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deceleration of an electron in an electric field

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Two charged, parallel, flat conducting surfaces are spaced d = 1.4 cm apart and produce a potential difference of 655 V between them. An electron is projected from one surface directly toward the second. What is the initial speed of the electron if its comes to rest just at the second surface?

This is what I've been trying to do:

qV=-W=-K=-1/2mv^2
655*-1.6*10^-19=-1/2(9.1*10^-31)v^2
I tried to get everything on one side and just square it to find the initial speed. But I got an incorrect answer.

Why does this not work?

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Solution Summary

A deceleration of an electron in an electric field is examined. The second surface electron speed is determined.

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The equations you used look ok. Let's try again:

The potential energy should decrease by e*655 V = ...

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