An electron is traveling at a speed of v_e = 2.20 times 10^6 m/s when it passes

Question

An electron is traveling at a speed of v_e = 2.20 times 10^6 m/s when it passes point A and encounters a region with an electric field that slows the electron down. If the electron comes to rest at point B, what is the electric potential difference between points A and B? (DeltaV = V_B - V_A = ?) b) A proton traveling with velocity v_p is brought to rest by the same magnitude of potential difference. What is the initial speed of the proton? (v_p = ?) For quick reference: m_e = 9.11times10^-31 kg, and m_p = 1.67 times 10^-27 kg.

Explanation / Answer

a) W = q*deltaV = 0.5*m*v^2

W = 1.6*10^-19*deltaV = 0.5*9.11*10^-31*(2.2*10^6)^2

delta V = 13.77 V = VB-VA

b) using

q*deltaV = 0.5*m*u^2

(1.6*10^-19)*13.77 = 0.5*1.67*10^-27*u^2

initial speed is u = 5.13*10^4 m/sec

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