Question

Electric charge is uniformly distributed alone a long straight wire of radius $1\text{mm}$. The charge per cm length of the wire is $Q$ coulomb another cylindrical surface of radius $50\text{cm}$ and length $1\text{m}$ symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is

• A. $\frac{Q}{{ϵ}_0}$
• B. $\frac{100Q}{{ϵ}_0}$
• C. $\frac{10Q}{\left(\pi {ϵ}_0\right)}$
• D. $\frac{100Q}{\left(\pi {ϵ}_0\right)}$

Answer 1

The correct option is B $\frac{100Q}{{ϵ}_0}$

We know that Electric flux through a closed surface is
$\Phi =\overrightarrow{E}.\overrightarrow{A}$
also
$\lambda =\frac{q}{l}=Q\text{coulomb}{cm}^{-1}=100Q\text{coulomb}{m}^{-1}$
We know that Electric field due to infinite long charged wire is
$E=\frac{\lambda }{2\pi {ϵ}_{0}r}$
and curved surface area of cylinder is $A$= $2\pi rl$
Since electric field vector and area vector are in same direction.
Therefore, the electric flux through the cylinder is
$=\frac{\lambda }{2\pi {ϵ}_{0}r}2\pi rl=\frac{\lambda l}{{ϵ}_0},(l=1m,\lambda =100Q\text{coulomb}{m}^{-1})=\frac{100Q}{{ϵ}_0}$
Hence the total electric flux through cylindrical surface is $\Phi =\frac{100Q}{{ϵ}_0}$