23. [0/1 Points] DETAILS PREVIOUS ANSWERS OSUNIPHYS1 22.4.WA.053. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.276 x 10'5 m/s, as shown in the figure. The distance between the two plates is 1.56 cm. The proton enters the capacitor halfway between the top plate and the bottom plate; that is, a distance r = 0.780 cm from each plate, as shown in the gure. The capacitor has a 2.80 x 10'4 MC uniform electric field between the plates that points downward from the top plate to the bottom plate. Neglecting gravitational forces, what horizontal distance does the proton traverse before the proton hits the bottom plate? x How does force relate to the electric field? Can you solve Newton's second law for acceleration? What kinematic equation can you use to determine the final velocity? m ma = RE a = eE m In magnitude a = eE m Now this is a constant acceleration therefore we can use kinematics Egh of motionAlong y- direction Using first Kinematics egh, Vy = ly + agt vy = 0 + CE m Vy = CE m Then Using third Kinematics Egh, weget, vy = My + zay ( By ) 2 ( ex ) ( 1 ) Vy = 2eEr 2 m Using qh 2 ) in Igh , we get zeEr = eEt m mof * = 28. m RE 2 X 0 7 8 X 1 0 X 1 . 6 7x 10 X 2 . BOX TO 2 = 0.762x 10 Bee 12 = 7.62 x 10 sec Now the horizontal distance ( d ) that proton traverse before the proton hits the bottom plate is given by Second Kinematics Egh, No acceleration along re -direction of = uut5 4 -2 d = 1 27 6 x 10 X 7. 62 X 10 m of of = 9. 72XO - m