Question

A ball of mass 100 g and having a charge of $4.9\times {10}^{-5}C$ is released from rest in a region where a horizontal electric field of $2.0\times {10}^{4}N{C}^{-1}$ exists. (a) Find the resultant forc eacting on the ball. (b) What will be the path of the ball? (c) Where will the ball be at the end of 2 s?

Answer 1

$M=100gm,q=4.9\times {10}^{-5},F_e=mg,f_e=eq,e=2\times {10}^{4}N/C$
So, the particle moves due to the resultant R
$R^2=F_g^2+f_e^2=(0.1\times 9.8{)}^2+(4.9\times {10}^{-5}\times 2\times {10}^4{)}^2=0.9604+96.04\times {10}^{-2}=1.9208So,R=1.3859N\tan \theta =1So,\theta ={45}^{\circ }$
Hence path is straight along resultant force at an angle ${45}^{\circ }$ with horizontal displacement (vertical)
$=\frac{1}{2}\times 9.8\times 2\times 2=19.6m$
Displacement (horizontal)
$=s=\frac{1}{2}.\times a\times t^2=\frac{1}{2}\times t^2\times 2\times 2=19.6m$
Net displacement
$=\sqrt{(19.6{)}^2+(19.6{)}^2}=\sqrt{768.932}=27.7m$