The vertical deflection of the electron is expressed as:

$\overline{){\mathbf{\u2206}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{v}}}_{\mathbf{0}\mathbf{y}}{\mathbf{t}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{a}}}_{{\mathbf{y}}}{{\mathbf{t}}}^{{\mathbf{2}}}}$

Acceleration in the vertical direction, a_{y} is:

$\overline{){{\mathbf{a}}}_{{\mathbf{y}}}{\mathbf{=}}\frac{\mathbf{q}\mathbf{E}}{\mathbf{m}}}$

Figure 17.44 shows an electron passing between two charged metal plates that create an 100 N/C vertical electric field perpendicular to the electron's original horizontal velocity. (These can be used to change the electron's direction, such as in an oscilloscope.) The initial speed of the electron is 1.00 × 10^{6} m/s, and the horizontal distance it travels in the uniform field is L = 10.0 cm.

**(a)** What is its vertical deflection?**(b)** What is the vertical component of its final velocity?**(c)** At what angle does it exit? Neglect any edge effects.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Electric Field concept. You can view video lessons to learn Electric Field. Or if you need more Electric Field practice, you can also practice Electric Field practice problems.