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How come the definition of voltage can be defined as the work required to push a positive charge from a location of lower potential to a location of higher potential (closer to source of electric field) such that the force that does the push is equal to but in opposite direction to the force due to the electric field?

If the forces are equal and in opposite direction, how does the push move the positive charge from lower potential to higher potential? Won't equal forces just cause the charge to remain in the same place, and thus no work is done by the push?

I can understand voltage from the viewpoint of potential difference between two locations, but when it is defined this way, it makes no sense to me.

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Won't equal forces just cause the charge to remain in the same place, and thus no work is done by the push?
You have forgotten that the charge could move on average at constant velocity.
I have added a word in your sentence How come the definition of voltage can be defined as the external work required to push a positive charge from a location of lower potential to a location of higher potential.

The equality of the work done by the electric field and the work done by the electric field means that the net work done on the charge (the system)is zero and the kinetic energy of the charge does not change.

I terms of the system of electric charge and the electric field, the external work done on the system increases the electric potential energy of the system, ie the electric charge is moved to a position where the system has more capability (potential) to do work.

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    $\begingroup$ Oh I see. Since if there was no opposite force, the charge's velocity will keep increasing due to the constant force by the emf. Since currents are moving at constant speed when voltage is constant it must have constant velocity. So when the charge moves from higher to lower potential, the external force will do negative work on the charge. The opposite of the negative work will be the work required move the charge back to the location with higher potential. Does this make any sense? Thank you. $\endgroup$ Oct 15, 2022 at 8:33
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This seems to be a fundamental Newton's-2nd-law misunderstanding.

Equal forces do not mean zero speed. They only mean zero acceleration,

$$\sum F=ma.$$

The point is not whether the speed is zero or not, but whether it changes or not. The definition of voltage as work done in moving a charge over a distance under force balance is thus perfectly valid. In fact, this is what you achieve every time you have steady current in a circuit - the charges are moving at constant (drift) speed, so zero acceleration. Any restrictions they encounter during that movement, such as frictions and resistances, is balanced out perfectly by the voltage so that the charge drift-speed is constant.

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I've never understood why textbook writers insist on bringing in an external force. Why not simply define as follows? $$V_A-V_B=\frac 1q(\text{Work done by electric field on q as q goes from A to B})=\int^{\vec{r_B}}_{\vec{r_A}\ \text{any route}}\vec E.d\vec r$$

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