linear charge density of current carrying wire of infinite length is `4(muC)/m`. Electric field intensity at distance 3.6 cm from wire is ............ `[therefore 1/(4piepsilon_(0)) = 9 xx 10^(9) SI]`

Options

A. `2 xx 10^(6)` V/m
B. `10^(6)` V/m
C. `10^(5)` V/m
D. `2 xx 10^(5)` V/m

Answer

A

Explanation

Electric field by current carrying wire at distance V:
`E = lambda/(2piepsilon_(0)).1/r =(2lambda)/(4piepsilon_(0)r) = (2klambda)/r`
`therefore E = (2 xx 9 xx 10^(9) xx 4 xx 10^(-6))/(3.6 xx 10^(-2))`
`therefore E = 2 xx 10^(6)` V/m