Work to Move a Charge (W = qΔV)

A potential difference is joules per coulomb. Multiply by the coulombs and you have the joules: $W = q\,\Delta V$. That is the entire content, and it is the bridge between the electrical world and the energy world — every time a charge crosses a voltage, energy changes hands, and this is the exchange rate. A 12 V battery pushing 2 C around a circuit does 24 J of work, whatever the circuit is made of.

The relation names its own unit. Take one elementary charge across one volt and the energy is $1.602 \times 10^{-19}$ J, which is the electron volt — the currency of atomic, nuclear and particle physics, precisely because the joule is absurdly large at that scale. An electron accelerated through the 20 kV of an old television tube arrives with 20 keV; the ionisation energy of hydrogen is 13.6 eV; a visible photon carries about 2 eV. Reading those numbers as "the voltage this particle fell through" makes a whole subject legible.

Signs carry real information here. Work is positive when something outside the system pushes the charge against the field, storing energy; negative when the field does the pushing and the charge gives energy up. An electron falling through a positive potential difference gains kinetic energy, and the $q$ in the equation is negative, so $W$ comes out negative — the field did the work, not you. The magnitude is the energy either way, and the sign only says which direction it flowed.

Three cautions. First, $\Delta V$ is a difference between two specified points; "the voltage at a point" is meaningless without saying against what. Second, this is the work done by or against the electric force only. A charge crossing a resistor also gives energy to the lattice as heat, and a charge crossing a battery's internal resistance loses some of the EMF before it ever reaches the terminals, so $q\Delta V$ at the terminals is not the full energy the source spent. Third, the relation assumes $\Delta V$ is fixed while the charge moves. It is, for a battery or a supply; it is emphatically not for a capacitor, whose voltage climbs as it fills, which is where the factor of a half in $\tfrac{1}{2}CV^2$ comes from and why that page is not this one.