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)   \(F = E/q\)