A pendulum bob of mass \(80mg\) carries a charge of \(2*10^{-8}C\) at rest in an horizontal uniform electric field of \(20,000V/m \). Find the tension in the thread of the pendulum and the angle it makes with the vertical.


Electric force $$ F=qE=2*10^{-8}*20,000=4*10^{-4}$$
Gravitational force $$ mg= 80*10^{-6}*9.8=7.84*10^{-4}N$$
Balancing the horizontal and vertical components of forces $$ T \sin \theta = F$$
$$ T\cos\theta=mg$$
where \(T\) is the tension in the thread
Now, $$ \tan\theta=\frac{F}{mg}=\frac{4*10^{-4}}{7.84*10^{-4}}=0.51$$
Now, \(\theta=27^\circ\)
To find the tension T
$$ T=\sqrt{F^2+(mg)^2}=\sqrt{(4*10^{-4})^2+(7.84*10^{-4})^2}=8.8*10^{-4}N$$