A semicircular wire is uniformly charged with linear charge density dependent on the angle `theta` from `y`-direction as `lambda=lambda_(0) |sin theta|`, where `lambda_(0)` is a constant. The electric field intensity at the centre of the arc is

Options

A. `(lambda_(0))/(2piepsilon_(0)R)(-hat(j))`
B. `(lambda_(0))/(4piepsilon_(0)R)(-hat(j))`
C. `(lambda_(0))/(6piepsilon_(0)R)(-hat(j))`
D. `(lambda_(0))/(2sqrt(2)piepsilon_(0)R)(-hat(j))`

Answer

b

Explanation

Lets take an element at angle `theta` with vertical, substending `d theta` angle electric field due to this element

`dE=(dq)/(R^(2))`
horizontal component `dE sin theta` will be cancelled out with opposite element, and `dE cos theta` components get added up
So `E= underset(theta=0 pi//2)overset(theta= pi//2)int dE cos theta=2 underset(theta=0)overset(theta= pi//2)intK((lambda_(0)sin theta))/(R ^(2))Rd theta cos theta`
`vec(E )= (lambda_(0))/(4pi epsilon_(0)R)(-hat(j))`