In an old television tube, a narrow beam of electrons leaves the gun and moves along the axis of…

In an old television tube, a narrow beam of electrons leaves the gun and moves along the axis of the tube at $2.4 \times 10^{7}\ \text{m/s}$ . It passes midway between two flat horizontal plates, each $3.0\ \text{cm}$ long and $1.0\ \text{cm}$ apart, as in the figure. Between the plates there is a uniform field of $1.2 \times 10^{4}\ \text{N/C}$ pointing straight up, and outside them there is no field. The plates are wide, so ignore the field at their edges. A flat screen stands $18\ \text{cm}$ beyond the far edge of the plates, at right angles to the axis. Take $e = 1.60 \times 10^{-19}\ \text{C}$ and $m_e = 9.11 \times 10^{-31}\ \text{kg}$ , and neglect gravity. Take up as positive in (c) and (e).

Side view of an electron beam. An electron moves to the right along a dashed axis at 2.4 times 10 to the 7 metres per second, midway between two horizontal plates 3.0 cm long and 1.0 cm apart. Small arrows between the plates point straight up and show the field E. A vertical screen stands 18 cm beyond the far edge of the plates; the gap is shortened in the drawing.

What is the size of the acceleration of an electron between the plates?
How long does an electron take to pass between the plates?
What is its vertical displacement as it leaves the plates?
At what angle to the axis does it leave the plates?
Where does the beam strike the screen, measured from the axis?
The field is raised slowly. Above what field does the beam strike a plate instead of leaving?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Between the plates the electron has a constant sideways acceleration and a steady speed along the axis: a projectile. After the plates it moves in a straight line to the screen.

Step 1. Acceleration (a). $a = \dfrac{eE}{m_e} = \dfrac{1.60 \times 10^{-19}\ \text{C} \times 1.2 \times 10^{4}\ \text{N/C}}{9.11 \times 10^{-31}\ \text{kg}} = 2.11 \times 10^{15}\ \text{m/s}^2$ . The field points up, so the force on the electron points down.

Step 2. Time (b). $t = \dfrac{l}{v_0} = \dfrac{0.030\ \text{m}}{2.4 \times 10^{7}\ \text{m/s}} = 1.25\ \text{ns}$ .

Step 3. Displacement at the exit (c). $y_1 = -\tfrac12 at^2 = -1.65\ \text{mm}$ , below the axis. It is less than $5.0\ \text{mm}$ , so the electron clears the lower plate.

Step 4. Angle (d). $v_y = at = 2.63 \times 10^{6}\ \text{m/s}$ , so $\tan\varphi = \dfrac{v_y}{v_0} = 0.1098$ and $\varphi = 6.26^\circ$ , pointing down.

Step 5. The spot (e). After the plates the electron goes straight. In $18\ \text{cm}$ it falls a further $18\ \text{cm} \times 0.1098 = 19.76\ \text{mm}$ , so the spot is at $-1.65\ \text{mm} - 19.76\ \text{mm} = -21.4\ \text{mm}$ . As a check, the straight line comes from the middle of the plates: $(1.5\ \text{cm} + 18\ \text{cm}) \times 0.1098 = 21.4\ \text{mm}$ below the axis.

Step 6. Hitting a plate (f). The beam hits the lower plate when the displacement at the exit reaches $5.0\ \text{mm}$ . The displacement grows in step with $E$ , so $E_{\max} = 1.2 \times 10^{4}\ \text{N/C} \times \dfrac{5.0\ \text{mm}}{1.647\ \text{mm}} = 3.64 \times 10^{4}\ \text{N/C}$ .

Answer: (a) $2.11 \times 10^{15}\ \text{m/s}^2$ . (b) $1.25\ \text{ns}$ . (c) $-1.65\ \text{mm}$ . (d) $6.26^\circ$ , downwards. (e) $-21.4\ \text{mm}$ . (f) $3.64 \times 10^{4}\ \text{N/C}$ .

Watch out. The electron is pushed against the field, so it goes down, not up. Gravity really can be left out: the electric acceleration is about $2.2 \times 10^{14}$ times $g$ .

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