AcademyElectric Charge and Fields
Academy
Coulomb Forces
Level 1 - Physics topic page in Electric Charge and Fields.
Principle
Coulomb's law gives the electric force between point charges as an inverse-square central interaction.
Notation
\(q_1,q_2\)
interacting point charges
\(\mathrm{C}\)
\(\vec r_1,\vec r_2\)
charge position vectors
\(\mathrm{m}\)
\(\vec r_{12}\)
vector from charge 1 to charge 2
\(\mathrm{m}\)
\(\hat r_{12}\)
unit vector from charge 1 to charge 2
1
\(k\)
Coulomb constant
\(\mathrm{N\,m^{2}\,C^{-2}}\)
\(\epsilon_0\)
permittivity of free space
\(\mathrm{C^{2}\,N^{-1}\,m^{-2}}\)
\(\vec F_{2\leftarrow1}\)
force on charge 2 by charge 1
\(\mathrm{N}\)
Method
Derivation 1: Build the vector separation
The force direction is set by the line joining the two charges. Define the displacement from the source charge to the charge being acted on.
Separation vector
\[\vec r_{12}=\vec r_2-\vec r_1\]
Distance
\[r=|\vec r_{12}|\]
Direction
\[\hat r_{12}=\frac{\vec r_{12}}{r}\]
Derivation 2: Attach magnitude and sign
The magnitude weakens as \(1/r^2\). The product \(q_1q_2\) supplies the sign: like charges repel and opposite charges attract.
Magnitude scale
\[F\propto\frac{|q_1q_2|}{r^2}\]
Coulomb constant
\[k=\frac{1}{4\pi\epsilon_0}\]
Vector force
\[\vec F_{2\leftarrow1}=k\frac{q_1q_2}{r^2}\hat r_{12}\]
A negative charge product reverses the direction of \(\hat r_{12}\).
Derivation 3: Add forces by superposition
Electric forces from multiple source charges are vector forces on the same object, so they add component by component.
Forces on charge 0
\[\vec F_{\mathrm{net}}=\vec F_{0\leftarrow1}+\vec F_{0\leftarrow2}+\cdots\]
Component form
\[\vec F_{\mathrm{net}}=(\sum F_x)\hat\imath+(\sum F_y)\hat\jmath\]
Rules
These are the compact force results from the construction above.
Coulomb constant
\[k=\frac{1}{4\pi\epsilon_0}=8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}}\]
Force magnitude
\[F=k\frac{|q_1q_2|}{r^2}\]
Vector force
\[\vec F_{2\leftarrow1}=k\frac{q_1q_2}{r^2}\hat r_{12}\]
Superposition
\[\vec F_{\mathrm{net}}=\sum_i \vec F_i\]
Examples
Question
Two charges
\[+3.0\,\mathrm{nC}\]
and \[-5.0\,\mathrm{nC}\]
are \[0.20\,\mathrm{m}\]
apart. Find the force magnitude.Answer
\[F=k\frac{|q_1q_2|}{r^2}=\frac{(8.99\times10^9)(3.0\times10^{-9})(5.0\times10^{-9})}{0.20^2}=3.4\times10^{-6}\,\mathrm{N}\]
The force is attractive because the charges have opposite signs.Checks
- Use distance squared, not distance.
- Direction comes from the charge signs and the line joining the charges.
- Forces from several charges are vectors; add components.
- The two charges exert equal-magnitude opposite-direction forces on each other.