Electric potential

Adding vectors is laborious, and the previous two lessons did a great deal of it for fields that a single number at each point could have described just as well.

The escape is a result already proved in mechanics: a force directed along the line to a fixed centre, with a magnitude depending only on distance from it, is conservative. The work it does between two points is independent of the route taken, which means a potential energy exists. Coulomb's force is exactly of that form, so everything mechanics says about gravitational potential energy carries across, with two changes: the force can be repulsive as well as attractive, and it is convenient to divide out the charge being moved.

The work is path independent

Take a fixed charge $Q$ at the origin and carry a test charge $q_{0}$ from a point at radius $r_{a}$ to one at radius $r_{b}$ along any path at all. Break the path into steps. Each step splits into a piece directly along the radius and a piece perpendicular to it, running along a sphere of constant radius. The force is radial, so it does no work at all on the perpendicular pieces: only the radial pieces count, and what they add up to is the net change in $r$ , whatever wandering happened in between.

So the work reduces to a single integral over $r$ . For like charges the force on the test charge points outward with magnitude $k Q q_{0} / r^{2}$ , and the work it does as the radius goes from $r_{a}$ to $r_{b}$ is

W = \int_{r_{a}}^{r_{b}} \frac{k Q q_{0}}{r^{2}} d r = k Q q_{0} (\frac{1}{r_{a}} - \frac{1}{r_{b}})

The answer depends only on the endpoints. Define the potential energy in the usual way, as minus the work done by the force, with the zero placed at infinite separation where the force vanishes:

U = \frac{k Q q_{0}}{r}

For like charges $U$ is positive and falls as they separate, which says that bringing them together costs work and that they will fly apart if released. For unlike charges $U$ is negative, which says the pair is bound: energy must be supplied to separate them to infinity.

Divide by the test charge and what remains belongs to the source and the point, exactly as dividing force by charge gave the field. That is the electric potential:

V = \frac{U}{q_{0}} = \frac{k Q}{r}

measured in joules per coulomb, called volts. A charge of 1 nC produces 8.99 V at a metre. The volt is the unit almost every practical measurement in electricity is made in, and it is worth registering that it is an energy per charge, not a force, not a field, and not a quantity of anything.

Why a scalar is worth having

The gain is immediate. Potentials from several charges add as ordinary numbers:

V = k \sum_{i} \frac{q_{i}}{r_{i}}

with the sign of each charge carried along and no components, no angles and no resolution into axes. For a continuous distribution the sum becomes $V = k \int d q / r$ , which is a scalar integral where the second lesson needed a vector one.

The ring is the case to compare. Every piece of a ring of radius $a$ carrying total charge $Q$ is the same distance $\sqrt{x^{2} + a^{2}}$ from a point on the axis, so

V = \frac{k Q}{\sqrt{x^{2} + a^{2}}}

with no cancellation argument, no cosine, and no worry about which way anything points. Compare the work needed for the field in the second lesson. The direction was recoverable all along, as the next section shows.

Example. A charge of $+ 5.0$ nC sits at the origin and a charge of $- 3.0$ nC at $x = 4.0$ cm. What is the potential at the point 3.0 cm directly above the origin?

That point is 3.0 cm from the first charge and, by Pythagoras on the 3, 4, 5 triangle, 5.0 cm from the second. Potentials add as numbers: $V = k (5.0 \times 10^{- 9} / 0.030) + k (- 3.0 \times 10^{- 9} / 0.050) = (8.9876 \times 10^{9}) (1.667 \times 10^{- 7} - 6.00 \times 10^{- 8}) = 959$ V. No geometry beyond the two distances was needed, which is the entire point.

Now you. A charge of $+ 4.0$ nC sits at the origin and a charge of $+ 6.0$ nC at $x = 8.0$ cm. What is the potential at the point 6.0 cm directly above the origin?

Answer

The distances are 6.0 cm and, from the 6, 8, 10 triangle, 10.0 cm. So $V = (8.9876 \times 10^{9}) (4.0 \times 10^{- 9} / 0.060 + 6.0 \times 10^{- 9} / 0.100) = (8.9876 \times 10^{9}) (1.267 \times 10^{- 7}) = 1138$ V.

Getting the field back

Potential would be a dead end if the field could not be recovered from it. It can, by differentiating.

Move a test charge a small distance $d s$ in some direction. The work done by the field is $q_{0} E_{s} d s$ , where $E_{s}$ is the component of $E$ along that direction, and by definition that work is $- q_{0} d V$ . Cancelling $q_{0}$ ,

E_{s} = - \frac{d V}{d s}

The field component in any direction is minus the rate of change of potential in that direction. In three dimensions this gives the three components as three partial derivatives, the operation called the gradient, but nothing in this course needs more than the one-dimensional form applied along a chosen axis.

Test it on the ring. Differentiating $V = k Q (x^{2} + a^{2})^{- 1 / 2}$ with respect to $x$ gives $d V / d x = - k Q x (x^{2} + a^{2})^{- 3 / 2}$ , so $E_{x} = k Q x / (x^{2} + a^{2})^{3 / 2}$ , which is exactly what the vector integral produced in the second lesson. One scalar integral and one derivative have replaced a symmetry argument and a component integral.

Two consequences of $E_{s} = - d V / d s$ are worth stating on their own. First, the field points from high potential to low, because of the minus sign, and a positive charge released from rest moves downhill in potential exactly as a mass falls downhill in height. Second, take a direction in which $V$ does not change: the field component along it is zero. The surfaces on which $V$ is constant, called equipotentials, are therefore everywhere perpendicular to the field. Around a point charge they are concentric spheres, and around any conductor in equilibrium they include the conductor's own surface, since the field inside is zero and no work is needed to move a charge anywhere within it. A conductor in electrostatic equilibrium is an equipotential volume, which is why one wire can be said to be at one voltage.

The units also reconcile here: volts per metre is the same as newtons per coulomb, since a joule is a newton metre.

The electron volt

A charge $q$ moved through a potential difference $V$ gains energy $q V$ , and for particles the natural unit is what an elementary charge gains across one volt:

1 eV = 1.602 \times 10^{- 19} J

It is a unit of energy, not of voltage, and it makes atomic numbers legible. The potential energy of the electron and proton in a hydrogen atom, at the Bohr radius of 52.9 pm, is $U = - k e^{2} / r = - 4.36 \times 10^{- 18}$ J, which is $- 27.2$ eV. The electron's kinetic energy is half the magnitude of that, by the virial result that holds for any inverse square orbit, so the total is $- 13.6$ eV, and 13.6 eV is exactly the energy needed to pull the electron off. That is the ionisation energy of hydrogen, measured spectroscopically to seven figures, and it drops out of a potential energy and a fact about orbits.

Example. An alpha particle, charge $+ 2 e$ and mass $6.645 \times 10^{- 27}$ kg, is accelerated from rest through a potential difference of 1.00 kV. How fast is it moving?

The energy gained is $q V = 2 (1.602 \times 10^{- 19}) (1000) = 3.204 \times 10^{- 16}$ J, which is 2000 eV, and all of it appears as kinetic energy. From $\frac{1}{2} m v^{2} = 3.204 \times 10^{- 16}$ , $v^{2} = 2 (3.204 \times 10^{- 16}) / (6.645 \times 10^{- 27}) = 9.64 \times 10^{10}$ , so $v = 3.11 \times 10^{5}$ m/s. Notice that the shape of the accelerating field never entered: only the endpoints matter, which is what path independence buys.

Now you. A proton, mass $1.673 \times 10^{- 27}$ kg, is accelerated from rest through 2.50 kV. How fast is it moving?

Answer

Energy gained is $(1.602 \times 10^{- 19}) (2500) = 4.005 \times 10^{- 16}$ J. Then $v^{2} = 2 (4.005 \times 10^{- 16}) / (1.673 \times 10^{- 27}) = 4.79 \times 10^{11}$ , so $v = 6.92 \times 10^{5}$ m/s.

Potential of a charged conductor, and why points spark

A conducting sphere of radius $R$ carrying charge $Q$ looks, from outside, like a point charge at its centre, so its surface sits at $V = k Q / R$ , and every point inside sits at the same value, since the interior field is zero. The potential does not drop to zero at the centre; it is flat there, which is what a zero field means.

Now connect two conducting spheres of different radii by a long thin wire. They form one conductor, so they must end at one potential, and charge flows until $k Q_{1} / R_{1} = k Q_{2} / R_{2}$ , that is $Q_{1} / Q_{2} = R_{1} / R_{2}$ . The larger sphere takes proportionally more charge. But surface charge density is $Q / 4 π R^{2}$ , so

\frac{σ_{1}}{σ_{2}} = \frac{Q_{1}}{Q_{2}} \cdot \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{2}}{R_{1}}

and the smaller sphere ends up with the higher density, and therefore, since $E = σ / ε_{0}$ just outside a conductor, the stronger field at its surface.

This is the quantitative version of the observation that charge concentrates at sharp points. A sharp region behaves like a small sphere, so it carries a high density and a high local field, and if that field exceeds about $3 \times 10^{6}$ V/m the air ionises and charge leaks away as a corona discharge. It is why high voltage equipment is built with fat rounded conductors and no sharp edges, and why a lightning rod is pointed: it is designed to leak, bleeding charge into the air and providing a preferred path when it does not.

The same number bounds what any isolated conductor can hold. A sphere of radius 10 cm reaches breakdown at $V = E R = (3 \times 10^{6}) (0.10) = 3 \times 10^{5}$ V, so 300 kV is roughly the ceiling for a 10 cm ball in air regardless of how much charge you try to push onto it. Van de Graaff generators get past that only by being large, and by being run in pressurised gas.

Example. Two conducting spheres, of radii 8.0 cm and 2.0 cm, are joined by a long wire and given 20 nC in total. Find the charge on each and the field just outside each.

Equal potentials require $Q_{1} / Q_{2} = R_{1} / R_{2} = 4$ , and the charges must sum to 20 nC, so $Q_{1} = 16$ nC on the large sphere and $Q_{2} = 4.0$ nC on the small one. The fields are $E_{1} = k Q_{1} / R_{1}^{2} = (8.9876 \times 10^{9}) (16 \times 10^{- 9}) / (0.080)^{2} = 2.25 \times 10^{4}$ V/m and $E_{2} = (8.9876 \times 10^{9}) (4.0 \times 10^{- 9}) / (0.020)^{2} = 8.99 \times 10^{4}$ V/m. The small sphere holds a quarter of the charge and carries four times the field, and it is the one that will spark first.

Now you. Two conducting spheres, of radii 9.0 cm and 3.0 cm, are joined by a long wire and given 24 nC in total. Find the charge on each and the field just outside each.

Answer

$Q_{1} / Q_{2} = 3$ , so $Q_{1} = 18$ nC and $Q_{2} = 6.0$ nC. Then $E_{1} = (8.9876 \times 10^{9}) (18 \times 10^{- 9}) / (0.090)^{2} = 2.00 \times 10^{4}$ V/m and $E_{2} = (8.9876 \times 10^{9}) (6.0 \times 10^{- 9}) / (0.030)^{2} = 5.99 \times 10^{4}$ V/m, three times larger.

What potential is not

Three confusions are worth heading off, because each of them survives into later lessons and does damage there.

Potential is not potential energy. $V$ belongs to a point in space whether or not anything is there; $U = q V$ belongs to a charge sitting at that point. The distinction is the same one as between $E$ and $F$ .

Only differences are physical. The choice of zero at infinity is a convention, convenient because it makes $V = k Q / r$ look tidy, and useless for an infinite line or an infinite plane, where the potential at infinity diverges and the zero has to be put somewhere finite instead. Nothing measurable changes: every experiment reports a difference. When a circuit is said to be at 5 V it means 5 V above whatever node was called ground.

And a high potential is not by itself dangerous, nor a low one safe. Shuffling across a carpet routinely puts a person at 10 kV or more relative to the room, and the resulting spark is startling rather than lethal because the total charge, and so the energy $q V$ , is tiny. What harms is energy delivered, and that is a question about how much charge is available and how fast it moves, which is the subject of the next two lessons.

Assembling charge costs energy

One more question closes the electrostatics half of the course. Bringing charges together against their mutual repulsion takes work, and that work must be stored somewhere.

Bring the first charge $q_{1}$ in from infinity to empty space: no other charge exists yet, so no work is done. Bring $q_{2}$ to a distance $r_{12}$ from it, moving against a potential $k q_{1} / r_{12}$ , at a cost of $k q_{1} q_{2} / r_{12}$ . Bring $q_{3}$ in and it must be pushed against the potential of both, costing $k q_{1} q_{3} / r_{13} + k q_{2} q_{3} / r_{23}$ . The total energy of the assembly is the sum over all distinct pairs,

U = k \sum_{pairs} \frac{q_{i} q_{j}}{r_{i j}}

and it does not depend on the order in which they were brought, which is another consequence of path independence.

That is a real, recoverable energy: release the charges and it appears as kinetic energy. Where was it while they sat still? The expression above locates it in the pairs, which is not a place. There is another way to compute the same number that locates it in the space between the charges instead, at a density fixed by the local field strength and nothing else. Working out which description is right, and finding the device that makes the question practical, is the next lesson.