In this example we will analyze the motion of an electron that is released in an

Question

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 uniform and has magnitude E=3.0×104N/C. Suppose the lower plate has positive charge, so that the electric field is vertically upward, as shown in (Figure 1) . (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=9.11×1031kg.

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 vy2=v0y2+2ayy. The electron’s initial velocity v0y is zero, so

vy2=2ayy.

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:

Fy==|q|E=(1.60×1019C)(3.0×104N/C)4.80×1015N

Newton’s second law then gives the electron’s acceleration:

ay=Fyme=4.80×1015N9.11×1031kg=+5.27×1015m/s2

We want to find vy when y=0.010m. The equation for vy gives

vy==2ayy=2(5.27×1015m/s2)(0.010m)1.0×107m/s

Finally, vy=v0y+ayt gives the total travel time t:

t=vyv0yay=5.9×106m/s01.76×1015m/s2=1.9×109s

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 5×1018 N , or about 5×1014 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.

QUESTION: In this example, suppose a proton (mp=1.67×1027kg) is released from rest at the positive plate. What is its speed just before it reaches the negative plate?

Explanation / Answer

Solution: Strength of the electric field between the plates, E = 3.0*104 N/C;

The mass of the proton mp = 1.67*10-27 kg;

The charge on proton, q = 1.6*10-19 C

The proton will travel from positive plate to negative plate as the positive charge accelerates in the direction of the electric field.

Force on the proton due the electric field is given by,

Fy = q*E

Fy = (1.6*10-19 C)*( 3.0*104 N/C)

Fy = 4.80*10-15 N

Thus the magnitude of force acting on the proton is same as that of the electron (but in opposite direction); due the different mass of proton, the acceleration will be different.

ay = Fy/mp

ay =4.80*10-15 N/1.67*10-27 kg

ay = 2.8743*1012 m/s2

Vertical distance covered by the proton while accelerating is y = 0.01 m.

Since the proton was released from the rest, vo = 0 m/s

vy2 = vo2 +2*ay*y

vy = [vo2 +2*ay*y]

vy = [(0m/s)2 + 2*(2.8743*1012 m/s2)*(0.01m)]

vy = 2.3976*105 m/s

With rounding to three significant figures, the answer becomes

vy = 2.40*105 m/s

Thus the speed of the proton just before it reaches the negative plate is 2.40*105 m/s

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