Calculating Work Done per Unit Charge in an Electric Field

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Homework Help Overview

The problem involves calculating the work done per unit charge when a charged particle is moved within an electric field created by a point charge. The context includes concepts of electric potential, electric field strength, and the relationship between work and charge.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the formula for work done in an electric field and question the dimensionality of work per unit charge. There are attempts to clarify the ambiguity in the problem statement and the relevance of the charge mentioned in the question.

Discussion Status

The discussion is ongoing with various interpretations of the problem being explored. Some participants have offered insights into dimensional analysis and questioned the clarity of the question. There is no explicit consensus on the correct interpretation yet.

Contextual Notes

Participants note potential ambiguities in the wording of the question and the provided answer choices. There are discussions about the implications of the distance being zero and how that affects the interpretation of the work done.

brake4country
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Homework Statement


A charged particle is moved from a great distance to a distance d from a point charge. At distance d, the electric field has a strength E and a potential V. Which of the following represents the work done per unit charge q?

Homework Equations


F = Eq; V = Ed, E = kq/r^2; U = Eqd, U = kq1q1//r

The Attempt at a Solution


In a previous thread, it was learned that the potential energy of a charged particle is denoted as U = Vq, where q is the charge being affected by the electric field. Now, a slightly different scenario is presented. The same particle is moved from an infinite distance to a distance d to a point source charge. The amount of work done here is the W = Eqd, a derivative of F=ma.

The main difference between the problem before and the problem here is the "distance". If we know the distance the charge was brought from far away to nearer to the point charge, then we can calculate the work done on the particle. This is the formula U =Eqd.

Is my thought processes correct here? Thanks in advance!
 
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brake4country said:
W = Eqd
That's only true for a constant field E acting in the same direction as the distance moved, d.
In general, ##W=\int\vec E.\vec {ds}##, where ##\vec{ds}## is the distance element vector.
It isn't entirely clear what the question is asking for . It says 'per unit charge', but there are two charges here.
What dimensionality would 'work per unit charge' have? Which, if any, of the offered answers has that dimensionality?
 
I see how the question is ambiguous. To answer your question, the dimensionality for work per unit charge is W/C = Fd/C = Eqd/C. The answer choices are (A) V, (B) Eqd, (C) Vq, and (D) Ed/q. A is wrong because the dimensionality for V is volt or J/C. C is wrong because Vq is the potential energy charge, which we are not asked for, and (D) is Force per charge, which is wrong because Force is not work although it is related via distance d. The best answer is B.
 
brake4country said:
I see how the question is ambiguous. To answer your question, the dimensionality for work per unit charge is W/C = Fd/C = Eqd/C. The answer choices are (A) V, (B) Eqd, (C) Vq, and (D) Ed/q. A is wrong because the dimensionality for V is volt or J/C. C is wrong because Vq is the potential energy charge, which we are not asked for, and (D) is Force per charge, which is wrong because Force is not work although it is related via distance d. The best answer is B.
I agree that's the only one which makes sense dimensionally, but it is clearly wrong. What would it give for the work done when d is zero?
So I tend to think the question has been badly worded - they want the work, not the work per unit charge. That suspicion is reinforced by the mention of q at the end of the sentence. If they wanted the work per unit charge, there would no relevance in mentioning q.
This leads me to the last option, but there seems to be at least one typo in it. Please check that you have posted it correctly.
 
First you want to fix the typo in equation 4. What is q1 anyway ?

You haven't mentioned the choices in your original post, even though they are essential. That way the question is not ambiguous at all.

Your dimensional reasoning is wrong. Work has the dimension kg m2/s2 , in short: Joule
(C) has the dimension of Work (=the dimension of potential energy change)
The dimension of (D) is not Force per unit charge. Think again.
 
Last edited:
I see that even Haru was lured into believing your equations were the choices. ?:)
 
BvU said:
I see that even Haru was lured into believing your equations were the choices. ?:)
Ah! Yes, I didn't notice the four later formulae were different from those in the OP. Thanks BvU.
Your list of dimensional representations finishes with Eqd/C, but q and d there are actual variables, not dimensionalities. Fix that and you should get some cancellation. What remains can be represented in a single concept.
 
haruspex said:
I agree that's the only one which makes sense dimensionally, but it is clearly wrong. What would it give for the work done when d is zero?
So I tend to think the question has been badly worded - they want the work, not the work per unit charge. That suspicion is reinforced by the mention of q at the end of the sentence. If they wanted the work per unit charge, there would no relevance in mentioning q.
This leads me to the last option, but there seems to be at least one typo in it. Please check that you have posted it correctly.
Yes, I agree that it was poorly worded but Exam Krackers is known for that. It should say "per unit charge." Thank you so much!
 
In post #5 I argue that the question with the choices as shown in your post #3 is not ambiguous at all. It is also properly worded.
Is it now clear to you which answer is correct, and why ?
 
  • #10
The answer choices are in post #3. Yes, I understand now. Thank you everyone.
 

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