A proton moves $0.10 \mathrm {~m}$ along the direction of an electric field of magnitude $3.0 \mathrm {~V} / \mathrm { m }$. What is the change in kinetic energy of the proton? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

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Quizplus · Accepted Answer

The change in kinetic energy of the proton is equal to the work done by the electric field, which is given by $ W = qEd $. Here, $ q = 1.60 \times 10^{-19} \, \mathrm{C} $, $ E = 3.0 \, \mathrm{V/m} $, and $ d = 0.10 \, \mathrm{m} $. Thus, $ W = (1.60 \times 10^{-19} \, \mathrm{C})(3.0 \, \mathrm{V/m})(0.10 \, \mathrm{m}) = 4.8 \times 10^{-20} \, \mathrm{J} $.

A Proton Moves 0.10 m0.10 \mathrm {~m}0.10 m Along the Direction of an Electric Field of Magnitude

A Proton Moves $0.10 \mathrm {~m}$ Along the Direction of an Electric Field of Magnitude