Electrostatics

The phone that feels a finger

• A touch screen knows where your finger is because the finger changes the electric field near the glass. Every capacitor, sensor and insulator is an electrostatics problem.
• For symmetric charges, one law turns the field into arithmetic. This lesson applies it to a charged sphere, inside and out.

Gauss's law

• Gauss's law 高斯定律 says the electric flux out of any closed surface equals the enclosed charge over $\varepsilon_{0}$: $\displaystyle\oint \mathbf{E} \cdot d\mathbf{A} = \dfrac{Q_{enc}}{\varepsilon_{0}}$.
• On a surface where the field is constant and normal, the integral is just $E \times A$. For a sphere of radius $r$: $E \cdot 4\pi r^{2} = \dfrac{Q_{enc}}{\varepsilon_{0}}$.

Inside a uniformly charged sphere

• A sphere of radius $a$ and charge density $\rho$. Inside, at radius $r$, the enclosed charge grows as $r^{3}$ while the area grows only as $r^{2}$.
• So $E \cdot 4\pi r^{2} = \dfrac{\rho \cdot \tfrac{4}{3}\pi r^{3}}{\varepsilon_{0}}$, giving $E = \dfrac{\rho r}{3\varepsilon_{0}}$: zero at the centre, rising linearly to the surface.

Outside: like a point charge

• Outside, all the charge is enclosed: $E = \dfrac{\rho a^{3}}{3\varepsilon_{0}r^{2}}$, the field of a point charge at the centre.
• At $r = a$ both formulas give $\dfrac{\rho a}{3\varepsilon_{0}}$, so the field is continuous at the surface.

Worked example: the numbers

• $\rho = 1\,\mu\text{C/m}^{3}$, $a = 2\,\text{cm}$: at the surface, $E = \dfrac{10^{-6} \times 0.02}{3 \times 8.854 \times 10^{-12}} = 753\,\dfrac{\text{V}}{\text{m}}$.
• At $r = 4\,\text{cm}$, twice the radius, the field is a quarter: 188 V/m. At $r = 1\,\text{cm}$, inside, it is half: 377 V/m.

Potential and capacitance

• The potential 电位 adds up the field from infinity: $V(r) = -\displaystyle\int_{\infty}^{r} \mathbf{E} \cdot d\mathbf{l}$. Capacitance 电容 is charge per volt, $C = \dfrac{Q}{V}$.
• An isolated sphere has $C = 4\pi\varepsilon_{0}a$, about 1.1 pF per centimetre of radius. That is why objects in open space barely couple.

A dielectric weakens the field

• Fill the space with a dielectric of relative permittivity $\varepsilon_{r}$ and the field drops by $\varepsilon_{r}$: $\varepsilon_{0}$ becomes $\varepsilon = \varepsilon_{0}\varepsilon_{r}$ in the law. With $\varepsilon_{r} = 4$ the surface value falls to 188 V/m.
• At a boundary between materials, the boundary conditions 边界条件 hold: tangential $E$ is continuous, and normal $D = \varepsilon E$ jumps only by free surface charge.

Three common slips

Recap