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Please eplain all the questions, this is last question that i can post 1.Two ide

ID: 1399302 • Letter: P

Question

Please eplain all the questions, this is last question that i can post

1.Two identical metal plates with small holes are aligned a distance L=5.8 cm apart. They have a potential difference of 115 Volts. An electron can enter through the hole in the negative plate and exit through the hole in the positive plate. If it enters with a velocity of 4620 km/s at the negative plate, with what velocity does it exit at the positive plate

2.A +67 C charge is placed 39 cm from an identical +67 C charge. If you assume the potential at infinity is zero, then what is the potential midway between the two charges? How much work is done to move a 0.60 C charge from a point midway between the charges to a point 10 cm closer to either of the charges?

3.An evacuated tube uses an accelerating voltage of 44.0 kV to accelerate electrons to hit a copper plate and produce x rays. Non–relativistically, what would be the maximum speed of these electrons?

Explanation / Answer

1)Work done in moving the electron from negative to positive charged plate is W = q*V = 1.6*10^-19*115 = 1.84e-17 J

apply work energy theorem

W = change in kinetic energy

1.84e-17 = 0.5*m*(v^2-(4620*4620*10^6))

m is the mass of the electron = 9.11*10^-31 kg

(v^2-(4620*4620*10^6)) = 1.84e-17/(0.5*9.11e-31) = 4e+13


v = sqrt((4e+13)+(4620*4620*10^6))

v = 7832.2 km/s

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2) potential at the mid point is V = 2*k*q/r

V = (9*10^9*67*10^-6)/(0.39/2) = 3.09*10^6 V

potential at a point which is closer to any one of the charge is V2 = k*q((1/0.1)+(1/0.29)) = (9*10^9*67*10^-6)*((1/0.1)+(1/0.29)) = 8.1*10^6 V

potential difference is dV = (8.1-3.09)*10^6 = 5.01*10^6 V

Work done W = q*dV = 0.6*10^-6*5.01*10^6 = 3.006 J

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3) apply 0.5*m*v^2 = q*V = (1.6*10^-19*44000) = 7.04e-15

v = sqrt(7.04e-15/(0.5*9.11*10^-31)) = 1.24*10^8 m/s