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Introductory Physics Homework Help
Electron brought to rest by the E-field, potential difference question
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[QUOTE="cookiemnstr510510, post: 6064512, member: 650627"] [h2]Homework Statement [/h2] An Electron with an initial speed of 500,000m/s is brought to rest by an electric field a)did the electron move into a region of higher or lower potential? I b) what was the potential difference that stopped the electron? [h2]Homework Equations[/h2] ΔV=ΔU/q ΔU=-W [h2]The Attempt at a Solution[/h2] a) my logic: if the electron is brought to rest that would mean the thing creating the electric field would be negative (since like charges repel each other). So I am imagining something like Electric Field.jpg attached. Since I know that ΔV=ΔU/q, an electron would have a very high potential energy when near a negative plate (like in my pic). That means the closer it travels in the direction of the charged plate it is gaining electric potential based on the equation ΔV=ΔU/q, since q is not changing. So I believe the electron traveled from a region of lower electric potential to higher electric potential. when the problem states "did the electron move into a region of higher or lower potential" they are saying electric potential, correct? just want to clarify b) ΔV=ΔU/q ΔU=-W W=ΔK=(1/2)mv[SUB]f[/SUB][SUP]2[/SUP]-(1/2)mv[SUB]i[/SUB][SUP]2[/SUP] combining these equations: ΔV=-[1/2)mv[SUB]f[/SUB][SUP]2[/SUP]-(1/2)mv[SUB]i[/SUB][SUP]2[/SUP]]/q the electrons is brought to rest so v[SUB]f[/SUB]=0 so the equation becomes: ΔV=(1/2)mv[SUB]i[/SUB][SUP]2[/SUP]/q→(1.1387x10[SUP]-19[/SUP])/(-1.602x10[SUP]-19[/SUP])=-0.712V Why do we have to use -e for electron? I know the charge on an electron is -1.602x10^-19, however I remember in the last few chapters when talking about electric fields etc sometimes we didn't have to use the negative value of the electron [/QUOTE]
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Electron brought to rest by the E-field, potential difference question
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