potential voltage and equipotential surfaces

Electric potential $V$ at a point is the potential energy of a unit positive charge placed there: $V = U/q$, measured in volts ($1\ \text{V} = 1\ \text{J/C}$). For a point charge $q$ at distance $r$, $V = k\dfrac{q}{r}$ with $k \approx 8.99\times10^{9}\ \text{N·m}^2/\text{C}^2$; unlike field strength, potential is a scalar, so contributions of several charges simply add: $V = \sum_i k q_i/r_i$.

The physically measurable quantity is the potential difference, or voltage, $V_A - V_B$: when charge $q$ moves from A to B, the electric field does work $W_{\text{field}} = q(V_A - V_B)$.

An equipotential surface joins points of equal potential, so moving a charge along it requires zero work. Field lines are everywhere perpendicular to equipotential surfaces and point toward decreasing potential, while the field magnitude equals the potential drop per unit length, $E = -\Delta V/\Delta r$ for small $\Delta r$. That is why field lines and equipotentials form a mutually perpendicular grid.

A point charge $q = +2.0\ \mu\text{C}$ sits in vacuum. Find the potential at distances $r_1 = 0.30\ \text{m}$ and $r_2 = 0.60\ \text{m}$ from the charge, and the voltage between these two points.

Charges $q_1 = +4.0\ \mu\text{C}$ at $x = 0$ and $q_2 = -4.0\ \mu\text{C}$ at $x = 0.80\ \text{m}$ are fixed on the $x$-axis. (a) Find the potential at $x = 0.20\ \text{m}$. (b) What work does the field do while moving charge $q_3 = +1.0\ \text{nC}$ from $x = 0.20\ \text{m}$ to $x = 0.40\ \text{m}$?

Along the $x$-axis the potential is given by $V(x) = (120\ \text{V/m}^2)x^2 - 40\ \text{V}$. (a) Find the field component $E_x$ at $x = 0.50\ \text{m}$. (b) Describe the equipotential surfaces. (c) Find the work done by the field while moving charge $q = +2.0\ \text{nC}$ from $x = 0.10\ \text{m}$ to $x = 0.50\ \text{m}$.

Work in the PhET simulation Charges and Fields (safe: no live circuits or voltages). (1) Place a single $+1\ \text{nC}$ charge and read the potential with the voltmeter at $0.5\ \text{m}$, $1.0\ \text{m}$, and $1.5\ \text{m}$; check that the values follow $V = kq/r$. (2) Add a $-1\ \text{nC}$ charge, switch on the equipotential display (or sample 8–10 grid points with equal voltmeter readings) and sketch one equipotential curve; verify that it crosses the field lines at right angles. (3) Choose points A and B, record $V_A$ and $V_B$, move a $+1\ \text{nC}$ test charge from A to B, and compare the work predicted by $W = q(V_A - V_B)$ with the work the simulation reports.

Show what equipotential surfaces look like for a single charge and for a dipole, and how they relate to field lines and to the sign of the charge.

A two-panel sketch. Left: a point charge $+q$ surrounded by dashed concentric circles (equipotentials labelled $V_1 > V_2 > V_3$) and radial field-line arrows pointing outward. Right: a dipole $+q$ and $-q$ with a grid of solid field lines and dashed equipotential curves crossing them at right angles; labels $V>0$ near $+q$ and $V<0$ near $-q$, a highlighted arrow $\vec{E}$ showing the direction of decreasing potential, and the caption $E = -\Delta V/\Delta r$.

A phone battery stores $4000\ \mathrm{mA\cdot h}=4.0\ \mathrm{A\cdot h}=1.44\times10^{4}\ \mathrm{C}$ and delivers it at about $3.85\ \mathrm{V}$. A charger marked $5\ \mathrm{V}$, $2\ \mathrm{A}$ transfers $10\ \mathrm{W}$. A full charge moves $1.44\times10^{4}\ \mathrm{C}$ through a potential difference of $3.85\ \mathrm{V}$, so the battery stores $W=qU\approx5.5\times10^{4}\ \mathrm{J}$; at $10\ \mathrm{W}$ that is a little over $1.5$ hours.

Here voltage is energy per unit charge: $U=W/q$, so each coulomb pushed through the battery gains $3.85\ \mathrm{J}$. The two terminals sit on two different equipotential surfaces of the circuit's field: sliding a charge along either terminal surface needs no work, and energy is spent only when moving it between those surfaces.

In a thunderstorm the cloud base can reach about $1\times10^{8}\ \mathrm{V}$ relative to the ground, and one stroke carries roughly $20\ \mathrm{C}$. The field does $W=qU=20\times10^{8}=2\times10^{9}\ \mathrm{J}=2\ \mathrm{GJ}$ of work. Since dry air breaks down near $3\times10^{6}\ \mathrm{V/m}$, a $10^{8}\ \mathrm{V}$ difference needs a gap of $d=U/E\approx33\ \mathrm{m}$; the discharge lasts about $100\ \mathrm{\mu s}$, giving a peak power near $2\times10^{13}\ \mathrm{W}$.

The average field between cloud and ground is $E=U/d$, the potential difference measured between two points divided by their separation. The cloud base and the ground act as two equipotential surfaces, and the leader follows field lines, which cross every equipotential surface at right angles.

A traction battery works at $400\ \mathrm{V}$ and drives $100\ \mathrm{A}$ into the motor, that is $40\ \mathrm{kW}$. A pack rated $60\ \mathrm{kW\cdot h}=2.16\times10^{8}\ \mathrm{J}$ therefore moves $q=W/U=2.16\times10^{8}/400=5.4\times10^{5}\ \mathrm{C}$ from one terminal to the other; at $100\ \mathrm{A}$ this takes $5.4\times10^{3}\ \mathrm{s}\approx1.5\ \mathrm{h}$. Inside the copper cables the potential falls by about $0.5\ \mathrm{V}$ per metre, a field of roughly $0.5\ \mathrm{V/m}$.

The relation $U=W/q$ fixes the energy each coulomb carries, and $E=-\Delta V/\Delta x$ links the fall of potential along a cable to the field inside the copper. The two battery terminals are equipotential surfaces $400\ \mathrm{V}$ apart, and the work of carrying a charge between them depends only on that difference, not on the route through the circuit.