An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude `2.0xx10^(4)N//C(Fig.a)`

Calculate the time it takes to fall through this distance starting from rest. If the direction of the field is reversed (fig .b) keeping its magnitude unchanged, calculate the time taken by a proton to fall through this distance starting from rest.

Explanation

The field is upward , so the negatively charged particle , electron experiences a downward froce of magnitude eE where E is the magnitude of the electric field.
The acceleration of the electron is ae `=(eE)/(me)`
Starting from rest , the time required by the electron to fall through a distance h
`t_(e)=sqrt((2h)/(ae))=sqrt((2hme)/(eE))=sqrt((2xx1.5xx10^(-2)xx9.11xx10^(-31))/(1.6xx10^(-19)xx2.0xx10^(-4)))=2.9xx10^(-9)S`.
The field is downward , and the positively charged proton experiences a downward force of magnitude eE . The acceleration of the proton is
`aP=(eE)/(m_(p))`
Hence , the time taken by the proton
`t_(p)=sqrt((2h)/(a_(p)))=sqrt((2hm_(p))/(eE))=sqrt((2xx1.5xx10^(-2)xx1.67xx10^(-27))/(1.6xx10^(-19)xx2.0xx10^(-4)))=1.3xx10^(-7)sec`.