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Question

An electron is projected, as shown in the figure, at a speed of 6×106ms1, at an angle of 450. It is given that E=2000Vm1 is directed upward, d=3cm and l=10cm. The electron will strike at :
10851_3f2cf59cb71940c9b246441213c54410.png
  1. No where
  2. Upper plate
  3. Mid point of the lower plate
  4. Opposite end of the lower plate

A
Opposite end of the lower plate
B
Mid point of the lower plate
C
Upper plate
D
No where
Solution
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The acceleration due to the electric field E is eEm=1.6×1019×20009.1×1031=3.2×10159.1 downward and is constant.
Hence, the motion is equivalent to a projectile motion.
T=2usinθa
range R=u2sin2θa=(6×106)2sin(2×450)3.2×10159.1=0.1m=10cm
the electron will strike the lower plate at the edge.
Max height reached: H=u2sin2θ2a=(6×106)2×(12)22×3.2×10159.1=0.025m=2.5cm
Since max height reached by electron is less than the separation between the plates, it will not strike the upper plate.

Hence it will touch the lower plate at the edge.

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