Problem 10-49 Electron deflection - Part 4 - B

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

\(F = qE = (1.6 \times 10^{-19} \; C)(2.2 \times 10^4\; N/C) = 3.52 \times 10^{15} \; N \\ a = F/m = (3.52 \times 10^{-15} \; N)/(9.1 \times 10^{-31} \; kg) = 3.87 \times \;10^{15} \; m/s^2\)

Correct!

There is no force acting in the \(x\)-direction so \(a_x = 0.\)

Therefore the time the electron is between the plates is:

(A)    \(4.0/(2 \times 10^7) = 2 \times 10^{-7 }s\)

(B)    \(4.0 \times 10^{-2}/(2 \times 10^7) = 2 \times 10^{-9} s\)

(C)    \(4.0 \times 10^{-3}/(2 \times 10^7) = 2 \times 10^{-10} s\)