Electrostatic Potential and Capacitance — Notes

The electrostatic force is conservative, so the work done in moving a charge between two points is independent of the path.

• Potential difference: $V_B - V_A = W_{A\to B}/q$, the work done by an external agent per unit positive charge (moved slowly, without acceleration).
• Potential at a point: work per unit charge to bring a test charge from infinity ($V_\infty = 0$). Unit volt ($1\ \text{V} = 1\ \text{J C}^{-1}$). A scalar.
• Point charge: $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$.
• System of charges: $V = k\sum q_i/r_i$, an algebraic (signed) sum.
• Dipole ($r \gg a$): $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$; on the axis $kp/r^2$, on the equatorial plane 0.
• Charged shell or conducting sphere (radius $R$): $V = kq/r$ outside; $V = kq/R$ everywhere on and inside.

An equipotential surface has the same potential everywhere on it.

• No work is done in moving a charge along it.
• $\vec E$ is perpendicular to it at every point and points towards lower potential.
• Equipotentials never intersect.
• Shapes: point charge → concentric spheres; uniform field → parallel planes perpendicular to $\vec E$; dipole → closed surfaces round each charge with the $V = 0$ plane midway.

Field from potential: $E = -\dfrac{dV}{dl}$ along the direction of greatest decrease. In a uniform field $E = \Delta V/d$. Closely spaced equipotentials mean a strong field. V m$^{-1}$ = N C$^{-1}$.

Two charges: $U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$ (work to bring them from infinity to separation $r_{12}$). Positive for like charges, negative for unlike (bound).

Three or more charges: sum over every pair once, e.g. $U = k\left(\dfrac{q_1q_2}{r_{12}} + \dfrac{q_1q_3}{r_{13}} + \dfrac{q_2q_3}{r_{23}}\right)$.

Charge in an external potential: $U = qV(\vec r)$. Two charges in an external field: $U = q_1V(\vec r_1) + q_2V(\vec r_2) + \dfrac{kq_1q_2}{r_{12}}$.

Dipole in a uniform field: $U(\theta) = -pE\cos\theta = -\vec p\cdot\vec E$. Minimum $-pE$ at $\theta = 0$ (stable); maximum $+pE$ at $180^\circ$ (unstable). Work to rotate from $\theta_1$ to $\theta_2$: $pE(\cos\theta_1 - \cos\theta_2)$.

Electron volt: $1\ \text{eV} = 1.6 \times 10^{-19}$ J, the energy gained by an electron through 1 V.

In electrostatic equilibrium:

1. $\vec E = 0$ inside the conductor;
2. just outside, $\vec E$ is normal to the surface, $E = \sigma/\varepsilon_0$;
3. there is no net charge inside; excess charge sits on the surface;
4. the potential is the same everywhere in and on the conductor;
5. a charge-free cavity has $\vec E = 0$ whatever the outside fields — electrostatic shielding (why a car or a metal cage protects people in a thunderstorm, and why sensitive instruments are kept in metal cases).

Charge density is larger where the surface is more sharply curved (pointed ends).

A dielectric is an insulator. In an external field:

• non-polar molecules (O₂, H₂) acquire induced dipoles;
• polar molecules (H₂O, HCl) have permanent dipoles that partly align with the field (opposed by thermal motion).

The material becomes polarised: dipole moment per unit volume $\vec P$, proportional to $\vec E$ for most materials. Bound charges appear on the surfaces facing the plates and produce a field opposite to the applied one, so the field inside is reduced: $E = E_0/K$, where $K$ (or $\varepsilon_r$) is the dielectric constant. $\varepsilon = K\varepsilon_0$.

A capacitor: two conductors with charges $+Q$ and $-Q$ separated by an insulator. $C = Q/V$, unit farad (μF, nF, pF in practice).

Parallel plates (area $A$, gap $d$): field between plates $\sigma/\varepsilon_0$ (sheet fields add; outside they cancel), $V = Qd/\varepsilon_0A$, so $C = \dfrac{\varepsilon_0A}{d}$. With a dielectric filling the gap: $C = \dfrac{K\varepsilon_0A}{d}$. Slab of thickness $t$: $C = \dfrac{\varepsilon_0A}{d - t + t/K}$.

Combinations:

Two in series: $C = \dfrac{C_1C_2}{C_1 + C_2}$; the smaller capacitor takes the larger voltage.

$U = \dfrac{Q^2}{2C} = \dfrac{1}{2}CV^2 = \dfrac{1}{2}QV$ (formula used without derivation; it is the area under the V–Q line). The energy resides in the field between the plates.

Inserting a dielectric (constant K):

Sharing charge between two capacitors conserves charge but always loses energy (heat and radiation in the wires). Two identical capacitors, one charged: half the energy is lost.