How Are Coulomb’s Law and Electric Field Intensity Related?

Coulomb’s law and electric field intensity describe closely related parts of electrostatics. Coulomb’s law gives the force between charges, while the electric field describes the force that a source charge produces in the space around it. This distinction lets you find the force on a charge without recalculating the field for every possible test charge.

Coulomb’s law: force between two charges

For two stationary point charges in vacuum, the magnitude of the electrostatic force is

$$F = k\frac{|q_1q_2|}{r^2},$$

where $q_1$ and $q_2$ are the charges, $r$ is the distance between them, and $k=1/(4\pi\varepsilon_0)\approx 8.99\times10^9\ \mathrm{N\,m^2/C^2}$. The force acts along the line joining the charges.

Like charges repel; opposite charges attract. The formula above gives the magnitude only. To include direction, use a vector form. If charge $q_1$ is at position $\mathbf r_1$ and charge $q_2$ is at position $\mathbf r_2$, the force on $q_2$ due to $q_1$ is

$$\mathbf F_{1\to2} = k\frac{q_1q_2}{|\mathbf r_2-\mathbf r_1|^3}(\mathbf r_2-\mathbf r_1).$$

The signs of the charges determine whether the force points toward or away from the source charge. In a material medium, the force depends on the medium’s permittivity; the vacuum expression is often a good approximation for air.

Electric field intensity: force per unit charge

The electric field at a point is defined by placing a small positive test charge $q_0$ there and dividing the force on it by $q_0$:

$$\mathbf E = \frac{\mathbf F}{q_0}.$$

The test charge is imagined to be small enough not to disturb the source charges. Electric field intensity is another name for electric field. Its SI units are newtons per coulomb ($\mathrm{N/C}$), equivalent to volts per metre ($\mathrm{V/m}$).

Combining this definition with Coulomb’s law gives the field due to a point source charge $q$:

$$\mathbf E = k\frac{q}{r^2}\hat{\mathbf r},$$

where $\hat{\mathbf r}$ points from the source charge to the location where the field is being calculated. A positive charge produces a field pointing away from it; a negative charge produces a field pointing toward it. The field is defined independently of the test charge. Once you know the field, the force on a charge $q_0$ placed there is simply $\mathbf F=q_0\mathbf E$.

Fields from several charges

Electric fields obey the superposition principle. Find the field from each source charge at the point of interest, then add the results as vectors:

$$\mathbf E_{\text{net}}=\sum_i \mathbf E_i.$$

Do not add field magnitudes unless all the fields point in the same direction. A negative source charge changes the direction of its field, not the rule for adding contributions.

Worked example: two source charges

A charge of $+2.0\ \mu\mathrm C$ is at $x=0$, and a charge of $-1.0\ \mu\mathrm C$ is at $x=0.30\ \mathrm m$. Find the field at $x=0.10\ \mathrm m$, then find the force on a $+0.50\ \mu\mathrm C$ charge placed there. Assume vacuum or air.

The point is $0.10\ \mathrm m$ from the positive charge. Its field points to the right, so

$$E_1 = (8.99\times10^9)\frac{2.0\times10^{-6}}{(0.10)^2} =1.798\times10^6\ \mathrm{N/C}.$$

The point is $0.20\ \mathrm m$ from the negative charge. Its field points toward that charge, also to the right:

$$E_2 = (8.99\times10^9)\frac{1.0\times10^{-6}}{(0.20)^2} =2.2475\times10^5\ \mathrm{N/C}.$$

Because both fields point right, their magnitudes add:

$$E_{\text{net}}=E_1+E_2=2.02275\times10^6\ \mathrm{N/C}\approx2.0\times10^6\ \mathrm{N/C}$$

to the right, rounded to two significant figures to match the input data. The force on the positive test charge is therefore

$$F=q_0E_{\text{net}}=(0.50\times10^{-6})(2.02275\times10^6)\approx1.0\ \mathrm N,$$

also to the right, rounded to two significant figures. A negative test charge at the same point would experience a force in the opposite direction, even though the electric field there would be unchanged.

Common mistakes

• Confusing force and field: Force depends on the charge placed in the field; the field is a property of the source charges and the location.
• Ignoring direction: Coulomb’s law’s magnitude formula does not tell you the direction by itself. Decide whether each interaction attracts or repels, or use vectors.
• Adding magnitudes automatically: Fields in opposite directions subtract; fields in the same direction add.
• Forgetting units or prefixes: Convert microcoulombs to coulombs ($1\ \mu\mathrm C=10^{-6}\ \mathrm C$) and use metres for distances.
• Applying the point-charge formula at the source: The ideal point-charge field is undefined at $r=0$; calculate it only at a distinct location.

Coulomb’s law is the pairwise force rule. Electric field intensity packages the effect of the source charges at each point, making it easier to calculate the force on any charge placed there.