A thin glass rod is bent into a semicircle of radius r. 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. `(2 Q)/(pi^(2)epsilon_(0)r^(2))`
C. `(4 Q)/(pi^(2)epsilon_(0)r^(2))`
D. `(Q)/(4 pi^(2)epsilon_(0)r^(2))`

Answer

A

Explanation

Take PO as the x-axis and PA as the y-axis. Consider two elements EF and E'F' pf width `d theta` at angular `theta` above and below PO, respectively

The magnitude of the field at P due to either element is
`dE=(1)/(4 pi epsilon_(0)) (rd theta xx Q//(pi r//2))/(r^2)=(Q)/(2 pi^(2)epsilon_(0)r^(2))d theta`
Resolving the field we find that the components along PO sum up to zero, and hence the resultant fields is along PB. Therefore, field at P due to pair of elements is `2d E sin theta`
`E=int_(0)^(pi//2) 2d E sin theta`
`=2 int_(0)^(pi//2) (Q)/(2 pi epsilon_(0)r^(2)) sin theta d theta (Q)/(pi^(2)epsilon_(0)r^(2))`.