Question

The bob of a simple pendulum has mass $2g$ and a charge of $q\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\text{V/m}$. At equilibrium, if the angle that the pendulum makes with the vertical is ${\tan }^{-1}\left(0.5\right)$, find $q$ (assuming $g=10m/s^2$)

• A. $3\mu C$
• B. $4\mu C$
• C. $5\mu C$
• D. $6\mu C$

Answer 1

The correct option is C $5\mu C$


$\tan \theta =\frac{qE}{mg}$

$q=\frac{mg\tan \theta }{E}=\frac{2\times {10}^{-3}\times 10\times 0.5}{2000}=5\times {10}^{-6}=5\mu C$