From Newton's second law:

$\overline{){\mathbf{F}}{\mathbf{=}}{{\mathbf{m}}}_{{\mathbf{e}}}{\mathbf{a}}}$

The force on electron moving under the influence of an electric field:

$\overline{){\mathbf{F}}{\mathbf{=}}{\mathit{q}}{\mathit{E}}}$

We'll also need to use the kinematic equation:

$\overline{){{{\mathbf{v}}}_{{\mathbf{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{{{\mathbf{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{a}}{\mathbf{\u2206}}{\mathbf{x}}}$

We get a by equating the first and second equations.

$\begin{array}{rcl}{\mathbf{m}}_{\mathbf{e}}\mathbf{a}& \mathbf{=}& \mathbf{q}\mathbf{E}\\ \mathbf{a}& \mathbf{=}& \frac{\mathbf{q}\mathbf{E}}{{\mathbf{m}}_{\mathbf{e}}}\\ & \mathbf{=}& \frac{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{30}\mathbf{\times}{\mathbf{10}}^{\mathbf{4}}\mathbf{)}}{\mathbf{9}\mathbf{.}\mathbf{11}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{31}}}\end{array}$

The electric field strength is 2.30×10^{4} N/C inside a parallel-plate capacitor with a 0.600 mm spacing. An electron is released from rest at the negative plate.

What is the electron's speed when it reaches the positive plate?

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 Fields in Capacitors concept. You can view video lessons to learn Electric Fields in Capacitors. Or if you need more Electric Fields in Capacitors practice, you can also practice Electric Fields in Capacitors practice problems.