Electric force,

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

Newton's second law,

$\overline{){\mathbf{F}}{\mathbf{=}}{\mathit{m}}{\mathit{a}}}$

UAM equations,

$\overline{)\mathbf{}{{\mathit{v}}}_{{\mathit{f}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{+}}{\mathit{a}}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{+}}{\frac{1}{2}}{\mathit{a}}{{\mathit{t}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{{{\mathit{v}}}_{{\mathit{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{}}{{{\mathit{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{+}}{\mathbf{2}}{\mathit{a}}{\mathbf{\u2206}}{\mathit{x}}}$

The magnitude of the electric force experienced by a proton or an electron is equal since they have the same charge.

The ratio of electrons acceleration to proton's acceleration,

$\begin{array}{rcl}{\mathbf{m}}_{\mathbf{e}}{\mathbf{a}}_{\mathbf{e}}& \mathbf{=}& {\mathbf{m}}_{\mathbf{p}}{\mathbf{a}}_{\mathbf{p}}\\ \frac{{\mathbf{a}}_{\mathbf{e}}}{{\mathbf{a}}_{\mathbf{p}}}& \mathbf{=}& \frac{{\mathbf{m}}_{\mathbf{p}}}{{\mathbf{m}}_{\mathbf{e}}}\end{array}$

The separation of the plates of the capacitors,

- Δx = ?
- v
_{f}= 45000 m/s - v
_{0}= 0 m/s - a = a
_{p}

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 45000 m/s.

What will be the final speed of an electron released from rest at the negative plate?

Express your answer to two significant figures and include the appropriate units

V=_________

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