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Question

A particle of mass 9×10−31kg having a negative charge of 1.6×10−19C is projected horizontally with a velocity of 106ms−1 into a region between two infinite horizontal parallel plates of metal. The distance between the plates is d=03cm, and the particle enters 0.1 cm below the top plate. The top and bottom plates are connected, respectively, to the positive and negative terminals of a 30 V battery. Find the components of the velocity of the particle just before it hits one of the plates

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Solution

An electron is a negatively charged particle, so it will be attracted by the positive plate with force F=eE. Hence, acceleration of electron along y-axis will be

a=Fm=eEm=eVmd {asE=Vd} (i)

So, from equation of motion, v2=u2+2as along the x-axis,

vx=v0=106(ms−1){asax=0} (ii)

And along the y-axis, v2y=2ay0|{asu=0ands=y0}

Now, as y0=1cm (given) and is given by equation (i), the electron will hit the top plate with the velocity

vy=√2y0eVmd=√2×1.6×10−19×20×1×10−39×10−31×3×10−3

=(4√2/3)×106=1.885×106ms−1 (iii)

So, the electron will hit the upper plate with the velocity

vx=106(ms−1) and vy=1.885×106ms−1

Note

In this problem, time taken by the electron to hit the plate

t=√2y0a=√2y0meE=√2y0mdeV

=√2×(2×10−3)×(9×10−31)×(3×10−3)1.6×10−19×30

=32√2×10−9s=1.06×10−6s

And in this time, electron will travel a horizontal distance

s=v0t=106×1.06×10−9=1.06×10−3m

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