A thin glass rod is bent into a semicircle of radius r. A charge +Q is uniformly distributed along the upper half and a charge -Q is uniformly distributed along the lower half, as shown in fig. The electric field E at P, the center of the semicircle, is

Options

A. `(Q)/(pi^(2)epsilon_(0)r^(2))`
B. `(2Q)/(pi^(2)epsilon_(0)r^(2))`
C. `(4Q)/(pi^(2)epsilon_(0)r^(2))`
D. `(Q)/(4pi^(2)epsilon_(0)r^(2))`

Answer

A

Explanation

`E_(Net) = 2E cos 45^(@) = sqrt(2)E`
`= sqrt(2) (2K lambda)/(r) sin 45^(@)`
`= (2sqrt(2)lambda)/(4pi epsilon_(0)r) xx (1)/(sqrt(2)) = (lambda)/(2pi epsilon_(0)r)`
`= (Q)/(2pi epsilon_(0)r((pir)/(2))) = (Q)/(pi^(2)epsilon_(0)r^(2))`