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An electron falls through a distance of ...

An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude `2.0 xx 104 N C^(1)` [Fig. 1.13(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.13(b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’.

Text Solution

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`t_(e)=sqrt((2s)/(a_(e)))=sqrt((2sm_(e))/(eE))`
`=sqrt((2xx1.5xx10^(-2)xx9.1xx10^(-31))/(1.6xx10^(-19)xx2xx10^(4)))=2.9xx10^(-9)` s
`t_(p)=sqrt((2s)/(a_(p)))=sqrt((2sm_(p))/(eE))`
`=sqrt((2xx1.5xx10^(-2)xx1.67xx10^(-27))/(1.6xx10^(-19)xx2xx10^(4)))=1.3xx10^(-7)` s
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