An electron falls through a distance of $1.5\; cm$ in a uniform electric field of magnitude $2.0 \times 10^{4} \;N C ^{-1} \text {[Figure (a)]} .$ The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Figure $(b)] .$ Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.
An electron falls through a distance of $1.5\; cm$ in a uniform electric field of magnitude $2.0 \times 10^{4} \;N C ^{-1} \text {[Figure (a)]} .$ The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Figure $(b)] .$ Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.
Solution In Figure $( a )$ the field is upward, so the negatively charged electron experiences a downward force of magnitude $e E$ where $E$ is the magnitude of the electric field. The acceleration of the electron is
$a_{e}=e E / m_{e}$
where $m_{e}$ is the mass of the electron.
Starting from rest, the time required by the electron to fall through a distance $h$ is given by $t_{e}=\sqrt{\frac{2 h}{a_{e}}}=\sqrt{\frac{2 h m_{e}}{e E}}$
For $e=1.6 \times 10^{-19} \,C , m_{e}=9.11 \times 10^{-31} \,kg$
$E=2.0 \times 10^{4} \,N\, C ^{-1}, h=1.5 \times 10^{-2} \,m$
$t_{ e }=2.9 \times 10^{-9}\, s$
In Figure $(b)$, the field is downward, and the positively charged proton experiences a downward force of magnitude $e E .$ The acceleration of the proton is $a_{p}=e E / m_{p}$
where $m_{p}$ is the mass of the proton; $m_{p}=1.67 \times 10^{-27}$ $kg$. The time of fall for the proton is
$t_{p}=\sqrt{\frac{2 h}{a_{p}}}=\sqrt{\frac{2 h m_{p}}{e E}}=1.3 \times 10^{-7} \;s$
Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of 'free fall under gravity" where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field:
$a_{p}=\frac{e E}{m_{p}}$
$=\frac{\left(1.6 \times 10^{-19} \,C \right) \times\left(2.0 \times 10^{4} \,N\,C ^{-1}\right)}{1.67 \times 10^{-27}\, kg }$
$=1.9 \times 10^{12} \,m s ^{-2}$
which is enormous compared to the value of $g\left(9.8\, m s ^{-2}\right),$ the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored in this example.
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$(B)$ the particle will hit $T$ if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal
$(C)$ time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu s$ as well as $\sqrt{\frac{5}{2}} \mu s$
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A uniform electric field, $\vec{E}=-400 \sqrt{3} \hat{y} NC ^{-1}$ is applied in a region. A charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $2 \sqrt{10} \times 10^6 ms ^{-1}$. This particle is aimed to hit a target $T$, which is $5 m$ away from its entry point into the field as shown schematically in the figure. Take $\frac{ q }{ m }=10^{10} Ckg ^{-1}$. Then-
$(A)$ the particle will hit $T$ if projected at an angle $45^{\circ}$ from the horizontal
$(B)$ the particle will hit $T$ if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal
$(C)$ time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu s$ as well as $\sqrt{\frac{5}{2}} \mu s$
$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu s$
- [IIT 2020]