A small charged particle of mass 3.5 times 10^9 kg is traveling rightward betwee
ID: 1636597 • Letter: A
Question
A small charged particle of mass 3.5 times 10^9 kg is traveling rightward between two plates separated by a distance d = 60 cm, as shown below. The electric field between the plates has magnitude 2.5 times 10^5 N/C and is directed rightward. The particle's kinetic energy at the left plate is three times its kinetic energy at the right plate, and the particle's speed at the right plate is 5 times 10^4 m/s. Ignore the effect of gravity. (a) Is the particle positively charged or negatively charged? Justify your answer briefly but clearly. (b) Find the charge (with correct sign) of the particle, as well as the potential difference (with correct sign) through which the particle has moved.Explanation / Answer
given mass of particel m = 3.5*10^-9 kg
distance between the plates, d = 0.6 m
electric field between the plates, E = 2.5*10^5 N/C , directed rightwards
KE at left plate = 3*KE at right plate
Speed at rigth plate , v = 5*10^4 m/s
so speed at left plate = u
0.5mu^2 = 3*0.5mv^2
u^2 = 3*v^2
u = v*sqroot(3) = 8.66*10^4 m/s
a. the electric field is directed rightwards, and the particle slows down as it moves towards right, so it is experiencing sa force towards left
this means the particle is -vely charged
b. let the charge on the particel be q
then aceleration of particle is a
using 2*a*d = v^2 - u^2
2*a *0.6 = (5*10^4)^2 - (sqroot(3)*5*10^4) = -2*25*10^8 = -50*10^8
a = -41.667*10^8 m/s/s
hence force on the particel = ma = qE
q = ma/E = 3.5*10^-9*a/2.5*10^5 = -5.833*10^-5 C
and, V = -E*d = -2.5*10^5*0.6 = -1.5*10^5 V