Question

A very long uniformly charged wire oriented along the axis of a circular ring of radius R rests on its centre with one of the ends (as shown in figure). The linear charge density on the wire is `lambda`. Evalute the flux of the vector `vecE` across the circle area.

Answer 1

Correct Answer - `(lambda R)/( 2 epsilon_(0))`
The component of electric field `E_y` will not contribute to flux as
angle between `E_y and vecA is 90^@`. Hence, flux due to `E_x` is
`dphi = E_x2piydy`
or `phi = int dphi = int_(0)^(R) (lambda/(4piepsilony) 2piydy`
`= lambda/(2epsilon_0) int_(0)^(R) dy = (lambdaR)/(2epsilon_0)`