A 1 kg ball is suspended in a uniform electric field with the help of a string fixed to a point. The ball is given a charge $\sqrt{5}$ coulomb and the string makes an angle $37^{0}$ with the vertical in the equilibrium position. In the equilibrium position the tension is double the weight of the ball. Find the magnitude of the electric field in N/C.

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Solution

The correct option is A 6
The force due to the electric field is the resultant of the tension $(T = 2 m g)$ in the string and the gravitational force $(m g)$ of ball.
therefore, $q E = \sqrt{(2 m g)^{2} + (m g)^{2} + 2 (2 m g) (m g) cos (180 - 37)} = m g \sqrt{4 + 1 + 4 cos 143} = 1.34 m g$
$E = \frac{1.34 \times 1 \times 10}{\sqrt{5}} = 6 N / C$

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