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_(0)R^(2))`
B. `q/(4 p epsilon_(0)R^(2))`
C. `q/(4 pi epsilonR^(2))`
D. `q/(2 pi epsilon_(0)R^(2))`

Answer

D

Explanation

From figure `dlj =Rdq`
Charge on `dl=lamda R d theta [lamda=q/(pi R)]`

Eelctric field at centre due to dl is `dE=(k . Lamda Ed theta)/(R^(2))`
We need to consider only the component dE cos q, as the component dE sin q will cancel out.
Total field at centre `=int_(0)^(pi//2) dE cos theta`
`=(2klamda)/R int_(0)^(pi//2)cos theta d theta =(2 k lamda)/R=q/(2 pi^(2) epsilon_(0)R^(2))`