Potential energy of point-charges: An electron is released from rest at a distance of 9.00 cm from a proton. If the proton is held in place, how fast will the electron be moving when it is 3.00 cm from the proton? (m_el= 9.11 × 10^-31 kg, e = 1.60 × 10^-19 C, k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

We can solve this problem using conservation of energy. Initially, the electron has only potential energy and no kinetic energy. When it moves towards the proton, the potential energy decreases and kinetic energy increases. The change in potential energy is given by:
ΔPE = kqQ/rf - kqQ/ri
where ri = 9.00 cm = 0.0900 m and rf = 3.00 cm = 0.0300 m are the initial and final distances between the proton and the electron, respectively. q = -e is the charge of the electron and Q = e is the charge of the proton. The Coulomb constant k = 1/4πε0 is given in the question.

Substituting the values, we get:
ΔPE = (8.99 × 109 N∙m2/C2) (-1.60 × 10-19 C)(1.60 × 10-19 C) / 0.0300 m - (8.99 × 109 N∙m2/C2) (-1.60 × 10-19 C)(1.60 × 10-19 C) / 0.0900 m
ΔPE ≈ -4.03 × 10-18 J

According to conservation of energy, the change in potential energy must be equal to the change in kinetic energy:
ΔKE = -ΔPE

The final kinetic energy of the electron can be calculated as follows:
KEf = 1/2 me vf2
where me is the mass of the electron and vf is the final velocity of the electron.

We can rearrange this equation to solve for vf:
vf = sqrt(2 KEf / me)

Substituting the values for KEf and me, we get:
vf = sqrt(2 (-ΔPE) / me)

Substituting the values for ΔPE, me, and solving for vf, we get:
vf ≈ 106 m/s

Therefore, the electron will be moving at a speed of approximately 106 m/s when it is 3.00 cm from the proton. The answer is B.

Potential Energy of Point-Charges: an Electron Is Released from Rest