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Q.

A charged ball of mass 9 kg is suspended from a string in a uniform electric field E→=(3i^+5j^)×105  N/C. The ball is in equilibrium with  θ=370.If direction of electric field is reversed, find the new equilibrium position of the ball. Give your answer in terms of angle made by string with vertical. Take  g=10 ms−2?

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a

tan−1(34)

b

cot−1(314)

c

cot−1(34)

d

tan−1(314)

answer is D.

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Detailed Solution

Tsinθ=3q×105Tcosθ=mg−5q×105Solve to get q=100 μC,  T=50 N .After the reversal of direction of electric field T'sinα=3q×105     or   T'cosα=mg+5q×105tanα=3q×105mg+5q×105=3×10−4×1059×10+5×10−4×105=314or  α=tan−1(314)
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