Two charged, parallel, flat conducting surfaces are spaced $\mathbf{d}\mathbf{}\mathbf{=}\mathbf{\text{1.00cm}}$apart and produce a potential difference$\mathbf{\Delta V}\mathbf{}\mathbf{=}\mathbf{}\mathbf{\text{625V}}$ between them. An electron is projected from one surface directly toward the second. What is the initial speed of the electron if it stops just at the second surface?

The separation between the plates,$\mathrm{d}=1.\text{0cm}$.

The potential difference, $\mathrm{\Delta V}=\text{625V}$.

According tothe law of conservation of energy can neither be created nor destroyed only converted from one form of energy to another.

Formulae:

The change in potential energy is define as,

$\mathrm{\Delta U}=\mathrm{\Delta V}\times \mathrm{e}$

The kinetic energy is,

$\mathrm{\Delta K}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2$

Where, $\mathrm{K}$ is kinetic energy, $\mathrm{m}$ is mass, $U$ is potential energy, and $\mathrm{v}$ is velocity. and $\mathrm{V}$ is a potential difference.

Apply the Law of Conservation of Energy to get the initial speed${\mathrm{V}}_{\mathrm{i}}\text{m/s}$at the first plate,

$\begin{array}{c}({\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}})=-\mathrm{\Delta U}\\ =-\mathrm{\Delta V}\times \mathrm{e}\end{array}$

Here,

The electron charge, $\mathrm{e}=1.6\times {10}^{-19}\text{C}$

Substitute known values in the above equation.

$0-\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2=-625\text{V}\times 1.6\times {10}^{-19}\text{C}$

Solving for ${\mathrm{v}}_{\mathrm{i}}$,

$\begin{array}{c}{\mathrm{v}}_{\mathrm{i}}=\sqrt{\frac{\left(625\text{V}\right)\times (1.6\times {10}^{-19}\text{C})\times 2}{9.1\times {10}^{-31}\text{kg}}}\\ =\sqrt{\text{219.78}\times {10}^{12}{\text{m}}^2/{\text{s}}^2}\\ =1.48\times 1{\text{0}}^{\text{7}}\text{m}/\text{s}\end{array}$.

Hence, the initial speed of the electron is $1.48\times 1{\text{0}}^{\text{7}}\text{m}/\text{s}$.