Question

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

• A. $\theta =tan^{-1}\frac{mg}{Eg}$
• B. $\theta =tan^{-1}\frac{m}{Eg}$
• C. $\theta =tan^{-1}\frac{Eq}{m}$
• D. $\theta =tan^{-1}\frac{Eq}{mg}$

Answer 1

Answer: (D)

Correct answer is (d) $\theta =tan^{-1}\frac{Eq}{mg}$