(9 marks ) TR 2. The following diagram shows a proton entering the region between two oppositely charged plates . The plates are 45.0 mm long and are separated by 17.5 mm . A potential difference of 0.208V generates an electric field between the plates. The proton is given an initial velocity of 1.35 x 10 4 m/s, in the positive x direction Use this information to determine the y-component of the displacement of the pro - ton when it leaves the region between the plates . (This displacement is how far the proton has moved above or below the x-axis the instant it leaves the region between the plates .) Be sure to begin your solution with a free body diagram and an analysis of the physics principles that apply. V = 0.208V 17.5 mm 45.0 mm Answer: -Fe Vi + Fe E = V/d = 0.208V /0.075m = 2.773 V/m Fe = 4.443 *10 ^-19 N [up ] Fe = ma a = 2.656 *10 ^8 m/s2 [up ] v=d/t t = d/v = 0.045 m / 1.35 *10 ^4 m/s = 3.333 *10 ^-6 s Y dy = (a*^2)/2 = 0.00148 m or 1.48 mm [above x-axis ]