Find the electric field at centre of semicircular ring shown in figure .
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Answer

` - ( 4 kq)/( pi R^2)`

Explanation

`vec E_1 =` is the field centre due to `+q`
`vec E_2 = ` is the field at centre due to `-q`
` E = ( 2 K lambda)/R sin 90^2`
`E_2 = E_1 = (2 K)/R (q/( pi R//2) . Sin 45^@`

`= 4 (JKq)/( pi R^2) xx 1/( sqrt 2) = 2 sqrt 2 (Kq)/( pi R^2)`
` vec E_0 = vec E_1 + vec E_2, vec E_1 _|_ vec E_2` and ` vec E_1 = vec E_2`
` E_0 = sqrt 2 E_1 = sqrt 2 + 2 sqrt (Kq)/( pi R^2 ) rArr vec E_0 = ( 4Kq )/( pi R^2) (- hat i)`.