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

ID: 1589118 • Letter: T

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

a) The bottom plate is positively chaarged and top plate is negatively charged, becuase electron faces replusion from top most plate and stops just before it reaches the top most plate.

b) At the top of the plate, the entire KE of electron changes to its electrical PE

KE of electron at the bottom of plate = 0.5mv^2 = 0.5 ( 9.1 x 10^-31)( 0.950 x 10^6)^2 = 4. 106 x 10^ -19 J

Therfore electrical PE of electron just before it reaches the top plate = 4. 106 x 10^ -19 J

c) I am using the electric field calculated in part (d) to calculate the potetial difference

V = Ed

V =  1.28 x 10^2 ( 2 x 10^-2) = 2.566 V apprx

The bottom plate is at higher potential

d) Electric field between plates

qEd = 1/2 mV^2

qE( 2 x 10^-2) = 4. 106 x 10^ -19

E = 1.28 x 10^2 N/C

Direction: Bottom to top

e) V= Qd/e0A

A = area of plate = 12 x 12 x 10^-4 = 144 x 10^-4 m^2

e0= 8.85 x 10^-12

Q = 2.566 x  8.85 x 10^-12 x 144 x 10^-4/ 2 x 10^-2 = 0.1635 x 10^-10 C apprx

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