कोई इलेक्ट्रॉन $2.0 \times 10^{4}\, N C ^{-1}$ परिमाण के एकसमान विध्यूत क्षेत्र में $1.5 \,cm$ दूरी तक गिरता है [चित्र $( a )]$ । क्षेत्र का परिमाण समान रखते हुए इसकी दिशा उत्क्रमित कर दी जाती है तथा अब कोई प्रोटोन इस क्षेत्र में उतनी ही दूरी तक गिरता है [ चित्र $( b )$ ]। दोनों प्रकरणों में गिरने में लगे समय की गणना कीजिए। इस परिस्थिति की 'गुरूत्व के अधीन मुक्त पतन' से तुलना कीजिए।

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.