Problem 10-49 Electron deflection - 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!

Positive charges move with the field so negative charges move against the field.

The magnitude of the force on the electron is given by:
 
(A)   $F = q/E$

(B)   $F = qE$

(C)   $F = E/q$