Problem 10-49 Electron deflection - Part 8 - A

An electron in an oscilloscope beam travels between the deflecting plates with an initial velocity of \(2.0 \times 10^7\; m/s\) parallel to the plates, which lie in a horizontal plane. The electric field is \(2.2 \times 10^4\; N/C\) downward, and the plates have a length of \(4.0\; cm.\) When the electron leaves the plates, (a) how far has it dropped or risen? (Specify which.) (b) what is its velocity (magnitude and direction)?
[Ans. (a) risen \(7.7 \times 10^{-3}\; m\)   (b) \(2.1 \times 10^7\; m/s\; \text{ at} \;21 ^\circ\) above horizontal]

Accumulated Answer

\(F = qE = (1.6 \times 10^{-19} \; C)(2.2 \times 10^4\; N/C) = 3.52 \times 10^{15} \; N \\ a = F/m = (3.52 \times 10^{-15} \; N)/(9.1 \times 10^{-31} \; kg) = 3.87 \times \;10^{15} \; m/s^2 \\ \text{Time of flight}\; = 2.0 \times 10^{-9} s \\ y = v_{0y^t} + ½\; a_y\; t^2 \\ v_{0y} = 0 \\ y = h = 0 + ½ (3.87 \times 10^{15}\; m/s^2)( 2.0 \times 10^{-9} s)^2 = 7.7X \times 10^{-3}\; m \; \text{ (answer to part (a))}\)

Correct!

In fact any of the three equations can be used. Let's use the first:

\(v_y = v_{0y} + a_{y^t} = 0 + (3.87 \times 10^{15} \; m/s^2)( 2.0 \times 10^{-9} \; s) = 0.77 \times 10^7\; m/s\)

The direction of the electron is given by:

(A)   \(\theta = \tan^{-1}v_x/v_y = \tan^{-1}(2.0/0.77) = 69^\circ\)

(B)   \(\theta = \tan^{-1}v_y/v_x = \tan^{-1}(0.77/2.0) = 21^\circ\)