The electric field intensity at the center of a uniformly charged hemispherical shell is `E_(0)`. Now two portions of the hemisphere are cut from either side, and the remaining portion is shown in fig. If `alpha=beta=pi//3`, then the electric field intensity at the center due to the remaining portion is

Options

A. `(E_(0))/(3)`
B. `(E_(0))/(6)`
C. `(E_(0))/(2)`
D. `E_(0)`

Explanation

consider the whole hemispere as three portion if electric field due to one portion is `E_(1)` then `2E_(1)sin30+E_(1)=E_(0)`
`2E_(1)=E_(0)`
`impliesE_(1)=(E_(0))/(2)`