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A thin glass rod is bent into a semicirc...

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

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)`

Text Solution

Verified by Experts

The correct Answer is:
D
`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)`
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