Potentia Difference and Electric Potential

[TheFireFox]

A proton moves 2.00 cm parallel to a uniform electric field of E = 200 N/C. (a) How much work is done on the proton by the field? (b)What change occurs in the potential energy of the proton? (c) What potential difference did the proton move through?


I really just need help with Part A, I should be able to get through the rest alright. I'm not entirely sure what the charge of the proton would be. I know that wok is equal to the change in potential energy, which is then equal to (kq1q2)/r...but I'm not sure what the chrages would be. Help?

 

[Delphi51]

The charge on the proton is the same as the charge on the electron (except it is positive), often called the "elementary charge". You can easily look up its value in your text or in Wikipedia.

V = (kq1q2)/r applies only two point charges. For this problem with the uniform E field, you really need F = qE, an electric field causes a force on a charge. From there you can find the work done quite easily.

In a uniform E field, E = V/d. This is often called the parallel plate formula because it tells how a potential across plates distance d apart causes a uniform electric field between the plates. You could use it backwards to find your part c.

Alternatively, use the definition of potential difference = (change in energy)/charge.

 

[nlightner]

As a scientist, it is important to always clarify and define all variables and units in order to accurately answer the question. In this case, we are given the distance (2.00 cm) and the electric field (E = 200 N/C), but we are not given the charge of the proton. Let's assume that the charge of the proton is q = +1.6 x 10^-19 C, which is the charge of a single proton in Coulombs.

(a) To calculate the work done on the proton by the electric field, we use the formula W = Fd, where W is work, F is force, and d is distance. In this case, the force acting on the proton is the electric force, which is given by F = qE, where q is the charge of the proton and E is the electric field. Substituting the values, we get W = (1.6 x 10^-19 C)(200 N/C)(0.02 m) = 3.2 x 10^-22 J.

(b) The change in potential energy of the proton can be calculated using the formula ΔPE = qΔV, where ΔPE is the change in potential energy, q is the charge of the proton, and ΔV is the change in potential difference. Since we are not given the potential difference, we can use the formula ΔV = Ed, where E is the electric field and d is the distance moved by the proton. Substituting the values, we get ΔV = (200 N/C)(0.02 m) = 4 V. Now, we can calculate the change in potential energy: ΔPE = (1.6 x 10^-19 C)(4 V) = 6.4 x 10^-19 J.

(c) The potential difference is the difference in electric potential between two points. In this case, the proton moved through a distance of 2.00 cm, so the potential difference is the change in potential energy divided by the distance: ΔV = ΔPE/d = (6.4 x 10^-19 J)/(0.02 m) = 3.2 x 10^-17 V.

In summary, the work done on the proton by the electric field is 3.2 x 10^-22 J, the change in potential energy is 6.4 x 10^-19 J, and the potential difference is