Electric Field

So far, the story of electrostatics has been about forces: one charge pushes or pulls another. But there is a deeper question lurking in the background. Suppose you place a charge $Q$ alone in empty space and then step back. Is the space around $Q$ truly empty, or has $Q$ done something to it? If you later bring a second charge $q$ to some point nearby and it immediately feels a force, how did that force get there?

Early physicists tackled this puzzle by introducing a powerful new idea: the electric field. Instead of thinking about force as a direct charge-to-charge interaction, they said that $Q$ fills the surrounding space with a field, and it is this field that grabs hold of any charge that enters it.

What the Electric Field Means Physically

Picture a campfire. Even when nobody stands near it, the fire heats the air around it. Walk into that zone and you feel warmth, not because the fire is reaching out and touching you, but because the air already carries the thermal energy. The electric field works in a similar way.

A charge $Q$ placed at some point in space modifies the space at every surrounding location. When another charge $q$ arrives at a point P, it does not need to “know” where $Q$ is or how far away it sits. It simply responds to the field that already exists at P. The field acts as an invisible intermediary (a go-between) that carries the influence of the source charge throughout all of space.

In physics, the word field describes any quantity, whether a number (scalar) or an arrow (vector), that has a well-defined value at every point in a region. Temperature throughout a room is a scalar field. Wind velocity throughout the atmosphere is a vector field. The electric field is a vector field: at every point in space, it has both a magnitude (how strong) and a direction (which way).

The Formula for the Electric Field of a Point Charge

Place a point charge $Q$ at the origin O in vacuum. The electric field it produces at a point located at position vector $\mathbf{r}$ (at distance $r$ from O) is:

$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}} \qquad \text{(1.6)}$

Here:

• $r$ is the distance from $Q$ to the point where you are evaluating the field
• $\hat{\mathbf{r}} = \mathbf{r}/r$ is the unit vector pointing from $Q$ (at the origin) toward the field point
• $\varepsilon_0$ is the permittivity of free space
• $\frac{1}{4\pi\varepsilon_0} \approx 9 \times 10^9\,\text{N m}^2\text{C}^{-2}$

Notice that this expression involves only $Q$ and the position $\mathbf{r}$. There is no second charge in the formula. The field is a property of $Q$ and the geometry of the surrounding space, existing at every point whether or not another charge is present.

Connecting Force and Field

Now suppose you place a charge $q$ at the point $\mathbf{r}$ where this field exists. What force does $q$ feel?

Multiply the field by $q$:

$\mathbf{F}(\mathbf{r}) = q\,\mathbf{E}(\mathbf{r}) \qquad \text{(1.8)}$

Substituting Equation (1.6) into (1.8):

$\mathbf{F} = q \times \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}} = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2} \hat{\mathbf{r}} \qquad \text{(1.7)}$

This is exactly Coulomb’s law. So the field picture and the direct force picture give the same answer. The electric field is simply a different (and often more convenient) way to package the same physics.

Equation (1.8) also tells us the SI unit of electric field. Since $\mathbf{E} = \mathbf{F}/q$, the unit is force divided by charge: newton per coulomb (N/C). An equivalent unit, volt per metre (V/m), will appear when you study electric potential in the next chapter.

Note that the charge $q$ also pushes back on $Q$ with an equal and opposite force, exactly as Newton’s third law demands. You can view this interaction either as $q$ feeling the field of $Q$, or as $Q$ feeling the field of $q$. Both viewpoints are equally valid.

Source Charge, Test Charge, and the Formal Definition

Two labels make the discussion clearer:

• Source charge ($Q$): the charge that creates the electric field you want to study
• Test charge ($q$): a small charge brought in to detect or measure the field

If $q$ has the value of one coulomb, then Equation (1.8) tells you that the electric field equals the force numerically. So you can think of the electric field at a point as the force that a unit positive charge would experience if placed there.

The Problem with a Real Test Charge

There is a practical catch. A real test charge $q$ is not passive: it exerts its own force on $Q$ and could push $Q$ out of position. If $Q$ moves, the field you are trying to measure changes, and your measurement becomes unreliable.

The solution is to imagine shrinking $q$ toward zero. As $q$ gets smaller, the force $\mathbf{F}$ it exerts on $Q$ also shrinks toward zero, leaving $Q$ undisturbed. The force on $q$ itself also drops toward zero, but the ratio $\mathbf{F}/q$ stays perfectly finite. That ratio is the electric field:

$\mathbf{E} = \lim_{q \to 0} \frac{\mathbf{F}}{q} \qquad \text{(1.9)}$

This limit is the formal, rigorous definition of the electric field. It ensures that the act of measuring the field does not alter what you are measuring.

How This Works in Real Life

You might wonder: if the test charge must be vanishingly small, how does anyone actually measure a field? In practice, the source charge is often held in place by other forces that are not part of the measurement. For example, when you bring a test charge near a charged metal plate, the charges on the plate are pinned to their positions by the interatomic forces within the metal. They do not budge when your tiny test charge approaches, so the field remains exactly as it was.

The Electric Field is Independent of the Test Charge

A key property deserves emphasis. Even though the field is defined through a test charge $q$, the result does not depend on $q$ at all. Here is why: Coulomb’s law says the force $\mathbf{F}$ on $q$ is proportional to $q$. When you divide $\mathbf{F}$ by $q$, the $q$ cancels out:

$\mathbf{E} = \frac{\mathbf{F}}{q} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}}$

No trace of $q$ remains. The electric field is entirely determined by the source charge $Q$ and the distance $r$. Whether you use a test charge of $1\,\mu\text{C}$ or $1\,\text{nC}$ makes no difference to the field value you obtain.

The field does, however, depend on position. Move to a different point in space (a different $\mathbf{r}$) and the field changes in both magnitude and direction. At every point in three-dimensional space, the electric field has a definite value.

Direction of the Field: Positive vs. Negative Source Charges

Fig 1.8: Electric field lines (a) due to a positive charge $Q$, radiating outward; (b) due to a negative charge $-Q$, pointing radially inward

The direction of $\mathbf{E}$ tells you which way a positive test charge would be pushed:

• Positive source charge ($Q > 0$): A positive test charge would be repelled, pushed directly away from $Q$. So the electric field points radially outward from $Q$ at every surrounding point.

• Negative source charge ($Q < 0$): A positive test charge would be attracted, pulled directly toward $Q$. So the electric field points radially inward toward $Q$ at every surrounding point.

In both cases the field is purely radial: every field vector lies along the straight line connecting $Q$ to the point in question.

Spherical Symmetry of the Field

Look at Equation (1.6) again. The magnitude of the field is:

$|\mathbf{E}| = \frac{1}{4\pi\varepsilon_0} \frac{|Q|}{r^2}$

This depends only on the distance $r$ from $Q$, not on which direction you go. Move to any point that is the same distance $r$ from $Q$ (any point on a sphere of radius $r$ centred on $Q$) and you find the same field strength.

The direction of the field, however, does change from point to point on the sphere: it is always radially outward (or inward), which means it points differently at the top, bottom, left, and right of the sphere.

So the electric field of a point charge has spherical symmetry: the magnitude is constant over any sphere centred on the charge, while the direction is everywhere radial. This symmetry will become very important when you study Gauss’s law later in this chapter.