Electric Charge and Electric Fields
Explain charge, conductors, insulators, Coulomb force, electric fields, electric potential, and field-line representations.
A lesson of Physics, in Electricity and Magnetism, about 10 minutes.
By the end of this lesson, you will be able to apply Coulomb's law, describe electric field lines, and calculate the electric field and force on charges.
Electric Charge
Charge is a fundamental property of matter. Like charges repel; unlike charges attract.
- Proton: $+e = +1.6 \times 10^{-19}$ C
- Electron: $-e = -1.6 \times 10^{-19}$ C
- Neutron: 0
Conservation of charge: net charge in a closed system is constant.
Quantization: charge comes in multiples of $e$.
Coulomb's Law
$$F = k\frac{|q_1||q_2|}{r^2}$$
$k = 8.99 \times 10^9$ N·m²/C² (Coulomb's constant = $1/4\pi\varepsilon_0$)
Direction: along the line connecting the charges. Repulsive for same sign; attractive for opposite sign.
The Electric Field
Electric field $\vec{E}$: force per unit positive test charge.
$$\vec{E} = \frac{\vec{F}}{q_0} = k\frac{Q}{r^2}\hat{r}$$
Units: N/C or V/m.
Direction: away from positive charges; toward negative charges.
Force on a charge in a field: $\vec{F} = q\vec{E}$
Electric Field Lines
Rules:
- Lines start on positive charges; end on negative charges
- Never cross
- Perpendicular to conductor surfaces
- Density of lines ∝ field strength
Example 1 — Coulomb's Law
Two charges $q_1 = 5\mu$C and $q_2 = -3\mu$C separated by 0.20 m. Find the force.
$F = 9\times 10^9 \times \frac{5\times 10^{-6} \times 3\times 10^{-6}}{(0.20)^2} = \frac{0.135}{0.04} = 3.375$ N (attractive)
Example 2 — Electric Field
Find $\vec{E}$ at 0.50 m from a $+4\mu$C charge.
$E = 9\times 10^9 \times \frac{4\times 10^{-6}}{0.25} = 144,000$ N/C, directed away from the charge.
Summary
Coulomb's Law: $F = kq_1q_2/r^2$; inverse-square; attractive for opposite charges.
Electric field: $E = kQ/r^2$; force per unit charge; direction away from $+$, toward $-$.
Field lines: start on $+$, end on $-$; density indicates strength.