Electric Potential: Solving Proton Displacement & Energy Conservation

[peaceandlove]

Homework Statement

A proton is released from rest in a uniform electric field of magnitude 1.8*10^5 V/m directed along the positive x axis. The proton undergoes a displacement of 0.6 m from the positively charged end of the field to the negatively charged end of the field. (a) Find the change in the electric potential if the proton moves from point A to point B. Answer in units of V. (b) Find the change in potential energy of the proton for this displacement. Answer in units of J. (c) Apply the principle of energy conservation to find the speed of the proton after it has moved 0.6 m, starting from rest. Answer in units of m/s.

Homework Equations

The Attempt at a Solution

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Nevermind, I figured out what I was doing wrong.

 

[AvroArrow]

For part (a) I got the correct answer, 9.0*10^3 V. For part (b) I got 3.24*10^-14 J and for part (c) I got 1.47*10^4 m/s.

 

[nlightner]

Great job figuring out the solution! Just to provide a response as a scientist, here is a breakdown of the solution for others who may be reading this:

(a) The change in electric potential (ΔV) is equal to the difference in electric potential between points A and B. This can be calculated using the equation ΔV = Ed where E is the electric field and d is the displacement. Plugging in the values given, we get ΔV = (1.8*10^5 V/m)(0.6 m) = 1.08*10^5 V.

(b) The change in potential energy (ΔPE) can be calculated using the equation ΔPE = qΔV, where q is the charge of the proton (1.6*10^-19 C) and ΔV is the change in electric potential calculated in part (a). Plugging in the values, we get ΔPE = (1.6*10^-19 C)(1.08*10^5 V) = 1.728*10^-14 J.

(c) According to the principle of energy conservation, the total energy of the system (kinetic energy + potential energy) remains constant. Since the proton starts from rest, its initial kinetic energy (KEi) is 0. At point B, the proton's final kinetic energy (KEf) will be equal to its total energy (TE) minus its potential energy (PEf). We can set up the equation TE = KEf + PEf and solve for KEf. Plugging in the values, we get KEf = TE - PEf = (1.728*10^-14 J) - (1.728*10^-14 J) = 0 J. This means that the proton's final kinetic energy is also 0, so its speed will also be 0 m/s.

Hope this helps! Keep up the good work in your scientific studies.