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

Question

Quizplus · Accepted Answer

The work done on a charged particle by an electric field is given by the equation W = qEd, where q is the charge of the particle, E is the magnitude of the electric field, and d is the distance the particle moves in the direction of the field. In this case, the proton has a charge of e = 1.60 × 10^-19 C, the electric field has a magnitude of E = 3.0 V/m, and the proton moves a distance of d = 0.10 m. 
So, the work done on the proton is: W = (1.60 × 10^-19 C)(3.0 V/m)(0.10 m) = 4.8 × 10^-20 J. 
Since the work done on the proton results in a change in kinetic energy, the change in kinetic energy is equal to the work done: 
ΔK = W = 4.8 × 10^-20 J. 
Therefore, the best choice is A.

A Proton Moves 0 E⃗\vec { E }E Field of Magnitude 3

A Proton Moves 0 $\vec { E }$ Field of Magnitude 3