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Below is a question from my physics textbook:

$X$ and $Y$ are two points in an electric field. The potentials at $X$ and $Y$ are $V_x$ and $V_y$ respectively where $V_x > V_y$. A small, positive test charge $q$ is placed at $X$.

Which of the following is the work done per unit charge by the electric field on the charge as the charge moves from point $X$ to point $Y$?

The answer reads $(V_x-V_y)$ and not $-(V_x-V_y)$. If the field did work to decrease the overall potential (as $V_x > V_y$), shouldn't the answer be negative?

The explanation reads:

If work done by the field is positive it means the charge will accelerate. In this case a positive charge moves from a position of high potential to a lower one (just like dropping a book in a gravitational field). So we need a positive answer.

This is insanely confusing. Which theory should be used here? In terms of the charge, or the potential of the charge.

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Since the electric field is a conservative field, we know that the work done by the field is equal to the negative change in potential energy $$W_\text{field}=-\Delta U$$

or, per unit charge

$$\frac{W_\text{field}}{q}=-\Delta V$$

Since the charge starts at $X$ and ends at $Y$, $\Delta V=V_y-V_x$. Therefore $$\frac{W_\text{field}}{q}=-(V_y-V_x)=V_x-V_y$$

The answer does not depend on how $V_x$ compares to $V_y$. We are told the charge moves from $X$ to $Y$, and that is all we need to know to determine the work done by the field. The given explanation is a nice consistency check, assuming the charge moves only due to the field alone, but it is not needed to actually answer the question. It is a very poor explanation, in my opinion, as it only argues why the work should be positive, but it does not necessarily explain why it should be the exact value of $V_y-V_x$.

I think you are getting confused with other explanations of work done by other forces; these are explanations I really do not like when I see them in introductory materials, as new students typically don't pick up the finer details and just get confused.

When a conservative field does work, we can either just talk about the work it does directly, or we can express that work in terms of a change in potential energy (as done above). When a conservative field does positive work the change in potential energy is negative and vice versa.

However, typically you will see explanations of "moving charges" and talk about the "work done on the charge by an external force". We additionally assume the charge is moved very slowly so that the external force is equal and opposite to the electric force at all instants in time. Then the work done by the external force is equal to the change in potential energy of the charge. And at this point you would be correct that $$\frac{W_\text{ext}}{q}=\Delta V=V_y-V_x$$

In this case the problem just asks about the electric field and makes no reference to an external force though. All you need is the start and end potentials to determine the work done by the field.

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  • $\begingroup$ Thank you for your coherent and focused explanation, now I understand. I have been dealing with "external force" previously in questions most of the time, so I was probably unfamiliar with the work done by the "field" itself $\endgroup$ – HirG Nov 18 '19 at 14:50
  • $\begingroup$ @HirG Yes, that usually does it :) Please make sure to up vote all useful answers and to accept an answer if it sufficiently answers your question. $\endgroup$ – Aaron Stevens Nov 18 '19 at 14:51
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The electric potential is a measure of how much work can be done by the field acting on a unit positive charge as it moves from the point in question to a chosen reference point. If the charge moves from one point to another at a lower potential, then part of that work has been done.

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If the field did work to decrease the overall potential (as Vx>Vy), shouldn't the answer be negative?

The electric field and therefore the electric potential acting on the test charge are not affected by the change of position of the charge $q$. The electric potential is not reduced by the movement of the charge! The field does not "work" on the potential!

The only thing that is reduced is the energy stored in the position of the charge. We shall call this energy electric energy of $q$. If the charge is at a position of high potential the electric energy of the charge is high, if it is at low potential the electric energy is low.

Think of it like a tennis ball on a high altitude and on a low altitude. Obviusly the one on top of the mountain has stored more energy.

According to Maxwell-Equations the charge will be moved to a lower potential, therefore losing energy.

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