Question

# An electron falls through a distance of 1.5cm in a uniform electric field of magnitude 2×104N/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.

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Solution

## ATQ, S=1.5cm=1.5×10−2m u=0m/sme : mass of electron =9.10×10−31Kgmp : mass of proton =1.672×10−27Kgfrom coloumb's force, →F=q→Eand given →E=2×104N/cforce proton →F(+)=1.6×10−19×2×104=3.2×10−19Nforce on electron, →F(−)=−1.6×10−19×2×104=−3.2×10−15NFrom Newton's Second law, →F=m→a∴ for proton, a(+)=→F(+)mp=1.913×1012m/s2for electron, a(−)=→F(−)me=−0.35×1016m/s2from equ of motion, S=ut+12at2⟶(1)ATQ, The direction of the field was reversed, hence we will ignore the sign convention for acceleration.∴ Substituting values in equ (1) for proton,1.5×10−2=12×1.913×1012×t21.568×10−14=t2⇒t=1.252×10−7secondsSubstituting values in equ(1) for electron,1.5×10−2=12×3.5×1015×t20.857×10−17=t2⇒t=2.927×10−19seconds

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