In short
The electric field at a point is the force that a small positive test charge q_0 would experience there, per unit charge:
The field is a property of space itself around source charges — it exists whether or not a test charge is there to feel it. For a point charge Q at the origin, the field at position \vec{r} is
pointing radially outward for Q>0 and inward for Q<0. Fields from multiple sources add as vectors (superposition): \vec{E}_\text{total} = \sum_i \vec{E}_i. The field lines of a charge configuration are directed curves tangent everywhere to \vec{E}; their density codes the field's magnitude. For a dipole of moment \vec{p} = q\vec{d}, the field on the axis at distance r \gg d is \vec{E}_\text{axial} = \tfrac{1}{4\pi\varepsilon_0}\,\tfrac{2\vec{p}}{r^3}, and on the equator at the same distance, \vec{E}_\text{eq} = -\tfrac{1}{4\pi\varepsilon_0}\,\tfrac{\vec{p}}{r^3} — twice as strong along the axis as in the plane perpendicular to it, and flipped in sign.
A thundercloud drifts over Powai at three kilometres altitude on a July afternoon. If you could stand underneath it with a pith ball on a fine thread and nothing else, the pith ball would swing off the vertical. If you had a sensitive electrometer you would measure, between the cloud base and the ground, an electric field of roughly 10 kV per metre — meaning every positive test charge of one coulomb you placed below the cloud would feel 10 000 newtons pulling it downward toward the positive ground. The charge is sitting three kilometres away, inside the cloud, and yet somehow it reaches down through three kilometres of damp Mumbai air and grabs hold of any charge in its path.
This is the central conceptual move of electrostatics. Instead of saying "the cloud exerts a force on the pith ball across three kilometres of space," you say "the cloud sets up an electric field everywhere in space, and whatever charge happens to be in the field feels a force." The cloud does its job once (sets up the field), and then the field does the rest. Remove the pith ball and the field is still there. Put a different charge in its place and the same field tells you the new force. The field is the messenger.
This shift from "action at a distance" to "fields everywhere in space" is not cosmetic. It is what lets you handle the cricket ball of charge that you will meet later — a continuous distribution, a ring, a cylinder, a plane, a cloud — without tracking forces between every pair of charges separately. You compute the field the source makes, and the field carries all the information. This article builds the electric field from scratch: starting with the test-charge definition, deriving the field of a point charge, stacking fields by superposition, working out the dipole (two opposite point charges), and finally drawing the picture with field lines. Three worked examples (a charge in an inkjet printer, a proton above a charged plate, a water molecule in an external field) anchor the formula. The going-deeper section handles continuous distributions, the limits of the point-charge idealisation, and how the field concept survives when relativity enters.
Defining the electric field — force per unit test charge
Suppose you have a fixed arrangement of charges — a source distribution, for short. Call them Q_1, Q_2, \ldots at positions \vec{r}_1, \vec{r}_2, \ldots. You want to know what force a new charge q_0 would feel if you placed it at some point \vec{r}.
By Coulomb's law and superposition, the total electrostatic force on q_0 is
Why: each source charge Q_i pulls or pushes q_0 along the line joining them. Coulomb's law gives the magnitude, the unit vector \hat{r}_{i\to 0} points from source i to the test charge, and the product Q_i q_0 decides whether the force is repulsive (+) or attractive (-). Superposition says all these forces simply add as vectors — a deep, experimentally verified property of the electric interaction.
Every term in the sum carries a factor of q_0. Pull that factor out:
Why: the force splits cleanly into two pieces — the test charge q_0, and a vector that depends only on where the source charges are and how big they are. That second piece has a physical meaning of its own.
Define the electric field at the point \vec{r} as the part of the expression that does not depend on the test charge:
Why: dividing the force by the test charge gives you a vector per unit charge — a quantity that belongs to the point in space rather than to the test charge. The q_0 \to 0 limit is a safety clause: if the test charge were large, it would itself disturb the positions of the source charges (by pushing or pulling them), and the field you would measure would not be the field of the original source distribution. Taking q_0 infinitesimally small avoids that self-interference.
With this definition, the force on any test charge becomes simply
where \vec{E} is the field of the surrounding sources evaluated at the test charge's location, and q is the charge (positive or negative) of the test particle. This is the everyday working form of the field concept: the field is the force per unit positive charge, and once you have the field at a point, any charge placed there feels q\vec{E}.
Units. Force divided by charge has units of newtons per coulomb (N/C). You will also see volts per metre (V/m), which is the same unit (1 V/m = 1 N/C) — the two names reflect the two uses of the field, forces in mechanics and potential differences in circuits. A dry-day atmospheric field near the Earth's surface is about 100 V/m. Inside a television picture tube the field is around 10^5 V/m. The field needed to ionise dry air — the dielectric strength — is about 3 \times 10^6 V/m, which is why lightning flashes need thunderclouds that build up enormous charge imbalances.
Sign convention. The field points in the direction a positive test charge would be pushed. A negative test charge placed at the same point feels a force in the opposite direction (because \vec{F} = q\vec{E} and q < 0). The field itself — its direction and magnitude — does not depend on what charge you use to probe it.
Field of a point charge
Take one point charge Q at the origin. What is the field at position \vec{r}?
Coulomb's law says the force on a test charge q_0 at \vec{r} is
where \hat{r} = \vec{r}/r is the unit vector from the source to the test point and r = |\vec{r}| is the distance.
Why: Coulomb's law for two point charges. The magnitude is k|Qq_0|/r^2; the direction is along the line joining them; and with the product Qq_0 keeping its sign, the formula above produces a repulsive force when Q and q_0 have the same sign (the force points in the +\hat{r} direction, away from Q) and an attractive force when they are opposite (the product is negative, the force points in -\hat{r}, toward Q).
Divide by q_0:
This is the field of a point charge. A few features to register:
- Radial. At every point in space, the field points along the line from Q to that point. If Q>0, the field points outward; if Q<0, the field points inward.
- Inverse-square. Double the distance from Q, and |\vec{E}| drops by a factor of four.
- No source at the field point. The field exists at every point in space around Q, but \vec{E} is not defined at r=0 (where Q itself sits) — the formula blows up there. This is a feature of the idealisation (point charges are not real; every real charge has some finite extent). In real problems you never need to evaluate the field at the source.
Superposition — adding fields as vectors
The field obeys the same linearity that Coulomb's force does. If you have two source charges Q_1 at \vec{r}_1 and Q_2 at \vec{r}_2, the field at a field point \vec{r} is just the vector sum of the two individual fields:
Why: superposition is built in at the force level (by experiment and by linearity of Coulomb's law), and the field is just force divided by a fixed test charge. If the forces add, the fields add.
For n source charges this becomes
where \hat{r}_i is the unit vector from source i to the field point \vec{r}. This one formula, applied to finitely many point charges or extended by integration to continuous distributions, is the workhorse of all of electrostatics.
The practical procedure. To find the field at a point due to a collection of charges:
- Pick a coordinate system and write out each source charge's position.
- For each source, compute the vector from that source to the field point, and its magnitude.
- Write down each source's field contribution \vec{E}_i at the field point (magnitude and direction).
- Break each \vec{E}_i into components along your coordinate axes.
- Add the components. The result is the total field.
This looks mechanical — and for discrete collections of charges, it is. The only trick is being careful with directions and signs. For continuous distributions (rods, rings, discs, spheres) you replace the sum by an integral and handle the bookkeeping with calculus; the physics is identical.
The electric dipole — two equal and opposite charges
A particularly important configuration is the electric dipole: two charges of equal magnitude and opposite sign, +q and -q, held a fixed distance d apart. You cannot get simpler than this (apart from a single point charge), and yet dipoles appear everywhere — a water molecule carries a permanent dipole; a radio antenna is a driven dipole; atoms and molecules polarise into induced dipoles in an external field; most real charged bodies look dipole-like when viewed from far away.
Dipole moment. Define the dipole moment vector
where \vec{d} points from the negative charge to the positive charge and has magnitude equal to the separation. The direction of \vec{p} is chosen by convention (negative to positive — the same convention that makes electric field lines leave + and enter -). The SI unit of dipole moment is the coulomb-metre (C·m); for a water molecule |\vec{p}| \approx 6.2 \times 10^{-30} C·m, a number you will see again when you study dielectrics.
Field on the axis of a dipole
Place the dipole along the x-axis, with -q at x = -d/2 and +q at x = +d/2. Find the field at a point on the axis at distance r from the midpoint, on the side of the positive charge (so at x = r).
Step 1. Distance from each charge to P.
The positive charge is at x = d/2, P is at x = r, so distance = r - d/2.
The negative charge is at x = -d/2, so distance = r + d/2.
Step 2. Field from +q at P, on the axis.
Why: +q repels a positive test charge at P, pushing it in the +\hat{x} direction (away from +q, i.e., outward along the axis).
Step 3. Field from -q at P.
Why: -q attracts a positive test charge, pulling it in the -\hat{x} direction (toward the negative charge). The magnitude uses |q| since we are writing E_- as the magnitude with the direction given separately.
Step 4. Net field.
Why: both fields lie along the axis, one in +\hat{x} and one in -\hat{x}. The net is the difference of their magnitudes, pointing in the direction of the larger field — which is E_+ because +q is closer.
Step 5. Combine the fractions over a common denominator.
Expand the numerator using (a+b)^2 - (a-b)^2 = 4ab with a = r, b = d/2:
The denominator is ((r-d/2)(r+d/2))^2 = (r^2 - d^2/4)^2. So:
Why: the identity (a+b)^2 - (a-b)^2 = 4ab is a clean algebraic simplification that pulls the d out into the numerator — exactly the factor that the dipole moment p = qd will absorb in the next step.
Step 6. In the dipole limit r \gg d, drop d^2/4 compared to r^2 in the denominator.
Why: the approximation r \gg d is what "far from the dipole" means — far enough that you cannot tell the two charges apart on the scale of your distance. In this limit, what remains depends only on p = qd, the dipole moment, not on q and d separately. That is why the dipole moment is the natural variable.
Putting the direction back in (along +\hat{x}, which is the direction of \vec{p}):
Field on the equatorial plane
Now put the field point on the perpendicular bisector of the dipole — the equatorial plane — at distance r from the midpoint.
Step 1. Distance from each charge to P.
By Pythagoras, both charges are at the same distance from P:
Why: P is on the perpendicular bisector, so the two charges are mirror images of each other with respect to P. The distances are equal.
Step 2. Magnitude of the field from each charge.
Why: by Coulomb, magnitude is kq/(\text{distance})^2. Both charges contribute the same magnitude because distances are equal.
Step 3. Directions and symmetry.
\vec{E}_+ points from +q away, along the line from +q to P, extended beyond P. \vec{E}_- points from P toward -q, along the line from P to -q. Decompose each into a component along the axis direction (call it \hat{x}, in the direction of \vec{p}) and a component perpendicular to the axis (along \hat{y}, the direction from O to P).
By symmetry, the perpendicular components of \vec{E}_+ and \vec{E}_- are equal in magnitude and opposite in sign — one points +\hat{y} and the other -\hat{y}. They cancel.
The axial components are equal in magnitude and both point in -\hat{x} (from +q side to -q side, opposite to \vec{p}). They add.
Why: the geometry has a mirror symmetry about the perpendicular bisector. Components perpendicular to that bisector (= along the dipole axis) survive; components along it cancel. The axial components both point "backward" — from the positive-charge side to the negative-charge side — because +q pushes P away to the -\hat{x} side and -q pulls P toward itself also on the -\hat{x} side.
Step 4. Compute the axial component.
The line from +q at (d/2, 0) to P at (0, r) makes an angle \theta with the axis, where \cos\theta = (d/2)/\sqrt{r^2 + d^2/4}. The axial component of \vec{E}_+ has magnitude E_+ \cos\theta and points in -\hat{x}.
Same for \vec{E}_-. Total axial field magnitude:
Simplify:
Why: the factor of 2 is for the two equal axial contributions; the \cos\theta picks out the axis projection; the denominator combines 1/(r^2+d^2/4) from Coulomb with 1/\sqrt{r^2+d^2/4} from \cos\theta into a 3/2 power.
Step 5. In the limit r \gg d, drop d^2/4.
Putting the direction (antiparallel to \vec{p}) back in:
The axial-equatorial comparison
Compare equations (5) and (6). At the same distance r from the dipole's centre:
- Axial field: magnitude \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2p}{r^3}, direction parallel to \vec{p}.
- Equatorial field: magnitude \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{p}{r^3}, direction antiparallel to \vec{p}.
The axial field is twice as strong as the equatorial field, and points in the opposite direction. Both fall off as 1/r^3, not 1/r^2 — the faster falloff is because the two charges' fields largely cancel at a distance, and what remains is a correction proportional to d.
This 1/r^3 falloff is the signature of a dipole. If you measure a static field that decays faster than 1/r^2, you are almost certainly looking at an object whose total charge sums to zero (a dipole, or more generally a neutral object with charge-separated structure). A water molecule is electrically neutral but strongly dipolar, which is why water is such a good solvent for ionic salts: the dipolar field pulls ions apart.
Electric field lines — the picture of the field
A single vector \vec{E} at each point is hard to draw over a whole region. The visual shortcut that Faraday invented is the electric field line: a continuous curve such that the electric field is tangent to it at every point, and the density of lines (lines per unit perpendicular area) is proportional to |\vec{E}|.
Six rules for drawing field lines
- Lines start on positive charges and end on negative charges. (Or they go to or come from infinity, which you can think of as the opposite sign "at infinity".) The starting and ending density is proportional to the charge magnitude: a +2q charge has twice as many lines leaving it as a +q charge.
- Lines never cross. If two lines crossed, the field at the crossing point would point in two different directions — impossible for a single well-defined vector field.
- The tangent to a line at any point gives the direction of \vec{E} at that point.
- The density of lines is proportional to |\vec{E}|. Where lines bunch up, the field is strong; where they fan apart, the field is weak.
- Lines are perpendicular to the surface of a conductor in electrostatic equilibrium. (A field with any tangential component on the conductor would drive charges along the surface until equilibrium was restored.)
- Lines do not form closed loops in an electrostatic field. (This is a statement of the conservative nature of the electrostatic force — lines going from + to - cannot loop back, since then a test charge carried around the loop would gain energy with no source.)
Some consequences of these rules worth internalising:
- Parallel straight lines = uniform field. In a region where all the field lines are parallel straight lines, equally spaced, the field is uniform — same magnitude and direction everywhere. This is what you get between two infinite parallel plates of opposite charge (the capacitor) — you will meet this again in applications of Gauss's law.
- Lines bending around a conductor. Near the surface of a conductor, lines bend so that they hit the surface perpendicularly. This rule alone is what makes the Faraday cage protect the inside from external fields — the external field lines cannot penetrate because they must terminate on the conductor's outer surface.
- A field map without arrows is ambiguous. When drawing lines, always indicate the direction with an arrowhead on at least one point of each line. A pattern of lines alone does not distinguish a positive from a negative configuration.
Watch a charge accelerate in a uniform field
In a uniform electric field \vec{E} = E_0\hat{x} (for instance, between two parallel plates of opposite charge), a free positive charge q experiences a constant force \vec{F} = q\vec{E}, so it accelerates uniformly: \vec{a} = q\vec{E}/m. The trajectory of a charge that enters the field with some initial velocity is a parabola — exactly like a projectile under gravity, with q\vec{E}/m playing the role of \vec{g}.
This is the physics behind the inkjet printer: a tiny ink droplet (mass \sim 10^{-11} kg) is given a controlled charge and shot through a uniform electric field of a few hundred volts per metre. The electric force deflects the droplet sideways by a precisely controlled amount, and the droplet lands at a specific pixel on the paper. By modulating the charge on each droplet, the printer places drops column by column as the print head sweeps across the paper.
Worked examples
Example 1: Field of two point charges along an axis
Two point charges sit on the x-axis: Q_1 = +4.0 nC at x = 0 and Q_2 = -2.0 nC at x = 0.30 m. Find the electric field at the point P = (0.50, 0) m.
Step 1. Distances from each charge to P.
r_1 = 0.50 - 0 = 0.50 m. r_2 = 0.50 - 0.30 = 0.20 m.
Why: both charges are to the left of P on the same axis. The distance is simply the difference of x-coordinates.
Step 2. Magnitude of each field at P.
Why: standard Coulomb formula for the magnitude. Use absolute values of the charges here — the direction is worked out separately in the next step.
Step 3. Determine the direction of each field at P.
\vec{E}_1: since Q_1 is positive and P is to the right of Q_1, \vec{E}_1 points in +\hat{x}.
\vec{E}_2: since Q_2 is negative and P is to the right of Q_2, \vec{E}_2 points toward Q_2, i.e., in -\hat{x}.
Why: the field of a positive charge points radially outward (away from the source); the field of a negative charge points radially inward (toward the source). At P, both fields lie along the axis, so only the sign matters.
Step 4. Add the fields as vectors.
Why: both fields lie along the \hat{x} direction, so the vector sum reduces to an algebraic sum of signed components. The negative sign means the net field at P points in -\hat{x}.
Result: The net electric field at P has magnitude \boxed{|\vec{E}|\approx 306\;\text{N/C}}, pointing in the -\hat{x} direction (toward the origin).
What this shows: Even though Q_1 is twice as large as |Q_2|, the closer negative charge dominates at P because the field falls off as 1/r^2. Distance matters more than magnitude when the distances differ enough.
Example 2: Electron suspended in a uniform field
In Millikan's oil drop experiment (and in the modern classroom Van de Graaff demonstration at any IIT), a small charged object can be held stationary in air if the upward electric force exactly balances gravity. Consider a single electron (q = -e = -1.6\times 10^{-19} C, m = 9.11 \times 10^{-31} kg). What uniform electric field (magnitude and direction) would suspend the electron motionless against gravity near Earth's surface?
Step 1. Write the equilibrium condition.
Why: Newton's first law for a stationary object — net force is zero. Rearranging gives the required field.
Step 2. Insert numbers for magnitude, with \vec{g} pointing down in -\hat{y}.
Why: plug in the known values. The denominator is the magnitude of the electron charge (the sign is handled by direction separately).
Step 3. Get the direction.
Since q = -e < 0, \vec{E} = -m\vec{g}/q = -m(-g\hat{y})/(-e) = -(mg/e)\hat{y}.
Why: substitute \vec{g} = -g\hat{y} and q = -e into the equilibrium equation. The two negatives cancel the \hat{y} but introduce another minus from the charge, leaving \vec{E} pointing in -\hat{y} — downward.
Result: A vanishingly weak, downward-pointing field of magnitude \boxed{|\vec{E}| \approx 5.6 \times 10^{-11}\;\text{N/C}} suffices to suspend an electron against gravity.
What this shows: The electric force between charges is so much stronger than the gravitational force that an almost immeasurably small field can hold an electron aloft. In Millikan's actual experiment, he used oil droplets (mass \sim 10^{-15} kg and carrying a handful of extra electrons) — still a minuscule field was required, because even oil droplets are featherlight on the scale of the electromagnetic force. This disparity — \sim 10^{39} between the electromagnetic and gravitational forces on the same pair of charged particles — is one of the deep unsolved mysteries of fundamental physics.
Example 3: Dipole moment of a water molecule
A water molecule (\text{H}_2\text{O}) has a permanent electric dipole moment of magnitude p = 6.2 \times 10^{-30} C·m, with \vec{p} pointing from the (slightly negative) oxygen toward the midpoint of the two (slightly positive) hydrogens. Find the electric field, along the axis of this dipole, at a distance r = 1.0 nm (10^{-9} m) from its centre — roughly the distance between two adjacent water molecules in liquid water. Compare this to the field one would need to orient a second water molecule significantly (say, 100 kV/cm = 10^7 V/m, roughly the breakdown strength of water).
Step 1. Use the axial dipole field formula, equation (5).
Why: we are asked for the axial field at distance r \gg d. The molecule's charge separation d is of order 10^{-10} m (an Ångström), and r = 10^{-9} m is ten times that — the far-field approximation is marginal but usable.
Step 2. Insert numbers.
Numerator: (8.99\times 10^9)(1.24\times 10^{-29}) = 1.115 \times 10^{-19} in the numerator.
Denominator: (10^{-9})^3 = 10^{-27}.
Why: careful with powers of ten — 10^{9}\times 10^{-30}/10^{-27} = 10^{9-30+27} = 10^{6}, multiplied by the prefactors 8.99 \times 2 \times 6.2 \approx 111 gives 1.1\times 10^8 N/C. Work the exponents first; plug in the prefactors second.
Step 3. Compare to the water-breakdown threshold 10^7 V/m.
Why: divide to find the ratio. The field a single water dipole exerts on its neighbour at r = 1 nm is about ten times the field that would electrically break down bulk water. This is why neighbouring water molecules interact so strongly — they live in each other's enormously strong near-fields.
Result: The axial field at r = 1 nm is \boxed{|\vec{E}_\text{axial}| \approx 1.1 \times 10^8\;\text{N/C}}.
What this shows: The water molecule's internal fields dominate the microscopic world of liquid water. This is why water forms hydrogen bonds (weak but numerous attractions aligning neighbours' dipoles with each other), why water has such a high dielectric constant (\varepsilon_r \approx 80), and why ions in water are effectively surrounded by shells of oriented water molecules. The macroscopic behaviour of water as a solvent — dissolving salts, conducting electricity through ions, supporting life's chemistry — traces back to this enormous dipolar near-field.
Common confusions
- "The field is what the test charge feels." Close, but the field exists whether or not a test charge is there. The field is a property of the space around the source charges, set up by them alone. A test charge dropped into the field feels \vec{F} = q\vec{E}, but the field does not go away when the test charge is removed.
- "Field lines are real." They are a visualisation, not a physical object. The field itself — the vector \vec{E} at each point — is real (in the sense that it carries energy, momentum, and can propagate as a wave). Field lines are a convenient way to draw the field; they are not more fundamental than the vectors they summarise. Two different line-drawings can represent the same field if they preserve the direction and the density relation.
- "A negative charge has a field pointing 'into' it, so the field is 'negative'." Fields do not have signs; they have directions. A positive charge's field points outward; a negative charge's field points inward. In both cases |\vec{E}| is a positive number — what differs is the direction, not the sign of the magnitude. The word "negative" belongs to the charge, not the field.
- "Superposition fails for large charges." Superposition of fields is linear: \vec{E}_\text{total} = \sum \vec{E}_i always, within classical electromagnetism. What can fail is the point-charge idealisation if a large test charge disturbs the source arrangement (pushes the source charges around) — this is why the definition uses the q_0 \to 0 limit. The superposition of fields from fixed source charges is exact.
- "Dipole field falls off as 1/r^2." No. A monopole (a single isolated charge) field falls off as 1/r^2, because that is Coulomb's law. A dipole (two equal and opposite charges) field falls off as 1/r^3. A quadrupole (four charges summing to zero with dipole moment zero) falls off as 1/r^4. Each term in the multipole expansion decays one power of r faster than the previous — because cancellations remove a power of 1/r at each order.
- "Field lines cross at the saddle point between two like charges." No — at the saddle point the field is zero (\vec{E} = 0), and the lines terminate or bend away. Field lines never cross, because a crossing would mean two directions at one point.
If you came here to understand the electric field and to compute fields of point charges and simple distributions, you have what you need. What follows is for readers going on to continuous distributions, the multipole expansion, and the conceptual status of the field in relativity.
Fields from continuous charge distributions
A real charged object (a ring, a disc, a rod, a sphere) is not a collection of point charges — at least not at our scale of resolution. You treat it as a continuous distribution with a local charge density:
- Linear density \lambda (C/m) for charge spread along a line.
- Surface density \sigma (C/m²) for charge spread on a surface.
- Volume density \rho (C/m³) for charge spread through a volume.
The field contribution from an infinitesimal piece dq at position \vec{r}' is, by Coulomb,
and the total field is the integral of all such contributions:
with dq = \lambda\,dl for a line, \sigma\,dA for a surface, or \rho\,dV for a volume. The integrations are almost always difficult by direct methods unless the distribution has symmetry.
Example: field on the axis of a uniformly charged ring. A ring of radius R carries total charge Q uniformly. You want the field on the ring's axis at distance x from the ring's centre. Each infinitesimal piece contributes a field of magnitude k\,dq/(R^2 + x^2); the components perpendicular to the axis cancel by symmetry (for every piece on one side of the ring, there is an equal piece on the other side contributing the opposite perpendicular component). Only the axial components survive:
Since \cos\theta and (R^2+x^2) are the same for every piece on the ring, pull them out of the integral:
Pleasing features to notice: at x=0 (centre of the ring), E_x=0 by symmetry — every piece is cancelled by the diametrically opposite piece. At x\gg R, (R^2+x^2)^{3/2} \approx x^3 and E_x \approx kQ/x^2 — the ring looks like a point charge of total charge Q, as it must.
The multipole expansion
Any localised charge distribution, viewed from far enough away, looks like a sum of terms, each falling off with a different power of r:
The monopole term is controlled by the total charge. If the distribution has net charge zero, the monopole term vanishes, and the leading contribution is the dipole term. If both the monopole and dipole moments are zero (by symmetry or by construction), the leading term is the quadrupole. And so on.
This is why an electrically neutral molecule (water, HCl, \text{CO}_2) still has a field — the monopole term vanishes but the dipole term does not. And it is why a very symmetric neutral molecule like \text{CO}_2 has a much weaker long-range field than water: \text{CO}_2 is linear and symmetric, so its dipole moment is zero; only the quadrupole survives, and quadrupolar fields fall off faster.
A conductor distorts the field it sits in
Drop a conductor into an external field. In electrostatic equilibrium:
- The field inside the conductor is zero. (If it were non-zero, charges would move until it became zero.)
- The field just outside the conductor is perpendicular to the surface. (Any tangential component would drive surface currents.)
- Charges accumulate on the conductor's outer surface with density \sigma = \varepsilon_0 E_\perp, where E_\perp is the field magnitude just outside.
These three facts — which you will derive from Gauss's law in applications of Gauss's law — spheres and planes — make conductors very useful: a hollow conducting box (a Faraday cage, the principle used in the metal bodies of Maruti cars and in sensitive electronic instruments at TIFR) entirely shields its interior from any external electrostatic field. The interior of the cage is at zero field regardless of what is happening outside.
The field in relativity — why the field concept is deep
In 19th-century physics, the field was a mathematical convenience. You could equivalently say "charges exert forces across space by Coulomb's law" or "charges set up fields and fields exert forces" — the two statements gave the same predictions for static charges.
But once charges start moving, the two views diverge. A charge that suddenly jerks does not instantaneously change the force on a far-away test charge; the disturbance propagates outward at the speed of light. The way you describe that — a disturbance in the field carrying energy and information at c — is not available in the action-at-a-distance picture. The field is real; it carries momentum and energy; a wave in it is a photon. This is Maxwell's insight, and it is what promotes the field from a calculational trick to a full physical entity.
The electric field is half of this story; the magnetic field \vec{B} is the other half, and the two are tied together by relativity — an observer moving through an electric field sees some of it as magnetic, and vice versa. You will meet the magnetic field in detail later; for now, think of the electric field as the "static" face of the electromagnetic field, and remember that the field concept is not optional. It is how electromagnetism works.
The test-charge limit, more carefully
The definition (1) uses \lim_{q_0 \to 0}. Why bother? Because a real finite test charge would itself polarise any conductor nearby, rearrange any nearby charge distribution, and generally perturb the source of the very field it is trying to measure. By taking the limit, you define the field of the original source configuration, not the configuration as perturbed by the probe.
In practice, if your test charge is small enough (say, an electron probing a macroscopic field of 10^5 V/m, the perturbation being imperceptible on the macroscopic scale), the limit is not needed — the measured field and the idealised field agree. But the mathematical definition is the cleanest, and in modern physics you define fields in a way that does not even reference test charges: through Maxwell's equations, which specify the field from its sources and boundary conditions, with no test particle anywhere in sight.
Where this leads next
- Electric Flux — the surface integral of \vec{E} through a surface, and the first half of the Gauss's law story.
- Gauss's Law — the powerful reformulation of Coulomb's law that lets you compute fields from symmetric charge distributions with astonishing ease.
- Applications of Gauss's Law — Spheres and Planes — field of a uniformly charged sphere, infinite plane, and two-plate capacitor.
- Electric Potential and Potential Difference — the scalar field whose gradient gives \vec{E}, and the cleaner way to handle energy in electrostatics.
- Coulomb's Law — the underlying force law from which everything in this article was derived.
- Superposition of Electrostatic Forces — the vector addition rule that generalises to continuous distributions and is the engine behind every integral field calculation.