Example, cont.

qp(Vi-Vf) = (1/2)mpvf2

We note that (Vi-Vf) = -DV since the change is normally final minus initial. Thus,

DV = -(1/2)mpvf2 / qp =

-(1/2)(1.67x10-27)(4.46x105)2 / 1.6x10-19 =

-1040 volts.

We see that the proton must fall down (DV is negative) through 1040 volts to reach a million miles/hour.

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