Particle's acceleration is given by:

$\overline{){\mathbf{a}}{\mathbf{=}}\frac{\mathbf{q}\mathbf{E}}{\mathbf{m}}}$

Let's take the distance from the negative plate that the electron travels to be **d**.

At the same time, the proton will have to travel a distance, 0.800cm - d.

In this problem, we can find d by using the kinematic equation:

$\overline{){\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{+}}{\frac{1}{2}}{\mathit{a}}{{\mathit{t}}}^{{\mathbf{2}}}}$

Here, v_{0} = 0 since the motion starts from rest.

Two parallel plates 0.800 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate.

Part A

How far from the negative plate is the point at which the electron and proton pass each other?

Express your answer with the appropriate units.

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What scientific concept do you need to know in order to solve this problem?

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