In short

The electric potential V at a point is the work done per unit charge in bringing a small positive test charge from infinity (where the potential is zero) to that point, against the electrostatic force:

V(\vec{r}) \;=\; -\int_\infty^{\vec{r}} \vec{E}\cdot d\vec{\ell} \;=\; \frac{W_{\infty \to \vec{r}}}{q_0}, \qquad [V] = \text{volt} = \text{J/C}.

For a point charge q at the origin, the potential at distance r is

V(r) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r} \;=\; \frac{kq}{r}.

The potential difference between two points is V_A - V_B = -\int_B^A \vec{E}\cdot d\vec{\ell}, equal to the work per unit charge to move from B to A. Unlike the electric field, the potential is a scalar — you add potentials of different sources like ordinary numbers. The natural energy unit on the atomic scale is the electron volt, 1 \text{ eV} = 1.602 \times 10^{-19} J, equal to the kinetic energy an electron gains when it falls through a potential difference of one volt.

Drop a marble into a bowl and it rolls to the bottom. Raise the marble by hand against the pull of gravity and you have stored energy in it — energy that the marble will give back the instant you let go. You never bother tracking the force vector at every point along the marble's path; instead, you carry in your head a single scalar number, the height, and you know that higher marble = more stored energy and lower marble = less. Gravity's whole bookkeeping collapses into one number per point in space.

The electric field is ready for exactly the same trick. Carrying a positive test charge through the field of some source charges takes work, and that work — like the work of lifting the marble — does not depend on which path you take, only on where you start and where you end. So there must be a scalar number attached to every point in space that plays the role of "height" for electricity, a number that tells you, in one stroke, how much work it takes to get a unit charge there from infinity. That number is the electric potential, V.

The payoff is enormous. Electric fields are vectors — three components per point, directions that have to be added head-to-tail with care. Electric potentials are scalars — one number per point, adding as plain arithmetic. Every problem you ever solve in electrostatics can be recast as "find the potential first, then recover the field by differentiation," and the recasting turns messy vector integrals into tidy scalar ones. This article builds the potential from scratch: the work-done-per-unit-charge definition, the derivation of V = kq/r for a point charge, the potential difference as the quantity that matters for doing work on a real charge, and the electron-volt — the energy unit without which almost all of modern physics, from Delhi Metro traction substations at 25 kV AC to the 0.7 V drop across a silicon junction inside an ISRO satellite power bus, would be unspeakable.

Why the electric force is conservative — and what that buys you

Before the potential can be a well-defined number at every point, one thing has to be true: the work done by the electric force in moving a charge from B to A must depend only on B and A, not on the path. A force with this property is called conservative. Gravity has it. Friction does not. You need to convince yourself that the static electric field does.

Start with a single source charge Q at the origin, and move a test charge q_0 from some point B to some point A along an arbitrary path. The work done by the electric force is

W_{B \to A} \;=\; \int_B^A \vec{F}\cdot d\vec{\ell} \;=\; \int_B^A q_0 \vec{E}\cdot d\vec{\ell}.

Why: work is the line integral of force along the path. Replacing \vec{F} with q_0\vec{E} is just using the definition of the electric field from Electric Field.

The field of the point charge is radial: \vec{E} = \tfrac{kQ}{r^2}\,\hat{r}. Write the path element as d\vec{\ell} = dr\,\hat{r} + r\,d\theta\,\hat{\theta} + \ldots — the bit along the radial direction plus bits perpendicular to it. The dot product \vec{E}\cdot d\vec{\ell} picks out only the radial component:

\vec{E}\cdot d\vec{\ell} \;=\; \frac{kQ}{r^2}\,dr.

Why: \hat{r} is perpendicular to \hat{\theta}, so \hat{r}\cdot\hat{\theta} = 0 — sideways motion does no work against a radial force. Only motion toward or away from the source costs energy.

So the work integral collapses to a one-variable integral:

W_{B \to A} \;=\; \int_{r_B}^{r_A} \frac{kQ q_0}{r^2}\,dr \;=\; kQ q_0 \left[-\frac{1}{r}\right]_{r_B}^{r_A} \;=\; kQ q_0\left(\frac{1}{r_B} - \frac{1}{r_A}\right). \tag{1}

Why: the integral of 1/r^2 is -1/r. Only the endpoints r_A and r_B survive — the zigzags in the middle, however elaborate, all contributed zero because they were perpendicular to the field. The work depends only on the starting and ending distances from the source.

That is the statement you wanted: the work to move q_0 from B to A in the field of a point charge depends only on r_A and r_B, not on the shape of the path. Because the field of any configuration is a sum of point-charge fields (superposition), and this no-path-dependence property is linear in the field, it holds for every static configuration. The electrostatic field is conservative.

A conservative field earns a scalar potential. Define the electric potential V(\vec{r}) by the rule

\boxed{\;V(\vec{r}) \;=\; -\int_\infty^{\vec{r}} \vec{E}\cdot d\vec{\ell}\;} \tag{2}

Why: the integral is the work done by the field in moving a unit positive charge from infinity to \vec{r}; the minus sign converts it to the work done against the field, which is the energy you had to supply. Choosing infinity as the reference point (V = 0 at infinity) is a convention — any reference point works, but infinity is convenient for isolated charge distributions, because the field dies off there.

The sign convention is worth dwelling on. The potential at a point is the energy you had to pump into a unit positive charge to drag it in from infinity against the field. If the source charges repel your test charge, you had to push — positive work done by you, positive V. If the source charges attract your test charge, the field did the pushing for you — negative work done by you, negative V. Positive sources make positive potentials around them; negative sources make negative potentials. Intuitive.

Deriving V = kq/r for a point charge

The single most used formula in electrostatics. Here is the one-line derivation.

Set up: point charge q at the origin, find the potential at distance r from it.

Step 1. Write the field along a radial line from infinity in to r.

\vec{E}(r') \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r'^2}\,\hat{r}, \qquad r' \in (r, \infty).

Why: this is the point-charge field from Electric Field, with r' the distance to a generic point along the integration path.

Step 2. Pick the simplest path: straight-line from infinity in along the radial direction. Then d\vec{\ell} = -dr'\,\hat{r} (the unit vector points outward, but you are moving inward, so the displacement is in the -\hat{r} direction as r' decreases from \infty to r).

Actually, there is a tidier way to handle the sign. Parameterise by r' going from \infty down to r, and let d\vec{\ell} = dr'\,\hat{r} with dr' < 0. The dot product:

\vec{E}\cdot d\vec{\ell} \;=\; \frac{kq}{r'^2}\,dr'.

Why: with d\vec{\ell} written as dr'\,\hat{r}, the two \hat{r}'s dot to 1, and both the field magnitude and the differential keep their signs. The limits of integration carry the direction.

Step 3. Plug into the definition (2):

V(r) \;=\; -\int_\infty^r \frac{kq}{r'^2}\,dr'.

Step 4. Evaluate the integral.

V(r) \;=\; -kq\left[-\frac{1}{r'}\right]_\infty^r \;=\; -kq\left(-\frac{1}{r} + 0\right) \;=\; \frac{kq}{r}. \tag{3}

Why: the antiderivative of 1/r'^2 is -1/r'. Evaluated at r gives -1/r; at \infty gives 0. The outer minus sign from the definition flips the result, leaving +kq/r.

\boxed{\;V(r) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r} \;=\; \frac{kq}{r}\;} \tag{4}

The potential of a positive point charge is positive and falls off as 1/r — more slowly than the field, which falls off as 1/r^2. This slower decay is one reason potentials are easier to work with: they do not collapse to zero as abruptly, so small errors in r hurt less. A negative point charge has negative potential V = -k|q|/r, and a test charge brought in from infinity gains kinetic energy (the field does positive work) as it falls toward it.

Potential V(r) around a positive and a negative point charge Two curves showing V = kq/r as a function of distance r. The positive-charge curve is a hyperbola in the upper half-plane; the negative-charge curve is its mirror below the axis. Both asymptote to zero as r grows. distance r from the source charge potential V(r) 0 V = +kq/r (positive source) V = −kq/r (negative source) source
The potential of a point charge falls off as 1/r, not 1/r². A positive source sits in a well of positive potential that slopes down toward zero at infinity; a negative source sits in a trough of negative potential that climbs up to zero at infinity. The slower decay is why potentials are the electrician's natural tool.

Potential difference — the number that does the real work

The potential at a single point depends on where you chose to put the zero. The potential difference between two points does not. It is the quantity that actually shows up when you compute the work done on a charge, the kinetic energy gained by an accelerated electron, the voltage read by a voltmeter.

Define the potential difference between points A and B as

\boxed{\;V_A - V_B \;=\; -\int_B^A \vec{E}\cdot d\vec{\ell}\;} \tag{5}

Why: subtracting V_B = -\int_\infty^B \vec{E}\cdot d\vec{\ell} from V_A = -\int_\infty^A \vec{E}\cdot d\vec{\ell}, the integrals from infinity to the common point cancel and you are left with the integral along any path from B to A. The reference point drops out — the difference is reference-free.

The physical meaning is clean. The work done by the electric field in moving a charge q from B to A is

W_{B\to A}^\text{field} \;=\; \int_B^A \vec{F}\cdot d\vec{\ell} \;=\; q\int_B^A \vec{E}\cdot d\vec{\ell} \;=\; -q(V_A - V_B). \tag{6}

And the work done by an external agent (you, a battery, the cell in your phone) moving the charge quasi-statically — just fast enough that the kinetic energy stays zero — is the negative of that:

\boxed{\;W_{B\to A}^\text{external} \;=\; q(V_A - V_B)\;} \tag{7}

Why: if the charge moves without gaining kinetic energy, the external force must exactly cancel the electric force at every instant, so the external work is -W^\text{field}. The potential difference, multiplied by the charge, gives you the energy an external agent must supply to move that charge between the two points.

This is the single most useful formula in circuits. A 1.5 V cell in your TV remote provides a potential difference of 1.5 V between its terminals: moving one coulomb of charge from the negative terminal (-) to the positive terminal (+) inside the cell, the chemistry of the battery does W = q(V_A - V_B) = (1)(1.5) = 1.5 joules of work. Outside the cell, as that coulomb flows back through the remote's circuitry, the electric field does 1.5 J of work on it — which is what powers the remote's LED and infrared emitter.

Sign bookkeeping, once and for all. A positive charge moves spontaneously from high potential to low potential (the field pushes it downhill). A negative charge moves spontaneously from low potential to high potential. If you want to push a positive charge uphill — from low to high potential — you must do positive work on it. Every current you have ever seen in a circuit is this rule in action.

The volt, the millivolt, the kilovolt, the megavolt

The SI unit of potential is the volt, defined as one joule per coulomb. Named after Alessandro Volta, but the name does not need to land — the size does. Useful benchmarks:

Sub-volt sub-units you will meet in electronics and biophysics:

Between 100 μV and 100 MV spans ten orders of magnitude — the same electric potential, the same definition, equally at home in the chemistry of a biological cell and in the discharge arc of a monsoon thundercloud.

Why the potential is a scalar — and why that changes everything

Let two point charges q_1 and q_2 sit somewhere in space. A third test charge q_0 at point P feels the vector sum of two Coulomb forces, one from each source. To find the total field at P, you have to draw two arrows and add them head-to-tail.

The potential is simpler. Superposition of forces \Rightarrow superposition of fields \Rightarrow superposition of potentials, but the last sum is scalar, not vector:

V(P) \;=\; V_1(P) + V_2(P) \;=\; \frac{kq_1}{r_1} + \frac{kq_2}{r_2},

where r_1 is the distance from q_1 to P and r_2 is the distance from q_2 to P. No angles, no cosines, no components. Add the two numbers. Done.

Adding potentials from two sources is scalar addition Two point charges q1 (positive, red) and q2 (negative, dark) at fixed positions; a test point P; distance lines labelled r1 and r2. A readout showing V(P) = kq1/r1 + kq2/r2 as a single scalar sum. q₁ = +3 nC q₂ = −2 nC P r₁ r₂ V(P) = kq₁/r₁ + kq₂/r₂ scalar sum — no direction, no vector arithmetic
The field at P is the vector sum of two separate arrows and requires care with components. The potential at P is the ordinary sum of two numbers. This one simplification — scalar instead of vector — is why potential-based methods are the default when there is more than one source charge.

There is a rule of thumb hidden here. Whenever a problem asks about energy or work on a charge, work with potentials; only switch to the field if the problem explicitly asks for a force or a direction. Most exam problems in Class 12 and JEE electrostatics are energy problems in disguise. Learning to see them that way cuts the algebra by a factor of three.

Worked examples

Example 1: Potential at the centre of three charges

Three point charges are placed at the vertices of an equilateral triangle of side 30 cm: q_1 = +4 nC, q_2 = +4 nC, and q_3 = -2 nC. Find the electric potential at the centroid of the triangle. Take k = 9.0 \times 10^9 N·m²/C².

Equilateral triangle of three point charges with the centroid marked An equilateral triangle of side 30 cm with two positive charges q1 and q2 at the base corners and a negative charge q3 at the top. The centroid is marked in the middle, at equal distance r from each vertex. q₃ = −2 nC q₁ = +4 nC q₂ = +4 nC centroid r r r
All three charges sit at the same distance $r$ from the centroid — the common centroid-to-vertex distance of an equilateral triangle.

Step 1. Find the distance r from each vertex to the centroid.

For an equilateral triangle of side a, the centroid-to-vertex distance is r = a/\sqrt{3}.

r \;=\; \frac{0.30 \text{ m}}{\sqrt{3}} \;=\; 0.1732 \text{ m}.

Why: the centroid lies two-thirds of the way from any vertex to the opposite midpoint along the median. The median of an equilateral triangle of side a has length \tfrac{\sqrt 3}{2}a, so the centroid sits at distance \tfrac{2}{3}\cdot\tfrac{\sqrt 3}{2}a = \tfrac{a}{\sqrt 3} from each vertex.

Step 2. Compute each individual potential at the centroid. Since all three distances are equal, factor out k/r.

V_i \;=\; \frac{k q_i}{r}, \qquad V \;=\; V_1 + V_2 + V_3 \;=\; \frac{k}{r}(q_1 + q_2 + q_3).

Why: this is the scalar-sum rule. The potential at a point from a cluster of charges is the algebraic sum of their individual potentials. Signs stay with the charges — the negative charge contributes a negative term.

Step 3. Substitute the numbers.

q_1 + q_2 + q_3 \;=\; (+4) + (+4) + (-2) \text{ nC} \;=\; +6 \text{ nC} \;=\; 6 \times 10^{-9} \text{ C}.
V \;=\; \frac{(9.0 \times 10^9)(6 \times 10^{-9})}{0.1732} \;=\; \frac{54}{0.1732} \;=\; 311.8 \text{ V}.

Why: the 10^9 from k and the 10^{-9} from the charges cancel — nanocoulombs and k = 9 \times 10^9 were engineered to produce nice volt-scale numbers. Always do this cancellation in your head before punching the calculator.

Result: The potential at the centroid is V \approx +312 V.

What this shows: Even though the electric field at the centroid is a non-trivial vector (the two positive charges pull outward while the negative pulls in, each along its own direction — you would need trigonometry to add the arrows), the potential is a single scalar that falls out of a one-line arithmetic sum. That is the whole point of the potential framework.

Example 2: Voltage required to accelerate an electron to a given speed

An ISRO cathode-ray tube needs electrons moving at 2.0 \times 10^7 m/s to draw a sharp image. Through what potential difference must the electrons be accelerated, starting from rest? Mass of an electron m_e = 9.11 \times 10^{-31} kg, charge e = 1.602 \times 10^{-19} C. Also express the answer in electron-volts.

Electron accelerated between two parallel plates An electron at rest at the negative plate is accelerated through a potential difference V to the positive plate on the right, where it has speed v. Arrows show the electric field pointing left from + to − and the force on the electron pointing right. − cathode + anode e⁻ at rest v E field F on e⁻ V_anode − V_cathode = V
An electron released at rest at the cathode falls "uphill" in the field (opposite to E, because it is negative) and gains kinetic energy equal to $eV$ when it reaches the anode.

Step 1. Write the energy balance.

The external field does work W = q(V_A - V_B) = (-e)(V_\text{cathode} - V_\text{anode}) = e(V_\text{anode} - V_\text{cathode}) = eV on the electron, where V = V_\text{anode} - V_\text{cathode} > 0. By the work-energy theorem, that work equals the gain in kinetic energy:

eV \;=\; \tfrac{1}{2}m_e v^2. \tag{8}

Why: the electron has charge -e, so moving it from the low-potential cathode to the high-potential anode is moving a negative charge uphill in potential — which is spontaneous for a negative charge (the field force on -e in a field pointing left is rightward). The external work by the field on the electron is positive; all of it becomes kinetic energy because the electron starts from rest.

Step 2. Solve for V.

V \;=\; \frac{m_e v^2}{2e}.

Step 3. Substitute the numbers.

V \;=\; \frac{(9.11 \times 10^{-31})(2.0 \times 10^7)^2}{2(1.602 \times 10^{-19})} \;=\; \frac{(9.11 \times 10^{-31})(4.0 \times 10^{14})}{3.204 \times 10^{-19}}
V \;=\; \frac{3.644 \times 10^{-16}}{3.204 \times 10^{-19}} \;=\; 1137 \text{ V} \;\approx\; 1.14 \text{ kV}.

Why: the powers of ten sort out as 10^{-31}\times 10^{14} / 10^{-19} = 10^{-31+14+19} = 10^{2}, which puts you in the hundreds-of-volts range — right where a CRT anode lives.

Step 4. Convert the kinetic energy to electron-volts.

The electron dropped through V \approx 1137 V of potential. By the definition of the electron-volt (see below), the kinetic energy gained is numerically equal to the voltage in volts, measured in eV:

\text{KE} \;=\; eV \;=\; 1137 \text{ eV} \;\approx\; 1.14 \text{ keV}.

Result: The electrons must be accelerated through about 1.14 kV, picking up 1.14 keV of kinetic energy in the process.

What this shows: The electron volt is a unit designed for exactly this calculation. If the question were phrased "an electron is accelerated through 1.14 kV, what is its kinetic energy?" the answer in eV is literally 1.14 keV — no multiplication by e, no power-of-ten gymnastics. Particle physicists work in MeV, GeV, and TeV for precisely this reason: their accelerators are voltage drops, and the natural unit of the output energy is the voltage itself.

The electron volt — the natural energy unit of modern physics

Joules are the wrong unit for anything involving single charges. A single electron accelerated through a single volt gains (1.602 \times 10^{-19})(1) = 1.602 \times 10^{-19} J — a number so small it is clumsy to write. Physics needs a smaller unit.

Define the electron volt as that amount of energy directly:

\boxed{\;1 \text{ eV} \;=\; 1.602 \times 10^{-19} \text{ J}\;} \tag{9}

It is the kinetic energy an electron picks up when it falls through a potential difference of exactly one volt, starting from rest. Equivalently (and more usefully for doing homework), it is the number you get by just writing down the potential difference in volts, as an energy in eV, when the charge is one electron.

Useful benchmarks in electron volts:

The eV ladder — meV, eV, keV, MeV, GeV, TeV — climbs through every branch of physics. And it works because the definition is directly tied to a voltage difference times a unit charge, which is the most natural unit of work a physicist can construct.

The cost of a joule, in rupees

Here is an ISRO-flavoured back-of-the-envelope. A lithium-ion battery pack on an Indian remote-sensing satellite stores on the order of 1 kWh = 3.6 \times 10^6 J per kilogram of mass. Launch cost to low Earth orbit, currently, is about ₹30 000 per kilogram (PSLV, rough figure — actual contracts vary). That makes the effective price of energy in orbit roughly

\frac{30\,000 \text{ rupees/kg}}{3.6 \times 10^6 \text{ J/kg}} \;\approx\; 8 \times 10^{-3} \text{ rupees per joule} \;\approx\; 8 \text{ paise per joule}.

Every time an onboard experiment accelerates an electron through 1 V, it spends 1.602 \times 10^{-19} J of stored energy — costing a fantastically tiny 1.3 \times 10^{-20} paise, which is irrelevant at the single-electron level but becomes a real design constraint when a detector has to accelerate 10^{15} electrons per second through kilovolt potentials. Every volt in the anode supply is a line item in the mission's mass budget, and therefore a line item in the mission's rupee budget. The potential difference is not an abstract textbook quantity; it is something you pay for.

Common confusions

If the definition V = -\int \vec E\cdot d\vec\ell and the formula V = kq/r for a point charge feel solid, you have the essentials. The rest of this section is for readers who want to see why the electrostatic field is conservative in the language of vector calculus, how the reference point is chosen for infinite charge distributions, and how the electron-volt collapses the ladder of energy scales in physics.

Why \nabla\times\vec E = 0 for electrostatics (the curl-free condition)

A vector field \vec F is conservative if and only if its curl is zero everywhere:

\nabla \times \vec F \;=\; 0 \quad\Longleftrightarrow\quad \vec F \;=\; -\nabla \phi \text{ for some scalar } \phi.

For a single point charge at the origin, \vec E = \tfrac{kq}{r^2}\hat r, and it is a standard exercise in vector calculus (Stokes' theorem, radial symmetry) to show that \nabla \times \vec E = 0 everywhere except at the origin, where the field is singular. Because the curl is a linear operator, the curl of a sum of point-charge fields is zero wherever none of the sources sits:

\nabla \times \vec E_\text{total} \;=\; \sum_i \nabla \times \vec E_i \;=\; 0.

That single line is the rigorous statement of "the electric work depends only on endpoints, not on the path." It is the electrostatic twin of \vec g = -\nabla\Phi_\text{grav} in classical mechanics, and the argument for both is the same.

This result breaks when the field is time-dependent — then Faraday's law adds a non-zero curl, \nabla\times\vec E = -\partial\vec B/\partial t, and the electric field is no longer conservative. A scalar potential in the electrostatic sense cannot exist for such fields. You have to use the combined scalar-and-vector-potential formulation of electromagnetism — which is a topic for the electromagnetic induction article. For now, remember: the entire edifice of "electric potential" assumes static charges. The moment a current changes with time, you need more than V.

Choosing the reference: why "infinity" sometimes fails

The formula V = -\int_\infty^{\vec r}\vec E\cdot d\vec\ell demands that the field at infinity falls off fast enough for the integral to converge. For a point charge, it falls like 1/r^2, and the integral of 1/r^2 from infinity inward converges — infinity is a fine reference.

But for a charge distribution that does not decay fast enough — say, an infinite uniformly charged line — the field falls only like 1/r, and the integral \int (1/r)\,dr diverges logarithmically. There is no finite V at infinity.

The fix is to choose a finite reference point r_0. For an infinite line of charge \lambda per unit length,

V(r) \;=\; -\int_{r_0}^r \frac{\lambda}{2\pi\varepsilon_0}\,\frac{dr'}{r'} \;=\; -\frac{\lambda}{2\pi\varepsilon_0}\ln\!\frac{r}{r_0}.

The choice of r_0 is a gauge choice — it shifts V everywhere by a constant, which drops out of all physically meaningful quantities (potential differences, forces on charges, energies). Only differences in potential matter. This is a general pattern in physics: whenever a theory has an arbitrary additive constant, choosing a reference is a matter of convenience, not of physics. The voltmeter on your multimeter obeys this rule — it only reads differences.

The eV as a speed-of-light calibration

One of the most useful things about electron volts is how they interlock with the rest-mass energies of elementary particles. The electron mass is 511 keV/c², the proton is 938 MeV/c², and combinations like the Higgs (125 GeV/c²) or the top quark (173 GeV/c²) are all quoted in this same unit, divided by c^2. When an electron is accelerated through a potential difference of 511 kV, its kinetic energy is of the order of its rest-mass energy, and the Newtonian \tfrac{1}{2}m_e v^2 formula starts to break down — you need the relativistic kinetic energy KE = (\gamma - 1)m_ec^2 with \gamma = 1/\sqrt{1 - v^2/c^2}.

A useful rule of thumb: non-relativistic electron dynamics is valid up to about 1 keV, questionable at 10 keV, and definitely relativistic beyond 100 keV. Worked Example 2 above was at 1.1 keV — right on the edge, but non-relativistic kinematics is still within about 0.5% of the correct answer. The moment a CRT runs at 20 kV (a standard old colour TV voltage), relativistic corrections matter at the 2% level, and at 100 kV they are essential.

Why engineers and physicists talk about the same thing differently

An electrical engineer works in volts, amperes, and watts — macroscopic units. A physicist working on single atoms works in electron volts, ångströms, and femtoseconds — microscopic units. The connection is the Faraday constant

F \;=\; N_A\,e \;=\; (6.022 \times 10^{23})(1.602 \times 10^{-19}) \;\approx\; 96{,}485 \text{ C/mol}.

Multiply an eV by N_A and you get roughly 96.5 kJ per mole — the typical energy scale for a chemical reaction, which is why bond energies are quoted sometimes in eV (per molecule) and sometimes in kJ/mol (per mole of molecules) and the two are about two orders of magnitude apart. This is the exact same potential difference rewritten for a laboratory chemist instead of a quantum physicist.

Where this leads next