A singly ionized particle of mass 5.00 times 10^-26 kg is accelerated from rest
ID: 1998937 • Letter: A
Question
A singly ionized particle of mass 5.00 times 10^-26 kg is accelerated from rest through a potential difference of 500 volts. The particle then enters horizontally midway between two charged parallel plates separated by d = 2.00 cm. A potential difference is created between the plates by charging the upper plate to +140 volts and the lower plate to +120 volts. A uniform magnetic field is created by electromagnets is a direction perpendicular to the page, pointing into the page with magnetic field strength of 120 mT. Determine the net acceleration acting on the charge due to the combined magnetic and electric fields as it starts to pass between the plates.Explanation / Answer
Speed v after accelerration through 500 V = sqrt(2*500*1.6e-19/5e-26) = 56568 m/s
Net force = qvB + qE
= [1.6e-19*56568*0.120 + 1.6e-19*(140-120)/0.02]
= 1.246*10^-15 N
acceleration = Force/mass = 1.246e-15/5e-26
= 2.49*10^10 m/s^2