In fig two equal positive point charge `q_(1)=q_(2)=2.0mu C`. Interact with a third point charge `Q=4.0 mu C`. The magnitude as well as direction of the net force on Q is

Options

A. `0.23 N` in the +x direction
B. `0.46 N` in the +x direction
C. `0.23 N` in the -x direction
D. `0.46 N` in the -x direction

Answer

B

Explanation

`F_("net")=2|F_(31)|cos alpha`
`=2 xx (1)/(4 pi epsilon_(0)) xx(2xx4xx10^(-12))/((0.5)^(2))xx(4)/(5)=0.46N`