# Potential difference in a electric field

I understand that the line integral of the electric field is path-independent. If the source is a point charge of $$q$$, line integral $$\int_{P_1}^{P_2} E\cdot ds$$; the integral from $$P_1$$, $$P_2$$ which is at $$r_1$$ and $$r_2$$ away from the source is $$\int_{P_1}^{P_2} E\cdot ds=\frac{q}{4\pi\epsilon_0}\left(\frac{1}{r_1}-\frac{1}{r_2} \right)$$ The line integral multiplied by the charge of the test charge moving through any path between the two points is $$q_0\int_{P_1}^{P_2} E\cdot ds=\frac{qq_0}{4\pi\epsilon_0}\left(\frac{1}{r_1}-\frac{1}{r_2} \right)$$ which is $$-\Delta U$$. So I get why the potential function is defined as $$\phi_{21}=-\int_{P_1}^{P_2} E\cdot ds$$ But why is this interpreted as

The work per unit charge done by an external agency in moving a positive charge from $$P_1$$ to $$P_2$$ in the field $$E$$. (The external agency must supply a force $$F_{ext}=-q_0E$$ to balance the electrical force $$F_{elec}=q_0E$$; hence the minus sign.)

What force is this external force? Is the external force excluding the force that makes the charge move along the path? Shouldn't the external force be different when the charge is moving in curves?

When thinking about moving charges from, let's say infinity to some point $$P$$, or from any point $$A$$ to any other point $$B$$, we do so in a way that doesn't change the velocity, or the "state of motion" of the charged particle. Since a force $$q_0 \vec E$$ exists already, an external agent should balance this force.