A particle of mass 6.7 x 10-27 kg and charge +3.2 x 10-19 C is placed near the top plate
an electric field similar to that shown in Figure 17.11. The potential difference across the
plates is 600V and the separation of the plates is 12 mm. Determine
(a) the force on the particle,
(b) the acceleration of the particle,
(c) the speed of the particle when it reaches the bottom plate.
(a) force = Eq = Vald = (600x 3.2 x 10-19)/12 x 10-3 = 1.6 × 10-14 N
(b) acceleration = force/mass = 2.3 x 10-12/6.7 x 10-27 = 2.4 x 1012m s-²
(c) v² = 0 + 2as = 2 x 2.4 x 1012 x 12 x 10-3 so speed v = 2.4 x 105 m s-1
Now it's your turn
6
An electron enters an electric field similar to that shown in Figure 17.12 with a
horizontal velocity of 7.0 x 107 m s-1. The horizontal length of the plates is 2.0c
electric field strength is 3.2 x 105 V m-1. Calculate the vertical displacement of
electron for the time it is between the plates.