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Question

An electric field is acting vertically upwards on a small body of mass 1gm and charge 1μC is projected with a velocity 10m/s at angle 45 with horizontal. If horizontal range is 2m then intensity of electric field is: (g=10m/s2)


A
20000N/C
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B
10000N/C
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C
40000N/C
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D
90000N/C
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Solution

The correct option is C 40000N/C
The horizontal range of a projectile is given by R=v2sin2θg

When the body is moving in the presence of electric and gravitational field, the body experiences both the weight and electric force, which can be called as apparent weight.
Electric force acting on the negatively charged body is in downward direction because the electric field is in upward direction.
Wapp=W+qE=mg+qE
Thus, the body feels a different g given by gapp=g+qEm

Range of the projectile, in the presence of electric field is R=v2sin2θg+qEm

Given that R=2 m, θ=45 and v=10 m/s

Thus, g+qEm=v2sin2θR=1002=50

Using g=10 m/s, we have E=40mq=40×1×1031×106=40000 N/C

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