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## Homework Statement

Two parallel plates having charges of equal magnitude but opposite sign are separated by 14.0 cm. Each plate has a surface charge density of 37.0 nC/m2. A proton is released from rest at the positive plate.

(a) Determine the magnitude of the electric field between the plates from the charge density.

(b) Determine the potential difference between the plates.

(c) Determine the kinetic energy of the proton when it reaches the negative plate.

(d) Determine the speed of the proton just before it strikes the negative plate.

(e) Determine the acceleration of the proton.

(f) Determine the force on the proton.

(g) From the force, find the magnitude of the electric field.

## Homework Equations

## The Attempt at a Solution

I have a few questions before I give you my solution to part a):

Are we allowed to assume that the two plates are in electrostatic equilibrium? If so, why?

If I attached a line to both planes, the line being perpendicular to both planes, the electric field will be constant across this entire line. Why is this so?

Here is my attempt at part a), so far:

To find the electric field at a point between the two planes, I'll apply Gauss's law. I'll use a cylinder for my Gaussian surface, where the axis of the cylinder will be parallel to the electric field. The surface area of a cylinder is, [itex]2 \pi r^2 + 2 \pi r h[/itex]; since none of the electric fields generated by the plane flows through the curved surface, there won't be any electric flux, and, consequently, we can remove the second term from our surface area equation. Additionally, since we only care about the electric field through one circular face of the cylinder, the one in between both planes, we can remove the 2 from first factor.

So, [itex]E(\pi r^2)=\frac{q_{enc}}{\epsilon_0}[/itex]. We don't know the amount of charge enclosed by the Gaussian surface; however, we do know the area that's enclosed: it's equal to the area of the circular face of the cylinder. We also know the amount of charge contained per unit area of the plane, [itex]\sigma[/itex]. Hence, [itex]q_{enc}=\sigma A_enc[/itex]

After substitution and simplification: [itex]E= \frac{\sigma}{\epsilon_0}[/itex]

The exact same process can be done for the other plane. So, for the negative plane, [itex]E=\frac{- \sigma}{\epsilon_0}[/itex]. But when I add them, the result is zero. Why is that?

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