The electrons in a particular beam each have a kinetic energy of 1.6*10 raised to -17 times j. What will be the magnitude and direction of the electric field that stops these electrons at a distance of 10cm?
Asked by Lahari nibhanupudi | 4th Feb, 2012, 03:44: PM
The K.E. of the electrons will convert in to P.E. .
So the change in K.E.= change in P.E.
Kf-Ki= q( Vf-Vi)
Kf=0, Ki=1/2 me v2
d=10 cm, q= charge on electron.
put the values and get the answer.
Answered by | 6th Feb, 2012, 04:22: PM
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