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 = $9.8\, m/s^{-2}$)

Answer 1

For equilibrium of bob
$ T\, cos\, \theta=mg$ and $T\, sin\, \theta=qE$
From these two equations we find
$T=\sqrt{(mg)^2+(qE)^2}\, and\, \theta=tan^{-1} \bigg(\frac{qE}{mg}\bigg)$
Substituting the values we have
$T=\sqrt{(80 \times 10^{-6} \times9.8)^2+(2 \times 10^{-8} \times 20000)^2}$
$ \, \, \, = 8.79 \times 10^{-4}\, N$
$\theta=tan^{-1} \bigg(\frac{2 \times 10^{-8}\times 20000}{80 \times 10^{-6} \times9.8}\bigg)=27^\circ$