An electron falls through a distance of $1.5 cm$ in a uniform electric field of magnitude $2 \times {10^4}N/C.$ Now the direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. Compute the time of fall in each case, neglecting gravity.

ATQ,

$S=1.5cm=1.5\times { 10 }^{ -2 }m$

$u=0m/s$

${ m }_{ e }$ : mass of electron $=9.10\times { 10 }^{ -31 }Kg$

${ m }_{ p }$ : mass of proton $=1.672\times { 10 }^{ -27 }Kg$

from coloumb's force, $\vec { F } =q\vec { E } $

and given $\vec { E } =2\times { 10 }^{ 4 }N/c$

force proton ${ \vec { F } }_{ (+) }=1.6\times { 10 }^{ -19 }\times 2\times { 10 }^{ 4 }=3.2\times { 10 }^{ -19 }N$

force on electron, ${ \vec { F } }_{ (-) }=-1.6\times { 10 }^{ -19 }\times 2\times { 10 }^{ 4 }=-3.2\times { 10 }^{ -15 }N$

From Newton's Second law, $\vec { F } =m\vec { a } $

$\therefore $ for proton, ${ a }_{ \left( + \right) }=\dfrac { { \vec { F } }_{ (+) } }{ { m }_{ p } } =1.913\times { 10 }^{ 12 }m/{ s }^{ 2 }$

for electron, ${ a }_{ \left( - \right) }=\dfrac { { \vec { F } }_{ \left( - \right) } }{ { m }_{ e } } =-0.35\times { 10 }^{ 16 }m/{ s }^{ 2 }$

from equ of motion, $S=ut+\dfrac { 1 }{ 2 } { at }^{ 2 }\quad \longrightarrow (1)$

ATQ, The direction of the field was reversed, hence we will ignore the sign convention for acceleration.

$\therefore $ Substituting values in equ (1) for proton,

$1.5\times { 10 }^{ -2 }=\dfrac { 1 }{ 2 } \times 1.913\times { 10 }^{ 12 }\times { t }^{ 2 }$

$1.568\times { 10 }^{ -14 }={ t }^{ 2 }$

$\Rightarrow \quad t=1.252\times { 10 }^{ -7 }seconds$

Substituting values in equ(1) for electron,

$1.5\times { 10 }^{ -2 }=\dfrac { 1 }{ 2 } \times 3.5\times { 10 }^{ 15 }\times { t }^{ 2 }$

$0.857\times { 10 }^{ -17 }={ t }^{ 2 }$

$\Rightarrow \quad { t }=2.927\times { 10 }^{ -19 }seconds$