sides of paper having a thickness of 0.015 mm. the dielectric constant of the paper is 2.5 and its breadth is 40 mm. the length of the foil used is

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

Answer 1

The correct option is C 13.4 mm

$\text{Expression for Capacitance(}C\text{)}$

$C=\frac{E_{0}A}{kd}$

$\text{Given:}$

$\begin{array}{c}C=2MF=2\times {10}^{-6}F\\ d=0.015mm=0.015\times {10}^{-3}m\\ k=2.5\\ b=40mm=40\times {10}^{-3}m\\ l=?\\ {ϵ}_0=8.85\times {10}^{-12}{C}^2/{Nm}^2\end{array}$

$\begin{array}{c}C=\frac{{ϵ}_{0}A}{kd}=\frac{{ϵ}_{0}lb}{kd}\\ \frac{Kcd}{b{\in }_0}=l\\ \frac{2.5\times 2\times {10}^{-6}\times 0.015\times {10}^{-3}}{40\times {10}^{-3}\times 8.85\times {10}^{-12}}=l\\ =13.4mm\end{array}$