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(C) The distance $d$ covered by a charged particle starting from rest in a uniform electric field $E$ is given by the equation of motion: $d = \frac{1

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More in the case of proton

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(C) The distance $d$ covered by a charged particle starting from rest in a uniform electric field $E$ is given by the equation of motion: $d = \frac{1}{2} a t^2$.Since the force $F = qE$ and $F = ma$,the acceleration is $a = \frac{qE}{m}$.Substituting this into the distance equation: $d = \frac{1}{2} \left( \frac{qE}{m} ight) t^2$.Solving for time $t$: $t = \sqrt{\frac{2dm}{qE}}$.Since the magnitude of charge $q$ is the same for both the electron and the proton $(e)$,and the distance $d$ and field $E$ are constant,we have $t \propto \sqrt{m}$.Because the mass of a proton $(m_p \approx 1.67 	imes 10^{-27} \ kg)$ is much greater than the mass of an electron $(m_e \approx 9.11 	imes 10^{-31} \ kg)$,the time taken by the proton will be greater than the time taken by the electron.

An electron falls through a small distance in a uniform electric field of magnitude $2 \times 10^4 \ N C^{-1}$. The direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. The time of fall will be

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