A charged ball of mass 9 kg is suspended from a string in a uniform electric field $\vec{E} = (3 \hat{i} + 5 \hat{j}) \times 10^{5} N / C$ . The ball is in equilibrium with $θ = 37^{\circ}$ . 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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$\tan^{- 1} (\frac{3}{4})$

$\cot^{- 1} (\frac{3}{14})$

$\cot^{- 1} (\frac{3}{4})$

$\tan^{- 1} (\frac{3}{14})$

answer is D.

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Detailed Solution

$Tsin θ = 3 q \times 10^{5}$ …(i)

$Tcos θ = mg - 5 q \times 10^{5}$ …(ii)

Solve to get $q = 100 μC, T = 50 N .$

After the reversal of direction of electric field

$T^{'} \sin α = 3 q \times 10^{5} or T^{'} \cos α = mg + 5 q \times 10^{5}$

$\tan α = \frac{3 q \times 10^{5}}{mg + 5 q \times 10^{5}}$

$= \frac{3 \times 10^{- 4} \times 10^{5}}{9 \times 10 + 5 \times 10^{- 4} \times 10^{5}} = \frac{3}{14}$

or $α = \tan^{- 1} (\frac{3}{14})$

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