# Problem 10-49 Electron deflection - Part 3 - 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 $F = qE = (1.6 \times 10^{-19} \; C)(2.2 \times 10^4\; N/C) = 3.52 \times 10^{15} \; N \\ a = F/m = (3.52 \times 10^{-15} \; N)/(9.1 \times 10^{-31} \; kg) = 3.87 \times \;10^{15} \; m/s^2$

Correct!

Now we need to determine the time during which the acceleration acts. Which of the following statements is correct?

(A)  The particle is accelerated in the $x$-direction.

(B)  The acceleration in the $x$-direction is zero.