Question

A pendulum bob of mass 80 mg and carrying a charge of
$2\times 10^{-8}C$ is at rest in a horizontal uniform electric field of
20,000 V/m. Find the tension in the thread of the pendulum
and the angle it makes with the vertical.
(Take g = $9.8m/s^{-2}$)

Answer 1

For equilibrium of bob
$Tcos\theta =mg$ and $Tsin\theta =qE$
From these two equations we find
$T=\sqrt{(mg{)}^2+(qE{)}^2}and\theta =ta{n}^{-1}\left(\frac{qE}{mg}\right)$
Substituting the values we have
$T=\sqrt{(80\times 10^{-6}\times 9.8{)}^2+(2\times 10^{-8}\times 20000{)}^2}$
$=8.79\times 10^{-4}N$
$\theta =ta{n}^{-1}\left(\frac{2\times 10^{-8}\times 20000}{80\times 10^{-6}\times 9.8}\right)=27^{\circ }$