Electrostatics • Topic 1 of 3

Electric Charge, Coulomb's Law & Field

Electric charge is the intrinsic property of matter that makes it experience a force in an electric field. There are two kinds — positive (deficiency of electrons) and negative (excess of electrons) — and like charges repel while unlike charges attract. The SI unit of charge is the coulomb (C).

Three basic properties govern charge. (i) Quantisation: charge always comes in integer multiples of the elementary charge, $q=ne$ where $e=1.6\times10^{-19}\ \text{C}$ and $n$ is an integer. (ii) Conservation: the net charge of an isolated system never changes — charge can be transferred but not created or destroyed. (iii) Additivity: the total charge of a body is the algebraic sum of all the charges on it.

Coulomb's law gives the force between two stationary point charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum:

$$F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$$ where $\frac{1}{4\pi\epsilon_0}=9\times10^{9}\ \text{N}\,\text{m}^2\,\text{C}^{-2}$ and $\epsilon_0=8.85\times10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ is the permittivity of free space. The force is directed along the line joining the charges. In a medium of relative permittivity (dielectric constant) $K$, the force reduces to $F=\frac{1}{4\pi\epsilon_0 K}\frac{q_1q_2}{r^2}$.

When several charges act on a charge, the net force is the vector sum of the individual forces — this is the principle of superposition: $\vec{F}=\vec{F}_1+\vec{F}_2+\dots$

The electric field $\vec{E}$ at a point is the force per unit positive test charge placed there: $\vec{E}=\frac{\vec{F}}{q}$, with SI unit $\text{N/C}$ (or $\text{V/m}$). The field of a point charge $q$ at distance $r$ is $E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$, pointing radially outward for positive $q$.

  • Field lines start on positive charges and end on negative charges; they never cross, and their density indicates field strength.
  • An electric dipole is two equal and opposite charges $\pm q$ separated by $2a$, with dipole moment $\vec{p}=q(2a)$ directed from $-q$ to $+q$.
  • On the axial line the field is $E_{axial}=\frac{1}{4\pi\epsilon_0}\frac{2p}{r^3}$; on the equatorial line $E_{eq}=\frac{1}{4\pi\epsilon_0}\frac{p}{r^3}$ (for $r\gg a$).

In a uniform field a dipole feels no net force but a torque $\tau=pE\sin\theta$ (i.e. $\vec{\tau}=\vec{p}\times\vec{E}$) that aligns it with the field. Finally, electric flux measures the number of field lines crossing a surface: $\phi=\vec{E}\cdot\vec{A}=EA\cos\theta$, a scalar with unit $\text{N}\,\text{m}^2\,\text{C}^{-1}$.

Field lines of an electric dipole (+q and -q)+q-q2ap (dipole moment, +q to -q reversed)
1
Worked Example
Two point charges $q_1=+3\ \mu\text{C}$ and $q_2=+5\ \mu\text{C}$ are 20 cm apart in vacuum. Find the force between them.
Solution
  1. Use $F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$ with $\frac{1}{4\pi\epsilon_0}=9\times10^{9}$.
  2. Substitute: $F=9\times10^{9}\times\frac{(3\times10^{-6})(5\times10^{-6})}{(0.2)^2}$.
  3. Numerator $=9\times10^{9}\times15\times10^{-12}=0.135$; divide by $0.04$.
  4. $F=\frac{0.135}{0.04}=3.375\ \text{N}$ (repulsive, both positive).

Answer: $F\approx3.38\ \text{N}$, repulsive.

2
Worked Example
How many electrons make up a charge of $-1\ \mu\text{C}$?
Solution
  1. By quantisation $q=ne$, so $n=\frac{q}{e}$.
  2. $n=\frac{1\times10^{-6}}{1.6\times10^{-19}}$.
  3. $n=6.25\times10^{12}$ electrons.

Answer: $n=6.25\times10^{12}$ electrons.

3
Worked Example
A charge $q=2\ \mu\text{C}$ placed at a point experiences a force of $0.4\ \text{N}$. Find the electric field there.
Solution
  1. Use $E=\frac{F}{q}$.
  2. $E=\frac{0.4}{2\times10^{-6}}$.
  3. $E=2\times10^{5}\ \text{N/C}$.

Answer: $E=2\times10^{5}\ \text{N/C}$.

4
Worked Example
Calculate the electric field at 30 cm from a point charge of $+5\ \mu\text{C}$ in vacuum.
Solution
  1. Use $E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$.
  2. $E=9\times10^{9}\times\frac{5\times10^{-6}}{(0.3)^2}$.
  3. $=9\times10^{9}\times\frac{5\times10^{-6}}{0.09}$.
  4. $E=5\times10^{5}\ \text{N/C}$, directed away from the charge.

Answer: $E=5\times10^{5}\ \text{N/C}$ (radially outward).

5
Worked Example
An electric dipole of moment $p=4\times10^{-9}\ \text{C·m}$ is placed in a uniform field $E=5\times10^{4}\ \text{N/C}$ at $30^\circ$ to the field. Find the torque on it.
Solution
  1. Use $\tau=pE\sin\theta$.
  2. $\tau=(4\times10^{-9})(5\times10^{4})\sin30^\circ$.
  3. $=(4\times10^{-9})(5\times10^{4})(0.5)$.
  4. $\tau=1\times10^{-4}\ \text{N·m}$.

Answer: $\tau=1\times10^{-4}\ \text{N·m}$.

6
Worked Example
A uniform field $E=2\times10^{3}\ \text{N/C}$ passes through a square of side 10 cm whose plane is tilted so its normal makes $60^\circ$ with the field. Find the electric flux.
Solution
  1. Use $\phi=EA\cos\theta$.
  2. Area $A=(0.1)^2=0.01\ \text{m}^2$.
  3. $\phi=(2\times10^{3})(0.01)\cos60^\circ=(20)(0.5)$.
  4. $\phi=10\ \text{N}\,\text{m}^2\,\text{C}^{-1}$.

Answer: $\phi=10\ \text{N}\,\text{m}^2\,\text{C}^{-1}$.

Key Points

  • Charge is quantised ($q=ne$, $e=1.6\times10^{-19}\ \text{C}$), conserved, and additive.
  • Coulomb's law: $F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$ with $\frac{1}{4\pi\epsilon_0}=9\times10^{9}\ \text{N}\,\text{m}^2\,\text{C}^{-2}$; force is along the line joining the charges.
  • Electric field: $\vec{E}=\frac{\vec{F}}{q}$; for a point charge $E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$. Forces and fields obey superposition.
  • Dipole moment $\vec{p}=q(2a)$ points from $-q$ to $+q$; axial field $\frac{2p}{4\pi\epsilon_0 r^3}$, equatorial field $\frac{p}{4\pi\epsilon_0 r^3}$.
  • In a uniform field a dipole feels torque $\tau=pE\sin\theta$ (no net force); electric flux $\phi=EA\cos\theta$.
Tap an option to check your answer0 / 4
Q1.The minimum possible charge that can exist freely is:
Explanation: Charge is quantised; the smallest free charge is the elementary charge $e=1.6\times10^{-19}\ \text{C}$.
Q2.If the distance between two point charges is doubled, the Coulomb force becomes:
Explanation: $F\propto\frac{1}{r^2}$, so doubling $r$ makes the force $\frac{1}{4}$ of the original.
Q3.The electric field due to a point charge varies with distance as:
Explanation: $E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$, an inverse-square dependence.
Q4.A dipole in a uniform electric field experiences:
Explanation: The equal and opposite forces cancel (no net force) but form a couple giving torque $\tau=pE\sin\theta$.