# Confusion in the sign of work done by electric field on a charged particle

In my book* it is given, "the work done to transport a charge $$q$$ through a potential difference $$\Delta V$$ is $$q \Delta V$$." Or mathematically, it can be written as follows:$$W_{\text{electric}}=q\Delta V.\tag{1}$$

We know that potential difference is defined as $$\Delta V=\frac{\Delta U}{q}$$ where $$\Delta U$$ is the potential energy difference as we move the charged particle from reference point (assumed to be at infinity) to the given location. From definition, we know that $$\Delta U=-W_{\text{electric}}$$ where $$W_{\text{electric}}$$ is the work done by the electric force. Now using this in the definition of potential difference, we get $$\Delta V=-\frac{W_{\text{electric}}}{q}$$ or: $$W_{\text{electric}}=-q\Delta V \tag{2}$$

Equations $$(1)$$ and $$(2)$$ for determining the work done by electric field give the same value but the result is of opposite signs. Equation $$(2)$$ must be incorrect, but the way I arrived at it seems to be totally correct. Then why do we obtain two different expressions for the same quantity?

I read this question/answer - How does one prove that Energy = Voltage x Charge?, but still my doubt about the sign inconsistency in the expression for work done by the electric field on a charged particle exists.

*Book: 'Concepts of Physics' by Dr. H.C.Verma; Page: 117.

• See the accepted answer of this previous question: physics.stackexchange.com/q/218829 Dec 14, 2019 at 6:44
• Are you sure it is the same $\Delta V$ in both cases? Are you sure the $\Delta$ is the same in both cases? Are you sure it is the same $V$ in both cases? Are you sure the curve that is integrated over is integrated over in the same direction in both cases?
– Emil
Dec 14, 2019 at 8:42
• Do you know if the work is done by the system or on the system in the two cases?
– Emil
Dec 14, 2019 at 8:49
• @Emil, "Are you sure it is the same $\Delta V$ in both cases":Yes. "Are you sure the $\Delta$ is the same in both cases":Yes. "Are you sure it is the same $V$ in both cases": Both $V$s represent electric potential. "Are you sure the curve that is integrated over is integrated over in the same direction in both cases?":I think yes. But if it's no then that explain the inconsistency. "Do you know if the work is done by the system or on the system in the two cases?": I'm sure it's the work done on the charges by the electric field. Dec 14, 2019 at 9:01
• @M.GuruVishnu I found an answer to a similar question, physics.stackexchange.com/a/216013/28273, if I understand it equations (1) and (2) have different forces in mind, taken from the same force pair.
– Emil
Dec 14, 2019 at 9:35

In equation 1 if $$q$$ is positive (a positive charge) and $$\Delta V$$ is positive (an increase in electrical potential) then that work is done by an external agent against the electric field and not by the electrical field. The work is positive because the direction of the force of the external agent is the same as the displacement of the charge.

At the same time the external agent is doing positive work the force of the electric field, which is opposite the displacement of the charge, is doing negative work taking the energy given to the charge by the external force and storing it as electrical potential energy of the electric field/charge system. That’s the electrical work of equation 2 and the reason it’s negative, assuming again the charge and change in potential are both positive.

The gravitational analogy is you, an external agent, do positive work of $$mgh$$ raising a mass $$m$$ and bringing it to rest a height $$h$$ while the force of gravity does an equal amount of negative work $$-mgh$$ taking the energy you gave the mass and storing it as gravitational potential energy of the mass/earth system

Hope this helps

I am pretty sure that $$W_{electric} = -q\Delta V$$. Electric field does positive work when it decreases the potential of a charge. Here is how I arrived at this:-

$$F_{electric} = qE$$ $$W_{electric} =\int F_{electric}dr = q\int Edr$$ $$dV = -Edr$$ Therefore $$W_{electric} = -q\Delta V$$

The definition for electric potential is work done by external force in moving a charge, per unit charge. In HCV the example no. 14 that is confusing you is talking about the work done by the battery force which is an external force.

• Thank you for your answer. Why is the work done by battery considered as a result of external force? I think it's an internal force because I consider forces other than electric to be external and here since the force exerted by the potential difference of the battery is electric in nature, and so it's internal. I know this is contradictory and depends on the choice of system. Are there any other consideration for this? Dec 14, 2019 at 12:32
• @M.GuruVishnu That is because the "battery force" is generally chemical in nature, in electrochemical batteries/cells it is a result of redox reactions in the battery. Page 176 of HCV has details on batteries and EMF and how exactly this works, there is also an entire chapter on electrochemistry. Dec 14, 2019 at 14:06