A particle fo mass `m` carries an electric positive charge `Q` and is subjected to the combined action of gravity `g` and a unifrom horizontal electric field of strength `E`,it is projected with speed `u` to the vertical plane parallel to the electrical field and at an angle `theta `to the horizontal .The maximum distance teh particle travels horizontally when it is at the same level as its starting point is

Options

A. `(u^(2)/(g))`
B. `(u^(2)(EQ+mg))/(mg^(2))`
C. `(u^(2))/(mg^(2))(EQ+sqrt((mg)+(EQ)^(2)))`
D. `(EQu^(2))/(mg^(2))`

Answer

C

Explanation

`x=ucosthetat+(1)/(2)(EQ)/(m)t^(2)y=usinthetat-(1)/(2)g t^(2)`
`t=(2usintheta)/(g)x=(u^(2))/(g) sin2theta+(1)/(2)(QE)/(m).(4u^(2)sin^(2)theta)/(g^(2))`

`(dx)/(d theta)=0` then `tan2 theta =(-mg)/(EQ)`
`x_("max")=(u^(2))/(mg^(2))(EQ+sqrt((mg)^(2)+(EQ)^(2)))`