An electron of mass m, charge e falls through a distance h meter in a uniform electric field E. Then time of fall

Question

An electron of mass m, charge e falls through a distance h meter in a uniform electric field E. Then time of fall (A) t = √(2hm/eE) (B) t = (2hm/eE) (C) t = √(2eE/hm) (D) t = (2eE/hm)

Tardigrade Admin · Accepted Answer

The force on electron =e E Due to motion, F=m a ∴ e E=m a a=(e E/m) Distance fallen, h =u t+(1/2) a t2 u =0 ⇒ h=(1/2) a t2 h =(1/2)((e E/m)) t2 ∴ t =√(2 h m/e E)

Q. An electron of mass $m$ , charge $e$ falls through a distance $h$ meter in a uniform electric field $E$ . Then time of fall

$t = \frac{2 hm}{e E}$

$t = \frac{2 hm}{e E}$

$t = \frac{2 e E}{hm}$

$t = \frac{2 e E}{hm}$

The force on electron $= e E$
Due to motion, $F = ma$
$∴ e E = ma$
$a = \frac{e E}{m}$
Distance fallen,
$h = u t + \frac{1}{2} a t^{2}$
$u = 0 \Rightarrow h = \frac{1}{2} a t^{2}$
$h = \frac{1}{2} (\frac{e E}{m}) t^{2}$
$∴ t = \frac{2 hm}{e E}$

Q. An electron of mass m, charge e falls through a distance h meter in a uniform electric field E. Then time of fall

t=eE2hm​​

t=eE2hm​

t=hm2eE​​

t=hm2eE​

The force on electron =eE Due to motion, F=ma ∴eE=ma a=meE​ Distance fallen, h=ut+21​at2 u=0⇒h=21​at2 h=21​(meE​)t2 ∴t=eE2hm​​

Q. An electron of mass $m$ , charge $e$ falls through a distance $h$ meter in a uniform electric field $E$ . Then time of fall

$t = \frac{2 hm}{e E}$

$t = \frac{2 hm}{e E}$

$t = \frac{2 e E}{hm}$

$t = \frac{2 e E}{hm}$

The force on electron $= e E$
Due to motion, $F = ma$
$∴ e E = ma$
$a = \frac{e E}{m}$
Distance fallen,
$h = u t + \frac{1}{2} a t^{2}$
$u = 0 \Rightarrow h = \frac{1}{2} a t^{2}$
$h = \frac{1}{2} (\frac{e E}{m}) t^{2}$
$∴ t = \frac{2 hm}{e E}$