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)