A insulating cylindrical rod of diameter `d` and length `l(l gt gt d)` has a uniform surface charge density such that the electric field just outside the curved surface of the cylinder at point `M` is `E_(0)`. Find the electric field due to charge distribuition at point `P (r gt gt l)`.

Options

A. `E_(0)(ld)/(2r^(2))`
B. `E_(0)(ld)/(4r^(2))`
C. `E_(0)(ld)/(3r^(2))`
D. `E_(0)(2ld)/(r^(2))`

Explanation

`(2lambda)/(2pi epsilon_(0)d)=E_(0)`
`lambda=pi epsilon_(0)dE_(0)`
So, `E_(p)=(E_(0)dl)/(4r^(2))`