Question

An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude `2.4xx10^(4) NC^(-1)` [Fig.1.12 (a)]`. The direction of the field is reversed keeping its magnitude unchagned and a proton falls through the same distance [Fig. 1.12 (b) ]. Complute the time of fall in each case. Contrast the situation (a) with that of free fall under gravity.

Answer 1

In Fig. 1.13(a) the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field. The acceleration of the electron is
`a_(e)=Ee//m`
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_(p)=sqrt((2h)/(a_(p)))=sqrt((2hm_(p))/(eE))`
`"For" e=1.6xx10^(-19)C,m_(e)=9.11xx10^(-31)kg,`
`E=2.0xx10^(4)NC^(-1),h=1.5xx10^(-2)m, `
`t_(e)=2.9xx10^(-9)s)`
In Fig. 1.13 (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE. The acceleration of the proton is
`a_(p)=eE//m_(p)`
where `m_(p)` is the mass of the proton : `m_(p)=1.67xx10^(-27)kg`. The time of fall for the proton is
`t_(p)=sqrt((2h)/(a_(p)))=sqrt((2hm_(p))/(eE))=1.3xx10^(-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)=(eE)/(m_(p))`
`((1.6xx10^(-19)C)xx(2.0xx10^(4)NC^(-1)))/(1.67xx10^(-27)kg)`
`1.9xx10^(12)ms^(-2)`which is enormous compared to the value of g (9.8 m s–2), 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