Why Birds Can Sit Safely on a High-Voltage Wire

A $9$ V battery and an $11$ kV power line differ in one key quantity — the electric potential difference they provide. Potential tells how much work the field does on each coulomb of charge, and it turns vector field problems into simpler additions of numbers.

This part covers electric potential, the potential of dipoles and charge systems, equipotential surfaces, and potential energy.

**The electric potential at a point is the work done per unit positive charge in bringing a test charge from infinity to that point without acceleration, the potential difference between two points is the work per unit charge between them, and a point charge $q$ produces $V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$ at distance $r$.

Definitions:**

$$V=\frac{W_{\infty \to P}}{q_0},V_B-V_A=\frac{W_{A\to B}}{q_0}$$

The unit is the volt, with $1$ V $=1$ J C${}^{-1}$.

Derivation. Bringing $q_0$ from infinity to distance $r$ against the field of $q$:

$$W=-{\int }_{\infty }^r\frac{1}{4\pi {\epsilon }_0}\frac{q{q}_0}{x^2}dx=\frac{1}{4\pi {\epsilon }_0}\frac{q{q}_0}{r}\Rightarrow V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$$

Worked example. The potential $0.09$ m from a charge of $+4\times 10^{-7}$ C:

$$V=\frac{9\times 10^9\times 4\times 10^{-7}}{0.09}=4\times 10^4\text{ V}$$

Bringing $+2\times 10^{-9}$ C from infinity to that point needs $W=qV=2\times 10^{-9}\times 4\times 10^4=8\times 10^{-5}$ J.

An everyday example. A rooftop water tank gives taps below a pressure difference, much as a potential difference pushes charge through a circuit.

The substance. **Potential falls as $\frac{1}{r}$ while the field falls as $\frac{1}{r^2}$**, and potential, being a scalar, needs no direction.

**Potentials are scalars, so the potential of a system of charges is the algebraic sum of each charge's potential, and a dipole produces $V=\frac{1}{4\pi {\epsilon }_0}\frac{p\cos \theta }{r^2}$ at a distant point, where $\theta$ is measured from the dipole axis.

System of charges:**

$$V=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\cdots +\frac{q_n}{r_n}\right)$$

Dipole potential, far from the dipole:

- On the axis, $\theta =0^{\circ }$: $V=\frac{1}{4\pi {\epsilon }_0}\frac{p}{r^2}$
- On the equatorial line, $\theta =90^{\circ }$: $V=0$

Worked example — two charges. Charges of $+3$ nC and $-1$ nC are $0.4$ m apart. At the midpoint:

$$V=9\times 10^9\left(\frac{3\times 10^{-9}}{0.2}-\frac{1\times 10^{-9}}{0.2}\right)=90\text{ V}$$

**Worked example — where is $V$ zero?** Between the charges, $\frac{3}{x}=\frac{1}{0.4-x}$, so $x=0.3$ m from the $+3$ nC charge.

An everyday example. Adding up the heights of the steps you have climbed needs no directions — potentials add as plain numbers in the same way.

The substance. Zero potential does not mean zero field — on a dipole's equatorial line $V=0$, yet $E$ is not zero.

**An equipotential surface has the same potential at every point, so no work is done moving a charge along it; the electric field is always perpendicular to such surfaces and points towards decreasing potential, with $E=-\frac{dV}{dr}$.

Properties:

- No work is done along an equipotential surface
- The field is perpendicular to the surface at every point
- Two equipotential surfaces never intersect
- Closely spaced surfaces mean a strong field

Shapes:

- Point charge — concentric spheres
- Uniform field — parallel planes perpendicular to the field
- Dipole** — curved surfaces around the charges, with $V=0$ on the equatorial plane

Field from potential:

$$E=-\frac{dV}{dr}$$

Worked example. Two parallel plates $2$ cm apart have a potential difference of $300$ V. The uniform field between them is

$$E=\frac{V}{d}=\frac{300}{0.02}=1.5\times 10^4{\text{ V m}}^{-1}$$

An everyday example. Contour lines on a hill-station map join points of equal height, and water runs downhill perpendicular to them — just as the field points perpendicular to equipotentials.

The substance. **V m${}^{-1}$ and N C${}^{-1}$ are the same unit**, both measuring field strength.

**The potential energy of two point charges is $U=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{12}}$, the work needed to assemble them from infinity, and a dipole in a uniform external field has $U=-pE\cos \theta =-\vec{p}\cdot \vec{E}$.

Two charges.** Bringing $q_1$ in needs no work; bringing $q_2$ in against the field of $q_1$ stores

$$U=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{12}}$$

which is positive for like charges and negative for unlike charges.

Worked example. Charges of $+4\mu$C and $-2\mu$C are $0.1$ m apart:

$$U=\frac{9\times 10^9\times (4\times 10^{-6})(-2\times 10^{-6})}{0.1}=-0.72\text{ J}$$

So $0.72$ J of work is needed to pull them completely apart.

Dipole in a uniform field. Rotating the dipole against the torque $pE\sin \theta$ gives

$$U(\theta )=-pE\cos \theta$$

- $\theta =0^{\circ }$: $U=-pE$, the minimum and stable position
- $\theta =180^{\circ }$: $U=+pE$, the maximum and unstable position

For $p=4\times 10^{-9}$ C m in $E=5\times 10^4$ N C${}^{-1}$, turning the dipole from aligned to opposite needs $2pE=4\times 10^{-4}$ J.

An everyday example. A compressed spring stores energy that is released when you let go, just as pushing two like charges together stores electrostatic potential energy.

The substance. For three or more charges, add the energy of every pair once — three charges give three pair terms.