Q. Electric charge is uniformly distributed along a long straight wire of radius $1\ mm.$ The charge per $cm$ length of the wire is $Q\ coulomb.$ Another cylindrical surface of radius $50\ cm$ and length $1\ m$ symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is
A) $\frac{Q}{\varepsilon_{0}}$
B) $\frac{100Q}{\varepsilon_{0}}$
C) $\frac{10Q}{(\pi\varepsilon_{0})}$
D) $\frac{100Q}{(\pi\varepsilon_{0})}$
Explanation
(b) Charge enclosed by cylindrical surface (length 100 $cm$) is $Q_{enc} = 100Q.$ By applying Gauss’s law $\phi = \frac{1}{\varepsilon_{0}}\left( Q_{enc.} \right) = \frac{1}{\varepsilon_{0}}(100Q)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.