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)/(4 pi epsi_(0)R^(2))`
B. `(Q)/(2 pi epsi_(0)R^(2))`
C. `(Q)/(( 4 pi epsi_(0)R^(2))`
D. `(Q)/(pi^(2) epsi_(0)R)`

Answer

D

Explanation

`E=1/(4 pi e_0) Q/R^2 (sin (pi//4))/(pi//4)`
`=Q/(sqrt2 pi^2 e_0 R^2)`
Now resultant field at
`P=sqrt2 E=Q/(pi ^2 e_0 R^2)`