A solid hemispherical uniformly charged body having charge `Q` is kept symmetrically along the `y`-axis as shown in the figure. The net electric field at (`d`, `0`,`0`) is

Options

A. `(1)/(4piepsilon_(0))(Q)/(d^(2))`
B. less then `(1)/(4piepsilon_(0))(Q)/(d^(2))`
C. ` gt (1)/(4piepsilon_(0))(Q)/(d^(2))` and `lt (1)/(4piepsilon_(0))(2Q)/(d^(2))`
D. `(1)/(4piepsilon_(0))(2Q)/(d^(2))`

Explanation

Due to solid hemispherical charge of `2Q` the field will be `(1)/(4piepsilon_(0))xx(2Q)/(d^(2))` along `x-`axis. So due to one hemisphere the component of field along `x-` direction will be `(1)/(4piepsilon_(0))xx(Q)/(d^(2))` . And hence the net field will be more than `(1)/(4piepsilon_(0))xx(2Q)/(d^(2))`