Electric force:

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

Newton's second law:

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

Uniform accelerated motion (UAM) equations:

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

Charge of a proton = +e = +1.6 × 10^{-19} C

ΣF = eE = ma

The electric field strength is 5.00×10^{4}N/C inside a parallel-plate capacitor with a 1.90mm spacing. A proton is released from rest at the positive plate. What is the proton's speed when it reaches the negative plate?

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