Problem 10-49 Electron deflection - Part 3 - C
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
\(F = qE = (1.6 \times 10^{-19} \; C)(2.2 \times 10^4\; N/C) = 3.52 \times 10^{15} \; N \)
No. Whenever unbalanced forces act there will be an acceleration. Newton's Laws do apply.