Problem 10-49 Electron deflection - Part 2 - 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

Correct!

The acceleration of the electron in the y-direction is:

(A)   \(a = F/m = (3.52 \times10^{15} N)/(9.1 \times 10^{-31} \; kg) = 3.87 \times 10^{15}\; m/s^2\)

(B)   \(a= m/F =(9.1 \times 10^{-31} \; kg)/ (3.52 \times 10^{15}\; N) = 2.58 \times 10^{-16}\; m/s^2\)

(C)    Newton's Law doesn't apply in electric fields so \(a = 0\)