Question

A pendulum bob of mass $80mg$ and carrying a charge of $4\times {10}^{-8}C$ is at rest in a horizontal uniform electric field of magnitude $20000V/m$. The tension in the thread of the pendulum and the angle it makes with the vertical are (Acceleration due to gravity $=10m{s}^{-2}$):

• A. $8\sqrt{2}\times {10}^{-4}N,{45}^o$
• B. $8\times {10}^{-4}N,{60}^o$
• C. $6\sqrt{2}\times {10}^{-4}N,{30}^o$
• D. $6\times {10}^{-4}N,{37}^o$

Answer 1

The correct option is A $8\sqrt{2}\times {10}^{-4}N,{45}^o$

Since bob is at rest, net force on it is zero.

Balancing horizontal forces, $qE=T\sin \theta$

Balancing vertical forces, $mg=T\cos \theta$

Solving above two equations,

$T=\sqrt{(mg{)}^2+(qE{)}^2}$

$T=\sqrt{(80\times {10}^{-6}\times 10{)}^2+(4\times {10}^{-8}\times 2\times {10}^4{)}^2}$

$T=8\sqrt{2}\times {10}^{-4}N$

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

$\tan \theta =\frac{4\times {10}^{-8}\times 2\times {10}^4}{80\times {10}^{-6}\times 10}$

$\tan \theta =1$

$\theta ={45}^o$