Question

An electric field `E` is produced between two parallel plates having a separation `d` as shown.
(a) With what minimum speed should an electron be projected from the lower plate in the direction of field so that it may reach the upper plate ?
(b) Suppose the electron is projected from the lower plate with the speed calculated in part `(a)`. The direction of projection makes an angle of `60^(@)`with the field. Find the maximum height reached by the electron.
(c ) After how much time , the electron again strikes the lower plate ?
(d) Horizontal distance travelled by the electron in time calculated in part `(c )`.
Charge on electron : `e` ,mass of electron : `m`, consider only electric force.

Answer 1

(a) Force on electron `F = eE`, in downward direction.
Let the electron be thrown with speed `u`.
Retardation of electron `a = (e E)/(m)`
The electron just reaches the upper plate i.e. its velocity at upper plate is zero.
`v^(2) = u^(2) - 2 as`
`0 = u^(2) - (2e Ed)/(m) rArr u = sqrt((2e Ed)/(m))`
Note : This problem is similar to motion under gravity ,here `g = eE//m`.
(b) This is the case of motion in a plane and similar to projectile motion , here `g = eE//m`.
`y`-direction : Velocity of electron at highest point `= 0`.
`v_(y)^(2) = u_(y)^(2) - 2 a_(y) s`
when `s = H_(max) ,v_(y) = 0`.
`0 = (u sin 30^(@))^(2) - 2 (e E)/(m) H_(max)`
`H_(max) = (u^(2) sin^(2) 30^(@))/(2 eE//m) = (2eEd)/(m) .(1)/(4) .(1)/(2e E//m)`
`= (d)/(4)`
(c) `O` to `A` : Displacement in y-direction is zero.
`y = u_(y)t - (1)/(2) a_(y) t^(2)`
`0 = u sin 30^(@) t - (1)/(2) a_(y) t^(2)`
`t = (2 u sin 30^(@))/(a_(y)) = 2 sqrt((2 eEd)/(m)) .(1)/(2) .(1)/(eE//m)`
`= sqrt((2 md)/(eE))`
(d) Distance `OA : x` -direction
`x = u_(x)t = u cos 60^(@) t`
`= sqrt((2 eEd)/(m)) .(1)/(2) sqrt((2md)/(eE)) = d`
Note : This case is similar to projectile motion , maximum height , time of flight and range are being asked.