In Physics 2 charge sat on small objects, and Coulomb's law treated each one as a point. Real charged things are spread out: charge along a wire, over a metal plate, through a block of plastic. To find their fields we will cut them into pieces so small that each piece is a point charge, and then add the pieces with an integral. First we need a way to say how much charge each piece holds.
Key idea
Why it works
| Density | Charge lives… | Units | A small piece | Even spread |
|---|---|---|---|---|
| $\lambda$ (lambda) | along a line: rod, wire, ring | C/m | $dq = \lambda\,dx$ | $\lambda = Q/L$ |
| $\sigma$ (sigma) | on a surface: plate, disk, shell | C/m² | $dq = \sigma\,dA$ | $\sigma = Q/A$ |
| $\rho$ (rho) | through a volume: ball, block | C/m³ | $dq = \rho\,dV$ | $\rho = Q/V$ |
Round objects need pieces shaped to match them. For a disk, use thin rings of radius $r$ and width $dr$: $dA = 2\pi r\,dr$. For a ball, use thin shells: $dV = 4\pi r^2\,dr$. Each ring or shell has one radius, so a density that depends on $r$ is constant across it.
Worked examples
Worked example: A charged shell
- The charge sits on a surface, so we want $\sigma$, in C/m².
- The area of a sphere is $A = 4\pi R^2 = 4\pi(0.050\ \text{m})^2 = 0.0314\ \text{m²}$.
- $\sigma = \dfrac{Q}{A} = \dfrac{6.0 \times 10^{-9}\ \text{C}}{0.0314\ \text{m}^2} = 1.91 \times 10^{-7}\ \text{C/m²}$, or about $191\ \text{nC/m}^2$.
Answer: $\sigma \approx 1.91 \times 10^{-7}\ \text{C/m²}$
Worked example: A rod that gets denser
- The density changes along the rod, so we cannot just multiply. A piece $dx$ at position $x$ carries $dq = bx\,dx$.
- Add the pieces: $$Q = \int_0^{0.50} bx\,dx = \frac{b}{2}x^2\Big|_0^{0.50} = \frac{40}{2}(0.50)^2.$$
- $Q = 5.0\ \text{nC}$. Check the units: $(\text{nC/m}^2)(\text{m}^2) = \text{nC}$.
- Sense check: the density runs from $0$ to $20\ \text{nC/m}$, an average of $10\ \text{nC/m}$ over $0.50\ \text{m}$, which is $5.0\ \text{nC}$.
Answer: $Q = 5.0\ \text{nC}$
Common mistake