Question

The figure below shows a proton entering a parallel-plate capacitor with a speed of 2.10 times 10^5 m/s. The proton travels a horizontal distance x = 5.20 cm through the essentially uniform electric field. The electric field of the capacitor has deflected the proton downward by a distance of d = 0.720 cm at the point where the proton exits the capacitor. (You can neglect the effects of gravity.) Using kinematics, find the vertical acceleration (including sign) of the proton in this electric field. Find the magnitude of the force on the proton. Find the strength of the electric field within the capacitor. Find the speed of the proton when it exits the capacitor.

Answer 1

b) x-component ofvelocity V v y-component ofvelocity vat 2dv The final velocity of the electron as it exits from the capacitor 2av -(2.1 *10 A5). Sqrt (1+4(0.0072/0.052) A2) 2.17903 * 10A5 m/s The magnitude ofelectric force isF Eq -(1.33 10 2 N/C) (1.602*10-19 C) = 2.13066 * 10^-17 N C)1.3310 2 N/O as previously calculated

D) The proton follows a parabola as in projectile motion The horizontal component of velocity remains 2.1 x 1045 m/s throughout flight because no accelerating force applies in horizontal direction. However, although the proton's initially vertical velocity, uy, its zero at the start, the proton is now moving with vertical velocity, vy on leaving the plates because it is under the influence of a we calculated earlier. But uy -zero vyA2 -2qEs/m vyA2 -2.335x 10A13 (m/s)2 The velocity on leaving the plates will be the vector sum of the vertical and horizontal velocities v - root of [2.335x 10 13+ (2.1 x 10A5)2] v 4.84 x 10-6 m/s