A particle of mass `m` and charge `q` is thrown in the vertical direction with a velocity `v_(0)` in a uniform horizontal electric field `E`. Assuming the gravity force to be negligible ,the equation of the path followed by the particle will be

Options

A. parabola ,`x = ((qE)/(2 mv_(0)^(2)))y^(2)`
B. hyperbola , `x = (qE)/(2 mv_(0)^(2) y)`
C. straight line ,`x = (qEy)/(2mv_(0)^(2))`
D. none of the above

Answer

1

Explanation

Displacement
`x : x = u_(x) t + (1)/(2) a_(x) t^(2) = 0 + (1)/(2) (qE)/(m) t^(2)` ……(i)
`y : y = v_(0) t` ……(ii)
Eliminating `t` between `(i)` and `(ii)`
`x = (1)/(2) (qE)/(m) ((y)/(v_(0)))^(2)`
`x = ((qE)/(2 m v_(0)^(2)))y^(2)`,parabola