Example from Physics, 18.4 Electric Potential

X-ray Tube (part 1)

Dentists use X-rays to image their patients’ teeth and bones. The X-ray tubes that generate X-rays contain an electron source separated by about 10 cm from a metallic target. The electrons are accelerated from the source to the target by a uniform electric field with a magnitude of about 100 kN/C, as drawn in Figure 18.26. When the electrons hit the target, X-rays are produced. (a) What is the potential difference between the electron source and the metallic target? (b) What is the kinetic energy of the electrons when they reach the target, assuming that the electrons start at rest?

This diagram shows a vertically oriented ellipse labeled “Electron source” on the left, and a trapezoid labeled “Target” on the right. Between them, in the middle section of the diagram, there are four tiny spheres labeled “Electrons”, each of which has a short, rightward pointing arrow attached to it. Five long, horizontal arrows point from the trapezoid to the ellipse, and a label between them says “Electric field”. Near the middle of the left edge of the trapezoid is a yellow circle from which emerge three long, orange, wavy lines, as well as a number of short yellow and red lines. The long wavy lines are labeled “X-rays”.

Figure 18.26 In an X-ray tube, a large current flows through the electron source, causing electrons to be ejected from the electron source. The ejected electrons are accelerated toward the target by the electric field. When they strike the target, X-rays are produced.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy for (a)
Use the equation $Δ V = - E (x_{f} - x_{i})$ to find the potential difference given a constant electric field. Define the source position as $x_{i} = 0$ and the target position as $x_{f} = 10 cm$ . To accelerate the electrons in the positive x direction, the electric field must point in the negative x direction. This way, the force $F = q E$ on the electrons will point in the positive x direction, because both q and E are negative. Thus, $E = - 100 \times 10^{3} N/C$ .
Solution
Solution for (a)

Using $x_{i} = 0$ and $x_{f} = 10 cm = 0.10 m$ , the equation $Δ V = - E (x_{f} - x_{i})$ tells us that the potential difference between the electron source and the target is
$Δ V = - E (x_{f} - x_{i}) = - (- 100 \times 10^{3} N/C) (0.10 m - 0) = + 10 kV.$
18.31

0 of 2 steps open

How did it go?

Strategy for (a)

Solution

Strategy for (a)

Solution