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. $\theta =ta{n}^{-1}\frac{mg}{Eq}$
• B. $\theta =ta{n}^{-1}\frac{m}{Eq}$
• C. $\theta =ta{n}^{-1}\frac{Eq}{m}$
• D. $\theta =ta{n}^{-1}\frac{Eq}{mg}$

Answer 1

The correct option is D $\theta =ta{n}^{-1}\frac{Eq}{mg}$

As the body is in equilibrium
Net horizontal forces $=0and$ Net vertical forces $=0$
$\Rightarrow Tcos\theta =mgandEq=Tsin\theta$

$\Rightarrow T=\frac{mg}{cos\theta }andT=\frac{Eq}{sin\theta }$

$\Rightarrow \frac{mg}{cos\theta }=\frac{Eq}{sin\theta }$

$\Rightarrow tan\theta =\frac{Eq}{mg}$

$\Rightarrow \theta =ta{n}^{-1}\frac{Eq}{mg}$