Question

A proton with an initial speed of 600,000 m/s is brought to rest by an electric field. Part A Part complete Did the proton move into a region of higher potential or lower potential? Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential. Because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of higher potential to a region of lower potential. Because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of lower potential to a region of higher potential. Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential. Previous Answers Correct Here we learn how to determine the distribution of the electric potential based on the movement of a charged particle. Part B Part complete What was the potential difference that stopped the proton? Express your answer with the appropriate units. ΔV = 1900 V Previous Answers Correct Here we learn how to use the law of energy conservation to find the potential difference needed to stop the movement of a charged particle. Part C What was the initial kinetic energy of the proton, in electron volts? Express your answer in electron volts.

Answer 1

Answer: Part A the right sentence is: Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential

Part B.  aproximatelly 98 times ΔV  V

Part C. the initial kinetic energy of the proton is 1.87 10^3 eV

Explanation: Part A. The field stops the proton so the lines of electric fild must be directed in opposite direction of its movement. This means that the proton moves to a higher potential. Part B The kinetic energy of the  is transformed  in electric potenctial for the proton.

Part C. Energy in J divide the charge of electron gives the energy  in eV.