A vertical of mass ' m ' and charge ' q' is thrown at a speed of 'u ' against a uniform electric field ' E' . How much distance will it travel before comming to rest .
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Let us consider a particle of mass m and of charge q travelling against the uniform electric field E then the force experience by the particle is
F=qE
ma=qE
a=qE/m …………(1)
we have:
v2=u2 +2as
0=u2 +2as
s=-u2/2a
from equation (1)
s=-u2/2(qE/m)
s=-u2m/2qE
Therefore the distance travel by the particle before coming to momentary rest is given by -u2m/2qE
Negative sign because the particle is retarding.
Hope this helps .
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