Question

An electron is released from rest on the axis of a uniform positively charged ring, 0.100 m from the ring's center. If the linear charge density of the ring is +0.100 µC/m and the radius of the ring is 0.300 m, how fast will the electron be moving when it reaches the center of the ring?

Answer 1

Using work-energy theorem to calcualte speed of electron:

dPE = dKE

q*dV = KEf - KEi

KEi = 0, SInce electron is released from rest

dV = Vf - Vi

Vi = potential at inital position

Vi = electric potential due to a uniformly charged ring at distance 'a' is = k*Q/sqrt (r^2 + a^2)

here r = radius of ring = 0.300 m

a = 0.100 m

Q = Charge on ring = 2*pi*r*lambda

lambda = linear charge density

Q = 2*pi*0.300*0.100*10^-6 C/m

Q = 1.89*10^-7 C

Now Vf = potential at the center of ring = kQ/r = k*2*pi*r*lambda/r = 2*pi*k*lambda

and q = e

So,

q*(Vf - Vi) = 0.5*m*v^2

v = electron's velocity at the center

v = sqrt (2*q*(Vf - Vi)/m)

v = sqrt (2*e*2*pi*k*lambda*(1 - r/sqrt (r^2 + x^2)))

Using known values

v = sqrt [2*1.6*10^-19*2*pi*9*10^9*0.100*10^-6(1 - 0.300/sqrt (0.100^2 + 0.300^2))/(9.1*10^-31)]

v = 1.01*10^7 m/sec

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