Electrostatics, parallel plates

[physics604]

1. A proton with kinetic energy of 2.1 x 10-17 J is moving into a region of charged parallel plates. The proton will be stopped momentarily in what region (attached diagram)?

a) Region K
b) Region L
c) Region N
d) Region M

Homework Equations

\DeltaE_k + \DeltaE_p = 0

The Attempt at a Solution

\DeltaE_k + \DeltaE_p = 0
1/2mv²= -2×Q\Deltav

v=√\frac{-2Q\Delta v}{m}
\Deltav = 2.1×10^-17 / 1.6×10^-19 = 131.25

I have an answer, but all the \Deltav in the regions are 100, so I don't get how I am supposed to know what region it's supposed to stop in.

 

[gneill]

Each region presents a separate ΔV to the moving proton. When the proton moves from one region to the next, it only sees the ΔV of that region, but it's experienced the energy change due to the previous one. So it loses some energy climbing the potential of the previous region before entering the next. At some point the proton won't have enough energy to completely traverse a region. In fact, it will stop and reverse direction.

You've calculated a total ΔV that will reduce the proton's KE to zero. Within which region will the total ΔV traversed reach this value?

 

[physics604]

Is it the second region (L) because 131 is between 100 and 200?

 

[gneill]

Is it the second region (L) because 131 is between 100 and 200?

That's a good reason