Find the force experienced by a semicircular rod having a charge q as shown in fig. Radius of the wire is R, and the line of charge with linear charge density `lambda` passes through its center and is perpendicular to the plane of wire.

Options

A. `(lambdaq)/(2pi^(2)epsilon_(0)R)`
B. `(lambdaq)/(pi^(2)epsilon_(0)R)`
C. `(lambdaq)/(4pi^(2)epsilon_(0)R)`
D. `(lambdaq)/(4piepsilon_(0)R)`

Answer

B

Explanation

`E=(lambda)/(2piepsilon_(0)R) dF=Edq=ERd thetalambda^(1)`
From symmetry `intdFcostheta =0F=intdFsintheta `
`=int_(0)^(pi)ERdthetalambda^(1)=(lambda)/(2piepsilon_(0)R)Rlambda^(1)int_(0)^(pi)sintheta d theta `
on solving `F=(lambdaq)/(pi^(2)epsilon_(0)R)`