A particle of mass 2.33 times 10^-26 kg and a magnitude of charge of 1.602 times

Question

A particle of mass 2.33 times 10^-26 kg and a magnitude of charge of 1.602 times 10^-19 C is accelerated from rest in the plane of the page through a potential difference provided by two parallel plates as shown to the right. The particle is injected through a hole in the right-hand plate into a region of space containing a uniform magnetic field of magnitude 1.25 Tesla oriented perpendicular to the plane of the page. The particle curves in a semicircular path of radius 2.54 cm and strikes a detector. a) Is the sign of the charge on the particle positive or negative? b) Is the direction of the magnetic field into the page or out of the page? c) Determine the potential difference of the two parallel plates.

Explanation / Answer

(a)

particle is attracted towards negative terminal and it speed increased this shows that particles would have positive sign

q i sposiitve

(b)

from the right hand rule, force is perpendicular to the velocity and directed towards the center of the semicircle

B would be out of page

(c)

v = Bqr/m

= ( 1.25 ) ( 1.6 * 10^-19 ) ( 2.54 * 10^-2)/2.33 * 10^-26 kg

=2.18 * 10^5 m/s

1/2 mv^2 = qV

V = 1/2 mv^2/q

= 1/2 * 2.33 * 10^-36 kg ( 2.18 * 10^5)^2/1.602 * 10^-19 C

= 3.47 * 10^7 V

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