Question

A particle of charge +15.5C and mass 7.69 × 10-5 kg is released from rest in a region where there is a constant electric field of +611 N/C. What is the displacement of the particle after a time of 4.38 × 10-2 s?

Answer 1

$\dpi{120} We\,\,know\,\,force(F)\,\,acting\,\,on\,\,a\,\,charge\,\,q\,\,over\,\, electric\,\, field\,\,(E)$

$F=q.E$

$For\,\, given\,\, problem\,\,q=+15.5\,C\,\,and\,\,E=+611\, N/C\\ \therefore\,\,force\,\, F=15.5\times611\,\,N\\ \Rightarrow\,\,\,\,\,\,\,\,\,\, \,\, \,\, \,\, \,\,\,\,\,\, =9470.5\,\,N$

$Now\,\, from\,\, Newton's\,\, second\, law\,\, force\,\,F=ma\\ Where\,\,m\,\, is\,\,mass\,\, and\,\,a\,\,is\,\, acceleration\\ \therefore \,\,9470.5=7.69\times10^{-5}a\,\, \,\, \,\, \,\, \,\,\,\,\,\,\,\, \,\,\,\,\,\, \,\, [given\,\,m=7.69\times10^{-5}\,\,kg ]\\ a=\frac{9470.5}{7.69\times10^{-5}}\,\,m/s^2\\ a=1.232\times10^8\,\,m/s^2$

$Now\,\, displacement\\ s=v_it+\frac{1}{2}at^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [Here\,\, initial\,\,velocity\,\,v_i=0\,\,and\,\,time\,\,t=4.38\times10^{-2}\,\,s]\\ =0+\frac{1}{2}\times(1.232\times10^8)\times(4.38\times10^{-2})^2\,\,\,\,m\\\\ =118176\,\,m$