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Q.

A mass m carrying a charge q is suspended from a string and placed in a uniform horizontal electric field of intensity E. The angle made by the string with the vertical in the equilibrium position is

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a

$θ = \tan^{- 1} \frac{Eq}{mg}$

b

$θ = \tan^{- 1} \frac{m}{Eq}$

c

$θ = \tan^{- 1} \frac{Eq}{m}$

d

$θ = \tan^{- 1} \frac{mg}{Eq}$

answer is D.

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Detailed Solution

$qE = mgtan θ$

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