Two adjacent parallel plate capacitors are used to deflect charged particles as
ID: 1399576 • Letter: T
Question
Two adjacent parallel plate capacitors are used to deflect charged particles as schematically displayed in Fig. 1. The relevant geometrical dimensions l and h are shown in the figure. The lower capacitor plates are grounded (zero potential) while the upper plates can be maintained at arbitrary controlling potentials V1 and V2, which are to be found in this problem given the following information. A positively charged particle of charge q and mass m enters the system at point A moving horizontally with velocity v. After executing some trajectory, the particle exits the system at point B, which has the same vertical position as point A. The velocity of the particle at point B is, however, not horizontal: the vertical component of the velocity there is found to be equal to vy. What are the potentials V1 and V2 used for the deflection? Gravitycan be neglected.
Explanation / Answer
The horizontal velocity component is not affected by the vertically-oriented E-field so it is constant.
the particle gains a downward velocity v1y in the 1st capacitor and double that value of upward velocity vfy in the 2nd capacitor
Since flight times through the two sections are equal, and -3 * the delta-v must be provided by the second section than by the 1st, V2/V1 = accel2/accel1 = -3.
F = Eq = Vq/h
a1y (acceleration in 1st capacitor ) = F/m = V1q/(hm)
t (per capacitor ) = v(x)/L
v1y (after flight through 1st capacitor ) = a1yt = V1qt/(hm)
vfy = -2v1y
vfy = -2V1qt/(hm)
V1 = -vfy*hm/(2qt), and V2 = -3V1