A charged particle ( mass m and charge q) moves along X axis with velocity `V_0` . When it passes through the origin it enters a region having uniform electric field `vecE=-E hatj` which extends upto x=d . Equation of path of electron in the region . `x gt d` is :

Options

A. `y=(qEd)/(mv_0^2)(x-d)`
B. `y=(qEd)/(mv_0^2)(d/2-x)`
C. `y=(qEd)/(mv_0^2)x`
D. `y=(qEd^2)/(mv_0^2)x`

Answer

B

Explanation

Let particle have charge q and mass .m.
Solve for (a,m) mathematically
`F_x = 0 , a_x = 0 , (v)_x` = constant
Time taken to reach at .P. = `d/(v_0) =t_0 ` (let) `" "...(i)`
(along -y axis), `y_0 = 0 +1/2 .(qE)/(m) . t_0^2 " "...(ii)`
`v_x = v_0`
(along -y axis), v = u + at
`tan theta=v_y/(v_x)=(qEt_0)/(m.v_0)" " (t_0=d/(v_0))`
`tan theta = (qEd)/(m.v_0^2)`
Slope `=(-qEd)/(mv_0^2)`
Now we have to find `eq^n` of straight line whose slope is `(-qEd)/(mv_0^2)` and it pass through point `rarr (d,-y_0)`
Because after `xgtd`
No electric field = `F_("net") = 0, vecv` = const.
`y=mx+c(m=(qEd)/(mv_0^2),(d-y_0))`
`-y_0=(-qEd)/(mv_0^2).d+cimpliesc=-y_0+(qEd^2)/(mv_0^2)`
Put the value -
`y=(-qEd)/(mv_0^2)x -y_0+(qEd^2)/(mv_0^2)`
`y_0=1/2.(qE)/m(d/v_0)^2=1/2(qEd^2)/(mv_0^2)`
`y=(-qEd)/(mv_0^2)-1/2(qEd^2)/(mv_0^2)+(qEd^2)/(mv_0^2)`
`y=(-qEd)/(mv_0^2)x+1/2(qEd^2)/(mv_0^2)`
`y=(qEd)/(mv_0^2)(d/2-x)`