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?

The K.E. of the electrons will convert in to P.E. .

 

So the change in K.E.= change in P.E.

 

(Kf-Ki)= Uf-Ui

 

Kf-Ki= q( Vf-Vi)

 

Kf-Ki=q(E* d)

 

Kf=0, Ki=1/2 me v²

 

^{d=10 cm, q= charge on electron.}

 

^{put the values and get the answer.}