A circular wire loop of radius R carries a total charge q distributed uniformly over its length. A small length `x (ltlt R)` of the wire is cut off. Find the electric field at the centre due to the remaining wire.

Answer

B, C

Explanation

if loop is complete, then net electric field at centre `C` is zero. Because equal and opposite pair of electric field vectors are cancelled.
If `PQ` portion is removed as shown in figure, then electric field due to portion RS is not cancelled.
Hence, electric field in ony due to the option `RS`.
`:. E=1/(4piepsilon_0) (q_(RS))/R^2=(1/(4piepsilon_0))((q/2piR)x)/R^2`
`=(qx)/(8pi^2epsilon_0R^3)`