Question

 An electron enters a parallel plate apparatus as seen in Figure 7 with an initial velocity of 2.05E6 m/s. The magnitude of the electric field strength across the plates is 237.7 N/C and the length of the plates is 7.08 cm. Determine the exit velocity of the electron.

Answer 1

Find the force:

$$F=Eq.$$

Find acceleration:

$$a=\frac Fm=\frac{Eq}{m}.$$

Determine the exit velocity:

$$v=\sqrt{v_0^2+2ad}=\sqrt{v_0^2+\frac{2Eqd}{m}}=3.18·10^6\text{ m/s}.$$