Practice Problem 17.5 em 11> In this example we will analyze the motion of an el

Question

Practice Problem 17.5 em 11> In this example we will analyze the motion of an electron that is released in an electric field. The terminals of a 100 V battery are connected to two large parallel, horizontal plates 1.0 cm apart. The resulting charges on the plates produce an electric field E in the region between the plates that is very nearly unifom and has magnitude E-3.0 x 10 N/C. Suppose the lower plate has positive charge, so that the electric field is vertically upward, as shown in (The thin pink arrows represent the electric field.) If an electron is released from rest at the upper plate, what is its speed just before it reaches the lower plate? How much time is required for it to reach the lower plate? The mass of an electron is me = 9.11 × 10 31 kg The thin arrows represent the uniform electric field cm 100 V SOLUTION SET UP We place the origin of coordinates at the upper plate and take the +y direction to be downward, toward the lower plate. The electron has negative charge, q =-e, so the direction of the force on the electron is downward, opposite to the electric field. The field is uniform, so the force on the electron is constant. Thus the electron has constant acceleration, and we can use the constant-acceleration equation t'y2-VOy2 + 2ayy. The electron's initial velocity woy is zero, so SOLVE The force on the electron has only a y component, which is positive, and we can solve E = F /q, to find this component F, 19 E = (1.60 x 10-19 C)(3.0 × 104 N/C) 4.80 × 10-15N = Newton's second law then gives the electron's acceleration 4.80×10 ily = +5.27 × 1015m/s2 a,-me-9.11x10-31 kg We want to find Uy when y = 0.010 m. The equation for uy gives v, v 2arV /2(527 × 101"m/s*)(0.010 m) 1.0×107 m/s = Finally, vyVOy + t gives the total travel time t y-0y 5.9x106 m/s-0 t=-a,-= 1.76x 1015 m/s2 1.9× 10 REFLECT The acceleration produced by the electric field is enormous; to give a 1000 kg car this acceleration, we would need a force of about 5x1018 N or about 5x1014 tons. The effect of gravity is negligible. Note again that negative charges gain speed when they move in a direction opposite to the direction of the electric field Part A - Practice Problem: In this example, suppose a proton (mp = 1.67 × 10 27 kg) is released from rest at the positive plate. What is its speed just before it reaches the negative plate? Express your answer in meters per second to three significant figures 2.1.6.10s m s

Explanation / Answer

Part A)
kinetic energy gained by the proton = workdone on the proton by electric force

(1/2)*m*v^2 = F*d

(1/2)*m*v^2 = q*E*d

v = sqrt(2*q*E*d/m)

= sqrt(2*1.6*10^-19*3*10^4*0.01/(1.67*10^-27))

= 2.40*10^5 m/s

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