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 $a = 0$