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

Question

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.

Answer 1

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

Answer 2

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.