Electrostatic Fields
Two charges feel each other across empty space. Coulomb measured that force; we recast it as a field that fills all of space, then learn to add up the fields of lines, sheets and clouds of charge — the foundation everything electrostatic stands on.
- Coulomb's law — the inverse-square force between two point charges.
- Electric field intensity \(\vec{E}\) — force per unit charge, the field a charge creates.
- The superposition principle for collections of point charges.
- The three continuous charge densities — line \(\rho_L\), surface \(\rho_S\), and volume \(\rho_v\).
- The classic results: the infinite line charge and the infinite sheet of charge.
- How electric field lines picture a field's direction and strength.
Coulomb's Law
Coulomb's law gives the force between two stationary point charges \(Q_1\) and \(Q_2\) separated by distance \(R\). The force is proportional to the product of the charges, inversely proportional to the square of their separation, and directed along the line joining them — repulsive for like charges, attractive for unlike.
Here \(\hat{a}_R\) is the unit vector from the source charge toward the charge feeling the force — built exactly as the distance vector of Chapter 1: \(\hat{a}_R = (\vec{r}_2 - \vec{r}_1)/|\vec{r}_2 - \vec{r}_1|\). The constant \(\varepsilon_0\) is the permittivity of free space; the grouping \(1/4\pi\varepsilon_0 \approx 9\times10^9\ \text{m/F}\) appears so often it is worth memorising.
Electric Field Intensity
Rather than speak of action at a distance, we say a charge fills space with an electric field. The electric field intensity \(\vec{E}\) at a point is the force a small positive test charge would feel there, per unit charge:
Once you know \(\vec{E}\) at a point, the force on any charge \(Q\) placed there is simply \(Q\vec{E}\). The field exists whether or not a test charge is present — it is a property of space set up by the source.
Superposition of Point Charges
Fields add as vectors. The principle of superposition says the total field of several charges is the vector sum of the fields each would produce alone. For \(N\) point charges \(Q_k\) at positions \(\vec{r}_k\), the field at observation point \(\vec{r}\) is:
The cube in the denominator is just \(R^2\) for magnitude times \(R\) for normalising the distance vector \((\vec{r}-\vec{r}_k)\) into a unit vector. Superposition is what lets us replace a sum over discrete charges with an integral over a continuous distribution — the move that powers the rest of this chapter.
The Line Charge
Real charge is rarely a tidy set of points. We describe spread-out charge with a density: charge per unit length \(\rho_L\) (C/m), per unit area \(\rho_S\) (C/m²), or per unit volume \(\rho_v\) (C/m³). Each tiny element \(dQ\) acts like a point charge, and we integrate its field:
For an infinite straight line of uniform density \(\rho_L\), symmetry kills every component but the radial one. The field points straight away from the wire and falls off as \(1/\rho\) — not \(1/\rho^2\), a slower decay because the line is endless:
Surface & Volume Charge
An infinite sheet of uniform surface charge \(\rho_S\) gives a startlingly simple result: the field is uniform, pointing away from the sheet, with a magnitude that does not depend on distance at all. Every field line that leaves the sheet stays parallel, so spreading never weakens it:
A volume charge \(\rho_v\) filling a region is handled the same way — integrate \(dQ = \rho_v\,dv\) over the volume. These integrals can be tedious, which is exactly the motivation for Gauss's law in Chapter 5: when symmetry is high, Gauss's law gives these same answers in one line.
| Distribution | Element \(dQ\) | Field magnitude | Decay |
|---|---|---|---|
| Point charge | \(Q\) | \(\dfrac{Q}{4\pi\varepsilon_0 R^2}\) | \(1/R^2\) |
| Infinite line | \(\rho_L\,dl\) | \(\dfrac{\rho_L}{2\pi\varepsilon_0 \rho}\) | \(1/\rho\) |
| Infinite sheet | \(\rho_S\,dS\) | \(\dfrac{\rho_S}{2\varepsilon_0}\) | uniform |
Electric Field Lines
Field lines (lines of force) make a field visible. They are drawn so that the tangent at any point gives the direction of \(\vec{E}\), and their density — how close together they are — represents the field's strength. They begin on positive charge and end on negative charge, and never cross.
Worked Examples
Problem. Find the force on \(Q_2 = 2\ \text{nC}\) due to \(Q_1 = 5\ \text{nC}\) separated by \(10\ \text{cm}\).
Solution. Apply Coulomb's law with \(1/4\pi\varepsilon_0 = 9\times10^9\):
Problem. Find \(\vec{E}\) at \(2\ \text{m}\) from a \(20\ \text{nC}\) point charge.
Solution. Use \(E = Q/4\pi\varepsilon_0 R^2\):
Problem. Charges \(+Q\) at \((0,0)\) and \(-Q\) at \((d,0)\). Find \(\vec{E}\) at the midpoint.
Solution. Both fields point toward \(-Q\) (in \(+x\)) and add:
Problem. An infinite line has \(\rho_L = 5\ \text{nC/m}\). Find \(\vec{E}\) at \(\rho = 0.5\ \text{m}\).
Solution. Apply \(E = \rho_L/2\pi\varepsilon_0\rho\):
Problem. Two infinite sheets carry \(+\rho_S\) and \(-\rho_S\). Find \(\vec{E}\) between and outside them.
Solution. Between, the two fields add; outside, they cancel:
This is the parallel-plate capacitor field we revisit in Chapter 8.
Problem. What field \(\vec{E}\) holds a \(1\ \text{g}\) bead carrying \(1\ \mu\text{C}\) against gravity?
Solution. Set the electric force equal to weight, \(QE = mg\):
Chapter Summary
\(\vec{F} = \dfrac{Q_1Q_2}{4\pi\varepsilon_0 R^2}\hat{a}_R\); like charges repel, unlike attract, force \(\propto 1/R^2\).
\(\vec{E}=\vec{F}/Q\); for a point charge \(\vec{E}=\dfrac{Q}{4\pi\varepsilon_0 R^2}\hat{a}_R\); force on any charge \(\vec{F}=Q\vec{E}\).
Total field is the vector sum of individual fields — discrete sums become integrals.
\(dQ=\rho_L\,dl=\rho_S\,dS=\rho_v\,dv\); integrate each element's point-charge field.
Line: \(\dfrac{\rho_L}{2\pi\varepsilon_0\rho}\) (\(1/\rho\)). Sheet: \(\dfrac{\rho_S}{2\varepsilon_0}\) (uniform).
Tangent gives direction, density gives strength; start on +, end on −, never cross.
Problems
For each item, identify the charge configuration first, then choose Coulomb's law, superposition, or a standard distribution result. Difficulty rises down the list.
- Find the force between \(+3\ \mu\text{C}\) and \(-3\ \mu\text{C}\) charges \(2\ \text{cm}\) apart.
- Compute \(\vec{E}\) at \((0,0,3)\) due to a \(10\ \text{nC}\) charge at the origin.
- Two \(+1\ \text{nC}\) charges sit at \((\pm1,0,0)\). Find \(\vec{E}\) at \((0,2,0)\).
- An infinite line charge \(\rho_L = 2\ \text{nC/m}\) lies on the \(z\)-axis. Find \(\vec{E}\) at \((3,4,0)\).
- Find \(\vec{E}\) above an infinite sheet with \(\rho_S = 10\ \text{nC/m}^2\).
- A charge \(Q\) is split into two equal parts placed a fixed distance apart. Show the force is maximised when the split is equal.
- Four equal charges \(+Q\) sit at the corners of a square of side \(a\). Find \(\vec{E}\) at the centre.
- A finite line charge of length \(2a\) and density \(\rho_L\) lies on the \(z\)-axis. Find \(\vec{E}\) on the perpendicular bisector at distance \(\rho\).
- A semicircular arc of radius \(a\) carries uniform \(\rho_L\). Find \(\vec{E}\) at the centre.
- Two infinite sheets with \(+\rho_S\) and \(+\rho_S\) face each other. Find \(\vec{E}\) between and outside.
- A ring of radius \(a\) carries total charge \(Q\). Derive \(\vec{E}\) on its axis at height \(z\), and find where it is maximum.
- Explain physically why an infinite line field decays as \(1/\rho\) while a point charge decays as \(1/R^2\).