An electron is released from rest at a distance of 7 cm from a proton, assume to

Question

An electron is released from rest at a distance of 7 cm from a proton, assume to be stationary. We want to determine how fast the electron will be moving when it is 5 cm from the proton. It is easiest to use poential and potential energy to solve this problem. consider well the correct sign of all quantities asjed for.

1) what is the electric potential 7 cm from the proton?

2) what is the electric potential 5 cm from the proton?

3) what is the electric potential of the electron when it is 5cm from the proton?

4) write an equaltion for the conservation of mechanical energy from this problem, involving the potential and kenetic energy at 7 cm and 5 cm from the proton.

5) what is the speed of the electron when it is 5 cm from the proton in m/s?

Explanation / Answer

at 7 cm: V = ke/r = 9.0E9*1.6E-19/0.07= 2.06E-8 V

PE = - k e^2/r = -9.0E9*1.6E-19^2/0.07= -3.29E-27 J

at 5cm
V = 9.0E9*1.6E-19/0.05=2.88E-8 V
PE = -9.0E9*1.6E-19^2/0.05 = -4.61E-27 J

conservation of energy

PE at 7cm = PE at 5cm + KE at 5cm

-3.29E-27 = -4.61E-27 + 0.5*9.11E-31*v^2

v= 53.8 m/s

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