Problem: The electric field strength is 2.30×104 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?

FREE Expert Solution

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:

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

We'll also need to use the kinematic equation:

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

We get a by equating the first and second equations.

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

Problem Details

The electric field strength is 2.30×104 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?