An infinitely long wire with linear charge density `+lambda` is bent and the two parts are inclined at angle `30^@` as shown in fig, the electric field at point P is :

Options

A. A. `lambda/(4pi in_0d)[(1+sqrt3)hati+2hatj]`
B. B. `lambda/(4pi in_0d)[(1-sqrt3)hati+hatj]`
C. C. `lambda/(4pi in_0d)[(1+sqrt3)hati-hatj]`
D. D. `lambda/(4pi in_0d)[(1+sqrt3)hati+hatj]`

Answer

D

Explanation

`E_(1)=E_(1)'=(lamda)/(4piin_(0)d)`
`E_(2)=(lamda)/(4piin_(0)((sqrt(3)d)/(2)))(sin60^(@)-sin0^(@))=(lamda)/(4piin_(0)d)`
`E_(2)'=(lamda)/(4pi in_(0)((sqrt(3)d)/(2)))(cos60^(@)+cos0^(@))=(sqrt(3)d)/(4pi in_(0)d)`
`therefore vecE=(E_(1)+E_(2)cos30^(0)+E_(2)'cos60^(0))hati+(E_(1)'+E_(2)'sin60^(0)-E_(2)sin30^(0))hatj`
`=((lamda)/(4piin_(0)d)+(sqrt(3)lamda)/(8piin_(0)d)+(sqrt(3)lamda)/(8piin_(0)d))hati+((lamda)/(4piin_(0)d)+(sqrt(3))/(8piin_(0)d)-(sqrt(3))/(8piin_(0)d))hatj`
`=(lamda)/(4piin_(0)d)[(1+sqrt(3))hati+hatj]`