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Question

A charged particle of mass m=1 kg and charge q=2 μC is thrown from the ground at an angle θ=45 with the horizontal, with initial speed 20 m/s. In the space a horizontal electric field of magnitude E=2×107 V/m exists. Find the range covered by the projectile on the horizontal ground. [g=10 m/s2]

A
100 m
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B
150 m
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C
170 m
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D
200 m
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Solution

The correct option is D 200 m

The path of the particle will be a parabola as it is a projectile. But, in this case due to the presence of the electric field in x direction some force will act on the particle and along x axis also the motion of the particle will be accelerated.

Using kinematic relation , y=uyt+12ayt2

Net displacement along y direction is zero (i.e) y=0.

0=uyT12gT2

T=2×20 sin4510=22 s

Along x direction, ax=Fm=qEm

ax=(2×106)(2×107)1=40 m/s2

Now, we can use the second equation of motion to find the motion of projectile along the x direction.

x=uxt+12axt2

Horizontal range of the particle will thus be

R=uxT+12axT2

R=(20 cos 45)(22)+12(40)(22)2

R=40+160=200 m

Hence, option (d) is the correct answer.

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