# 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=_________

###### FREE Expert Solution

Electric force,

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

Newton's second law,

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

UAM equations,

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 = ?
• vf = 45000 m/s
• v0 = 0 m/s
• a = ap ###### Problem Details

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=_________