# Problem: A proton with an initial speed of 800000 m/s is brought to rest by an electric field.Part A. Did the proton move into a region of higher potential or lower potential?Part B. What was the potential difference that stopped the proton?ΔV = ________ VPart C - What was the initial kinetic energy of the proton, in electron volts?Ki =_________ eV

###### FREE Expert Solution

We are required to identify the effects of change in potential on the motion of a positively charged particle and perform calculations involving energy.

Law of conservation of energy:

$\overline{){{\mathbf{K}}}_{{\mathbf{0}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{0}}}{\mathbf{+}}{{\mathbf{W}}}_{{\mathbf{nc}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}}$, where Wnc is the work done by non-conservative forces such as friction.

Kinetic energy:

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{mv}}}^{{\mathbf{2}}}}$

86% (500 ratings) ###### Problem Details

A proton with an initial speed of 800000 m/s is brought to rest by an electric field.

Part A. Did the proton move into a region of higher potential or lower potential?

Part B. What was the potential difference that stopped the proton?
ΔV = ________ V

Part C - What was the initial kinetic energy of the proton, in electron volts?
Ki =_________ eV