A point charge 'q' is placed at the centre of left circular end of a cylinder of length l=4cm and radius R=3 cm shown. Then the electric flux through the curved surface of the cylinder is

Options

A. `(q)/(5epsilon_(0))`
B. `(2q)/(5epsilon_(0))`
C. `(3q)/(5epsilon_(0))`
D. `(4q)/(5epsilon_(0))`

Answer

B

Explanation

The electric flux through the right end plane surface
`phi_(PS)=(q)/(4piepsilon_(0))xx2pi(1-costheta)=(q)/(2epsilon_(0))(1-(4)/(5))`
`phi_(PS)=(q)/(10epsilon_(0))`
Hence the electric flux through the curved surface of the cylinder is
`phi_(CS)=(q)/(2epsilon_(0))=phi_(PS)`
`=(q)/(2epsilon_(0))-(q)/(10epsilon_(0))=(4q)/(10epsilon_(0))`
`phi_(CS)=(2q)/(5epsilon_(0))`