3.3.6 Applications of the electrostatic potential

Energy = charge × voltage

By Eq. (3.3.11) the work done on a charge between two points is fixed by the potential difference, whatever the path. Consider an electron (charge $-e$) released at rest from a heated cathode at potential $V=0$ and accelerated towards an anode at $V=V_a>0$ (Figure 3.3.19). The work done by the field on the electron is the negative of the work an external agent would have to do, $-q(V_B-V_A)=-(-e)(V_a-0)=e{V}_a$, and it all goes into kinetic energy:

Only the potential difference matters: the shapes of the electrodes determine the trajectory and the focusing of the beam, but not the final speed. Between plane electrodes a distance $d$ apart the potential rises linearly, $V=V_{a}x/d$, and the field $E=V_a/d$ is uniform (Problem 3.3.1). It points from the anode to the cathode, and the force $-eE$ on the electron points towards the anode. A negative charge "rolls uphill" in $V$: for it, the landscape is turned upside down.

The electron volt

The energy an electron gains when it passes through a potential difference of one volt is called one electron volt:

A 25 kV tube produces 25 keV electrons: in this unit the energy is the accelerating voltage. Typical accelerating voltages are about 25 kV in a colour cathode-ray tube, 1–30 kV in a scanning electron microscope, 80–300 kV in a transmission electron microscope and 30–150 kV in a medical X-ray tube. In an X-ray tube the whole energy $e{V}_a$ of one electron can be converted into a single photon, so the shortest emitted wavelength is ${\lambda }_{\min }=hc/(e{V}_a)$ (the Duane–Hunt law). For $V_a=100\mathrm{kV}$ this is about 12 pm.

Raise the accelerating voltage $V_a$ and note that the electron arrives faster but is deflected less by the same plate voltage: the transit time between the plates shrinks.

E = −∇V under our feet

When a fault current $I$ flows into the ground through an earth electrode, it spreads out through the soil. In this application $\rho$ denotes the resistivity of the soil (typically 10–1000 $\Omega$m), not a charge density. In the soil the current density and the field are related by Ohm's law, $J=E/\rho$. In steady state no charge accumulates anywhere in the (homogeneous) soil, so $\nabla \cdot J=0$, hence $\nabla \cdot E=0$, and with $E=-\nabla V$:

The potential in the soil obeys Laplace's equation again, and the solutions of Problems 3.3.1–3.3.3 apply. (A steady current field is curl-free just like an electrostatic one, so the potential concept carries over unchanged.)

For a hemispherical electrode of radius $a$ the current leaves radially and crosses hemispheres of area $2\pi r^2$, so $J=I/(2\pi r^2)$ and

with remote earth as the reference, $V(\infty )=0$. This is the $1/r$ solution of Laplace's equation found in Problem 3.3.3.

Definitions

The quantities used in earthing design follow directly from Eq. (3.3.19) (Figure 3.3.20):

• the earth potential rise $V(a)=\rho I/(2\pi a)$, the voltage of the electrode and of everything bonded to it with respect to remote earth, which during a fault can reach tens of kilovolts (the ratio $R_g=V(a)/I=\rho /(2\pi a)$ is the resistance of the electrode),
• the touch voltage $V(a)-V(r)$, which appears across the body of a person whose hand is on the earthed structure while the feet stand at distance $r$, and
• the step voltage $V(r)-V(r+1\mathrm{m})$, which appears between the two feet, about one metre apart, of a person standing at distance $r$.

Both the touch and the step voltage are largest where $|\nabla V|$ is largest, i.e. next to the electrode. For $a=0.5\mathrm{m}$, $\rho =100\Omega$m and $I=1\mathrm{kA}$ the electrode rises to 31.8 kV. A person touching it with the feet 1 m from its edge ($r=1.5\mathrm{m}$) is exposed to 21.2 kV, and a person merely standing with the feet at $r=1\mathrm{m}$ and $r=2\mathrm{m}$ from its centre to 7.96 kV between the feet, falling to 0.53 kV for $r=5$–6 m (Worked example 3.3.2). Figure 3.3.21 reads these numbers off the potential curve: touch and step voltages are simply vertical distances on it.

Remedy

Standards for substation earthing (such as IEEE Std 80) set limits for both quantities. The remedy is equipotential bonding: a buried earthing mesh under the whole station brings everything a person can touch or stand on to nearly the same potential, i.e. it makes $|\nabla V|$ small where people are. A surface layer of high-resistivity crushed rock further limits the current that can flow through the body.