Charges `q` is uniformly distributed over a thin half ring of radius `R`. The electric field at the centre of the ring is

Options

A. `q/(2pi^2 epsilon_0R^2)`
B. `q/(4pi^2 epsilon_0R^2)`
C. `q/(4pi epsilon_0R^2)`
D. `q/(2pi epsilon_0R^2)`

Answer

A

Explanation

`dl = Rdtheta`
dq=(Charge on dl) `= lamda R dtheta`
`[lamda = q/(piR)]`
`dE = (Kdq)/(R^2) = (KlamdaR"d" theta)/(R^2)`
Total field on the centre –
`E=2int_0^(pi//2)(KlamdaRcostheta d theta)/R^2=(2Klamda)/R [sin theta]_0^(pi//2)`
`=(2Klamda)/R = (lamda)/(2pi epsilon_0R)=q/(2pi^2 epsilon_0R^2)`