Electric Field

A charge does not need a partner to influence space — it sets up an electric field $\vec E$ that exists everywhere around it. The field is the messenger: any test charge placed in the field feels a force.

Concept

The electric field at a point is defined as the force per unit positive test charge: $\vec E = \lim_{q_0\to 0} \frac{\vec F}{q_0}.$ The limit is needed so the test charge does not disturb the original distribution. Units: $N/C$ or equivalently $V/m$.

Field of a point charge. A charge $q$ at the origin produces a field at position $\vec r$ given by $\vec E(\vec r) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat r.$

• For $q > 0$, field lines point radially outward.
• For $q < 0$, field lines point radially inward.

The field is a vector, so for a collection of charges use superposition: $\vec E_\text{net} = \sum_i \vec E_i = \sum_i \frac{kq_i}{r_i^2}\hat r_i.$

Derivation

Start from Coulomb's law. The force on a test charge $q_0$ at $\vec r$ due to source $q$ at origin is $\vec F = \frac{kqq_0}{r^2}\hat r.$ Divide by $q_0$: $\vec E = \frac{\vec F}{q_0} = \frac{kq}{r^2}\hat r.$ This is independent of the test charge — the field is a property of the source distribution, not of what is placed in it.

Symmetry check. A point charge has spherical symmetry, so the field must point radially. Its magnitude depends only on $r$, not on angles.

Worked Example

A charge of $+5\,nC$ sits at the origin. What is the electric field at the point $(0.3\,m, 0, 0)$?

$E = \frac{(9\times10^9)(5\times10^{-9})}{(0.3)^2} = \frac{45}{0.09} = 500\,N/C.$ Direction: along $+\hat x$ (away from the positive source).

Now add a second charge $-5\,nC$ at $(0.6\,m, 0, 0)$. At the midpoint $(0.3\,m, 0, 0)$, both fields point in the $+\hat x$ direction (away from $+q$ and towards $-q$). Each has magnitude $E_0 = k|q|/(0.15)^2 = 2000\,N/C$, so $E_\text{net} = 4000\,N/C$ along $+\hat x$.

Common Confusions

• The electric field is not a force. It is force per unit charge, with units $N/C$.
• Field exists even where no test charge is. Placing a charge merely "samples" the pre-existing field.
• Field of a point charge falls as $1/r^2$, but field of a dipole falls as $1/r^3$. Don't confuse the two.
• Field at the location of the source is undefined (it blows up). This is a known idealization issue with point particles.

Key Takeaways

• $\vec E = \vec F / q_0$, with $q_0 \to 0$ in principle.
• For a point source $E = kq/r^2$, directed away from $+q$ or towards $-q$.
• Field obeys vector superposition.
• Units: $N/C = V/m$.