What is the meaning of external work in electric potential? We learned that  $$-Work = \Delta U = q(Va-Vb).$$ However, we write that $$W = q(Va-Vb)$$ in this question. Why don't we write the minus sign in the equation? How does external work affect the minus sign ? Can you help me to understand these questions and explain them clearly? I really appreciate any help you can provide.

 A: When the potential energy function is introduced, it is because
the work is being done by a conservative-force...
in this case, the electrostatic force.
Thus, "Work" in "-Work = delta U" refers to the work done by the electrostatic force.

*

*Why the minus sign?
From the work-energy theorem, the work-done is separated into parts that
"don't depend on the path" (that's the work done by conservative forces) and those that "do depend on the path" (that's the work done by nonconservative forces)
\begin{align}
W_{net}&=\Delta K\\
W_{cons}+W_{noncons}&=\Delta K\\
\end{align}
We introduce an "energy of position" (energy of configuration) called the potential energy.  We choose to express this as $W_{cons}=-\Delta U$ so that we can write
\begin{align}
W_{net}&=\Delta K\\
W_{noncons}&=\Delta K - (-\Delta U) = \Delta (K+U)
\end{align}
and refer to the $(K+U)$ are the "total mechanical energy".
It would be weird to use "$W_{cons}=+\Delta U$" because "$(K-U)$" doesn't look like a "total". Thus, we use  $W_{cons}=-\Delta U$.

In the context of your question, "W = q(Va-Vb)" the "work" there refers to an external agent applying forces (to balance the electrostatic force... so that the displacement is done "slowly" in a quasi-equilibrium way... practically speaking, "moved with a tiny constant velocity").
[For clarity, it might be good to specify the initial and final positions.]
A: In the first equation the work is that done by the electric field. In the second equation the work is that done by an external agent. The work done by the electric field is the negative of the work done by the external agent.
The gravity analogy is when I lift an object a height $h$ I, the external (to the object/earth system) agent do positive work (since my force is in the same direction as the displacement of the object) of $mgh$. At the same time the gravitational field does negative work of $-mgh$ (since the direction of its force is opposite the displacement of the object) taking the energy I gave the object and storing it as gravitational potential energy.
Hope this helps.
