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 E⃗ 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=9.11×10−31kg. 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 E⃗ =F⃗ ′/q′ to find this component:

Fy==|q|E=(1.60×10−19C)(3.0×104N/C)4 .80×10−15N

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

ay=Fyme=4.80×10−15N9.11×10−31kg=+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)(0.01 0m)−−−−−−−−−−−−−−−−−−−−−−−√1.0×107m /s

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

t=vy−v0yay=5.9×106m/s−01.76×1015m/s 2=1.9×10−9s

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=1.67×10−27kg) is released from rest at the positive plate. What is its speed just before it reaches the negative plate?