Please show all work As in problem 1, two large oppositely charged parallel plat

Question

Please show all work

As in problem 1, two large oppositely charged parallel plates separated by a distance of 2.0 mm create a uniform electric field between them of magnitude 50.000 N/C. The positive plate is on the left and the negative plate is on the right as shown above. A proton is released from rest at the midpoint between the two plates. As the proton moves only under the influence of the electric field, does its electric potential energy increase, decrease, or stay the same? Does the proton move toward a region of higher or lower electric potential? Find the magnitude of the change in potential between the point where the proton starts (midway between the plates) and the point where it hits one of the plates. Use your answer to part (c) to find the kinetic energy of the proton just before it hits the plate. Use your answer to part (d) to find the speed of the proton just before it hits the plate.

Explanation / Answer

The electric field E in a parallel plate capacitor is constant (independent of position of the point between plates). The potential is V = E*x (x is the distance from the lower potential plate (negative) to the point).

The proton is positive, it means moves toward negative potential (towards lower potential). The potential energy of the proton is

Ep = +e*V. It means it decreases over time

c)

Variation of potential is

Delta(V) = E*x/2 = 50000*10^-3 =50 V

d)

Ec = e*delta(V) = 1.6*10^-19*50 =8*10^-18 J (=50 eV)

e)

mv^2/2=Ec

v = sqrt(2*Ec/m) = sqrt(2*8*10^-18/1.67*10^-27) =97881.7 m/s

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