Question 4: Use F = ma to solve for the motion of the electron

Answer to Question 4

We drop an electron of charge q = - e with zero initial speed v0 near the center of the ring at a distance y0 from the origin. The initail conditins at t=0 are as follow:

i) y[t=0] = y0 where y0<< R = radius of the ring.

ii) v[t=0] = v0 = 0 where v0 is the initial speed of the electron

1) F = m a

2) a=dv [t] / dt = dy[t] ^2 /dt = y[t]''

q=-e

3) F= q Ep = - e Ep

4) Ep = k Q y / R ^3

5) w0 = 2Pi / T= k Q e / (m R^3) = 1

Substituting equations ( 2,3,4 and 5) into 1; the differential equation for the motion of the electron through the ring at any desired time becomes

6) y[t]'' + y[t] = 0 ,

with solution

7) y[t] = y0 cos[t].

The velocity of the electron is the derivative of y[t] with respect to t or

8) y[t]' = v[t] = - y0 sin[t]

The acceleration of the electron is the derivative of the velocity with respect to t or

9) y[t]'' = a[t] = - y0 cos[t]

Equations (7, 8 and 9 ) are satisfied by the equation ( 6, i and ii )

( End of Answer to Q4 )

• Answer to Question 1 (Conceptual outcome)
• Answer to Question 2 (Electric potential )
• Answer to Question 3 (Potential and field near the center)
• Answer to Question 4 (Electron's equation of motion)
• Answer to Question 5 (Explain physical results)

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