Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Home exercise 3.1.3

Interactive: Electron Beam Deflection

Home exercise 3.1.3

Cathode-ray tube

cf. Application 3.1.4

The electrostatic deflection system of a cathode-ray oscilloscope is depicted below. Electrons from a heated cathode are given an initial velocity $v_{0} = v_{0} a_{z}$ by a positively charged anode. The electrons enter at $z = 0$ into a region of deflection plates where a uniform electric field $E_{d} = - E_{d} a_{y}$ is maintained over a width $w$ . Ignoring gravitational effects, find the vertical deflection of the electrons on the fluorescent screen at $z = L$ .

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 3-3.

Try it yourself first.

Strategy

This is mechanics with the Coulomb force $F = q E$ : a constant force between the plates (uniformly accelerated motion, as in projectile motion), no force beyond them (straight-line motion). We solve Newton's law in the two regions and join the solutions where the electron leaves the plates.

Inside the plates

The electric field between the plates is $E_{d} = - E_{d} a_{y}$ . The charge of the electron is $- e$ , so by Coulomb's law, $F = q E$ , the force acting on the electron inside the plate system is

F = (- e) (- E_{d} a_{y}) = e E_{d} a_{y} = m \frac{d v _{y}}{d t} a_{y},

where $m = m_{e}$ is the electron mass. Solving for the vertical velocity,

\frac{d v _{y}}{d t} = \frac{e E _{d}}{m} \Rightarrow v_{y} (t) = \frac{e E _{d}}{m} t + C_{1} .

The electron has no initial vertical velocity, so

v_{y} (0) = 0 \Rightarrow C_{1} = 0 \Rightarrow v_{y} (t) = \frac{e E _{d}}{m} t = \frac{d y}{d t} .

Integrating once more, $y = \frac{e E _{d}}{2 m} t^{2} + C_{2}$ , and since the electron is at $y = 0$ at $t = 0$ ,

C_{2} = 0 \Rightarrow y = \frac{e E _{d}}{2 m} t^{2} .

There is no force along $z$ , so the electron keeps its axial velocity $v_{0}$ . Eliminating $t = z / v_{0}$ shows that the path inside the plates is the parabola $y = e E_{d} z^{2} / (2 m v_{0}^{2})$ . Denoting by $t_{1}$ the time the electron spends inside the plates,

v_{0} t_{1} = w \Rightarrow t_{1} = \frac{w}{v _{0}},

so the vertical position and the vertical velocity of the electron when it leaves the plates are

d_{1} v_{y} (t_{1}) = \frac{e E _{d}}{2 m} (\frac{w}{v _{0}})^{2}, = \frac{e E _{d}}{m} (\frac{w}{v _{0}}) .

Between the plates and the screen

Beyond the plates there is no field and the electron moves on a straight line. Denoting by $t_{2}$ the time it takes to travel from the plates to the screen,

v_{0} t_{2} = L - w \Rightarrow t_{2} = \frac{L - w}{v _{0}},

so the additional vertical displacement it gains is

d_{2} = v_{y} (t_{1}) t_{2} = \frac{e E _{d}}{m} (\frac{w}{v _{0}}) (\frac{L - w}{v _{0}}) .

Total deflection

The total vertical deflection of the electron on the screen is

d_{0} = d_{1} + d_{2} = \frac{e E _{d}}{2 m} (\frac{w}{v _{0}})^{2} + \frac{e E _{d}}{m} (\frac{w}{v _{0}}) (\frac{L - w}{v _{0}}) = \frac{e E _{d}}{m} (\frac{w}{v _{0}^{2}}) (\frac{w}{2} + L - w),

Answer

d_{0} = \frac{e E _{d}}{m} (\frac{w}{v _{0}^{2}}) (L - \frac{w}{2}) [m]

Interpretation

The exit angle satisfies $tan α = v_{y} (t_{1}) / v_{0} = e E_{d} w / (m v_{0}^{2})$ , so $d_{0} = (L - w /2) tan α$ : seen from the screen, the electron appears to come from the middle of the plates. The deflection is proportional to $E_{d}$ , i.e. to the plate voltage, which is why the screen can be used as a voltmeter. With the distance $L_{2} = L - w$ measured from the end of the plates the result reads

d_{0} = \frac{e E _{d} w}{m v _{0}^{2}} (\frac{w}{2} + L_{2}),

the form used in Application 3.1.4 and in Worked example 3.1.3.

Check

Units: $e E_{d} / m$ is an acceleration [m/s²] and $w (L - w /2) / v_{0}^{2}$ is [s²], so $d_{0}$ is in metres. Limits: $E_{d} = 0$ gives no deflection, $w \to 0$ (no plates) gives none either, and a faster electron (larger $v_{0}$ ) spends less time between the plates and is deflected less, as $1/ v_{0}^{2}$ .

Change the plate voltage, the plate length and the accelerating voltage: the spot on the screen moves in proportion to the plate voltage and falls back as $1/ v_{0}^{2}$ when the electrons are made faster.

InteractiveLoading simulation…

Home exercise 3.1.3: Cathode-ray tube