Problem: 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 unitsV=_________

Electric force,

$\boxed{\mathbf{F}\mathbf{=}\mathbf{q}\mathbf{E}}$

Newton's second law,

$\boxed{\mathbf{F}\mathbf{=}\mathit{m}\mathit{a}}$

UAM equations,

$$\begin{gathered} \boxed{\mathbf{ }{\mathit{v}}_{\mathit{f}}\mathbf{ }\mathbf{=}\mathbf{ }{\mathit{v}}_{\mathbf{0}}\mathbf{ }\mathbf{+}\mathit{a}\mathit{t} \\ \mathbf{∆}\mathit{x}\mathbf{=}\mathbf{ }\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}\mathit{t} \\ \mathbf{∆}\mathit{x}\mathbf{=}\mathbf{ }{\mathit{v}}_{\mathbf{0}}\mathit{t}\mathbf{+}\frac{1}{2}\mathit{a}{\mathit{t}}^{\mathbf{2}} \\ \mathbf{ }{{\mathit{v}}_{\mathit{f}}}^{\mathbf{2}}\mathbf{=}\mathbf{ }{{\mathit{v}}_{\mathbf{0}}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{2}\mathit{a}\mathbf{∆}\mathit{x}} \end{gathered}$$

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
• v₀ = 0 m/s
• a = a_p