Question

# A charged particle of mass 1.0 g is suspended through a silk thread of length 40 cm in a horizontal electric field of intensity 4.0×104NC−1 as shown in the figure. If the particle stays in equilibrium at a distance of 24 cm from the wall, find the net charge on the particle?

A

q=1.8×107C

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B

q=3.2×107C

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C

q=3.2×107C

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D

None of these

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Solution

## The correct option is A q=1.8×10−7C Let’s draw the free body diagram of the charged particle- The situation is shown in figure. The forces acting on the particle are (i) The electric force F = qE horizontally leftwards, (ii) The force of gravity mg downward and (iii) The tension T along the thread As the particle is at rest, these forces should add to zero. T cos θ = mg and T sin θ = F or, F = mg tan θ From the figure, Sinθ=2440=35 Thus, tan θ=34 From (i), q(4.0×104NC−1)=(1.0×10−3kg)(9.8m−s−2)(34) Giving q=1.8×10−7C

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