Two square parallel plates, each of side 12.0 cm, are separated by a distance of

Question

Two square parallel plates, each of side 12.0 cm, are separated by a distance of 2.00 cm. The plates have equal but opposite charges placed on them. An electron is released near one of the plates with an initial speed of 0.950 x 10^6 m/s directed toward the other plate. The electron slows down due to the electrostatic force on it and comes to a stop just before it reaches the other plate. Ignore gravity in this problem. Which plate is positively charged and which is negatively charged? Justify your answer. What is the electrical potential energy of the electron when it reaches the top plate? (Take the potential energy to be zero at the bottom plate.) What is the potential difference between the top and bottom plates? Which plate is at a higher potential? What is the magnitude and direction of the electric field between the plates? What is the magnitude of the charge on each plate?

Explanation / Answer

PART A

lower plate is positively charged because if we charged it negaitively than there is no force to stop electron on upper plate. and upper plate is -ve charged

PART B

U=KQq/r

here r become zero so

U become infinite

PART C and D

by vf2=vi2 + 2aS

(0.95*106)2=02 +2a*0.02

a=-2.25625*1013m/s2

F=ma=9.1*10-31*2.25625*1013=2.0532*10-17N

F=eE

E=F/e=2.0532*10-17N/1.6*10-19=128.325N/C

V=E/d=6416.25 volts

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