A charge `+Q` is uniformly distributed along the circular arc of radius `R` as shown in the figure.The magnitude of the force experienced by the point charge `+q` place at the centre of curvature is `(k=(1)/(4piepsilon_(0)))`

Options

A. `(kQqsin""(theta)/(2))/(R^(2)theta)`
B. `(2kQqsin""(theta)/(2))/(R^(2)theta)`
C. `(2kQqcos""(theta)/(2))/(R^(2)theta)`
D. `(2kQqtan""(theta)/(2))/(R^(2)theta)`

Answer

B

Explanation

The focus on the pt,change `q` due to the differential element is `dF=(kqdq)/(R^(2))=(klambdaR(dphi)q)/(R^(2))`
`F=(kqlambda)/(R)int_(-theta//2)^(theta//2)cosphid phi=(2kqlambdasin""(theta)/(2))/(R)`
but `lambda=(Q)/(Rtheta)rArr F=(2kQqsin""(theta)/(2))/(R^(2)theta`