How does one charge push another across empty space? In Fields: Forces at a Distance you met the idea of a field. Here we make it precise for electric charges, starting from Coulomb's law.
Why it works
A charge $Q$ changes the space around it, and any other charge placed there feels a force. To describe that change without naming the second charge, place a small test charge $q$ at a point and divide the force on it by $q$. Since the force is proportional to $q$, the ratio is the same for any test charge: it belongs to the point, not to the test charge. For a point charge, Coulomb's law gives $F = k|Qq|/r^2$, so $E = F/q = k|Q|/r^2$.
Key idea
$\vec E = \vec F/q$, the force per coulomb, in $\text{N/C}$. A point charge makes $E = k\dfrac{|Q|}{r^2}$, pointing away from a positive charge and toward a negative one. A charge $q$ in a field feels $\vec F = q\vec E$: along $\vec E$ if $q > 0$, opposite if $q < 0$.
Worked examples
Worked example: The field of a point charge
Find the electric field $0.20\,\text{m}$ from a $+4.0\,\text{nC}$ charge.
- $E = k\dfrac{|Q|}{r^2} = (8.99 \times 10^{9})\dfrac{4.0 \times 10^{-9}}{(0.20)^2} = 899\,\text{N/C}$.
- The charge is positive, so the field points directly away from it.
Answer: $899\,\text{N/C}$, away from the charge
Worked example: An electron in a field
An electron sits where the field is $500\,\text{N/C}$ pointing east. Find the force on it.
- $F = |q|E = (1.60 \times 10^{-19}\,\text{C})(500\,\text{N/C}) = 8.0 \times 10^{-17}\,\text{N}$.
- The electron is negative, so the force points opposite the field: west.
Answer: $8.0 \times 10^{-17}\,\text{N}$, west
Worked example: Changing the test charge
A $2.0\,\text{nC}$ test charge at point P feels $6.0 \times 10^{-6}\,\text{N}$. What force would a $4.0\,\text{nC}$ test charge feel there, and what is the field?
- $E = F/q = \dfrac{6.0 \times 10^{-6}}{2.0 \times 10^{-9}} = 3000\,\text{N/C}$.
- The field stays $3000\,\text{N/C}$; the larger test charge feels $F = qE = (4.0 \times 10^{-9})(3000) = 1.2 \times 10^{-5}\,\text{N}$, twice the force.
Answer: $1.2 \times 10^{-5}\,\text{N}$; $E = 3000\,\text{N/C}$ either way
Common mistake
Thinking the field at a point depends on the test charge placed there. Doubling the test charge doubles the force, but the field, force per coulomb, stays the same. The field is made by the other charges and is there even when nothing is placed at the point. Check: $Q$ in $k|Q|/r^2$ is the charge making the field, never the test charge.