The Electric Field

Stand near a bonfire and you feel heat without touching a single flame — the fire has changed the space around it. Coulomb's law, for all its precision, stays silent about how one charge reaches another across empty space. Charge $A$ pushes charge $B$ instantly, through nothing? Newton called the same idea in gravity "so great an absurdity" that no competent thinker could accept it. Physics' repair job is the most productive idea in this entire subject: the electric field. The move splits one mysterious long-range push into two local, checkable steps. First, a charge modifies the space around it — it dresses every point with a vector $\vec{E}$ that says "if a positive charge sat here, it would be pushed this way, this hard." Second, any charge that actually shows up simply reads the vector at its own location and feels $\vec{F} = q\vec{E}$ . Charge $B$ never needs to know charge $A$ exists; it only consults the field where it stands. For exams, the field is the workhorse variable of the next thirty modules — field lines, flux, Gauss's law, potential, capacitors all speak field language, and "find $E$ , then find the force" is a two-mark pattern on every board. For physics, it is bigger: shake a charge and the field ripples outward at the speed of light as light itself. The vector you learn to compute today is the same object that carried this page's photons to your eye.

Key takeaways

The electric field at a point is the force per unit charge a small positive test charge would feel there: $\vec{E} = \vec{F}/q_0$ , measured in N/C (identical to V/m).
A point charge $q$ creates $E = kq/r^2$ at distance $r$ — inverse-square, pointing away from positive charges and toward negative ones.
Once $\vec{E}$ is known, the force on any charge is $\vec{F} = q\vec{E}$ : positive charges pushed along the field, negative charges pushed against it.
The test charge must be small, so it does not shove the source charges around and distort the very field it measures.
The field exists at every point of space, charge present or not — it is a property of the place, prepared in advance by the sources.
Key idea: sources make the field; the field moves charges — every electrostatics calculation walks these two steps.

The core insight

The electric field is the force per unit positive test charge at a point: $\vec{E} = \vec{F}/q_0$ , so that any charge $q$ placed there feels $\vec{F} = q\vec{E}$ — the field converts a mysterious action-at-a-distance into a local, two-step story.

Where this lives today

The two-step dance — compute the field once, then bill every charge locally — is the architecture of modern technology. Your phone's touchscreen holds a groomed pattern of fields above a grid of electrodes and locates your fingertip by the disturbance it makes. Inkjet printers steer microscopic droplets with deflecting fields recalculated thousands of times a second. Hospital X-ray tubes and the 27-kilometre Large Hadron Collider are both, at heart, machines for making enormous $\vec{E}$ and letting $\vec{F} = q\vec{E}$ do the accelerating — the LHC's cavities hold oscillating fields of megavolts per metre, timed so every passing proton bunch always feels a forward push. Plasma-simulation codes that design fusion reactors advance billions of particles per step using nothing deeper than $a = qE/m$ . And the conceptual upgrade Faraday started is now measured fact: fields genuinely detach from their sources and carry energy across space — sunlight pressing on your skin left the Sun eight minutes before you felt it.

The physical picture

Think of the field as force-readiness saturating space. Before any second charge arrives, a source charge has already prepared, at every point around it, an answer to a hypothetical question: which way would a positive charge here be pushed, and how hard? Near the source the prepared arrows are long; twice as far away they are four times shorter; at every point they aim radially — outward from a positive source like the spray of a fountain, inward toward a negative one like water circling a drain. A test charge dropped anywhere does not create this arrangement. It reads it.

Figure

A positive point charge dresses space with force-readiness arrows — long nearby, quarter-length at double distance, radial everywhere

Two details of the definition are engineered on purpose, and both carry marks. The test charge is positive by convention — that single choice fixes every direction in the subject: fields point away from $+$ , toward $-$ , and a negative charge feels a force opposite to $\vec{E}$ . And the test charge must be small — not because small charges are tidier, but because a big probe would repel or attract the source charges, shifting them and rewriting the very field it came to measure. Board answers that define $\vec{E}$ without saying "small (test) positive charge" routinely drop a mark for exactly this reason. Why bother with the field at all, when Coulomb's law already gives forces? Because the field splits every problem into two independent halves. Solve the hard half once — what field does this arrangement of sources make? — and the easy half, $\vec{F} = q\vec{E}$ , answers force questions for any charge placed anywhere, forever, without touching Coulomb's law again. Every tool coming up — field lines, flux, Gauss's law, potential — is a statement about $\vec{E}$ . You are not learning a new convenience; you are learning the language the rest of electromagnetism is written in.

Defining and using the field

The definition turns the bonfire intuition into an equation. Place a small positive test charge $q_0$ at the point of interest, measure the electric force $\vec{F}$ on it, and divide:

\vec{E} = \frac{\vec{F}}{q_0} \qquad (1)

where: $q_0$ — a positive test charge small enough not to disturb the sources; $\vec{E}$ — the field at that point, in newtons per coulomb (N/C). Dividing by $q_0$ strips the probe out of the story: Coulomb's law makes $\vec{F}$ proportional to $q_0$ , so the quotient is the same whatever probe you use — which is exactly what entitles us to call it a property of the point, not of the measurement. (You will also meet the unit V/m once potential arrives in the next chapter; $1\ \mathrm{N/C} = 1\ \mathrm{V/m}$ exactly.) Read backwards, the definition becomes the force law — the version you will use daily:

\vec{F} = q\vec{E} \qquad (2)

where: $q$ carries its sign. A positive charge is pushed along the field; a negative charge — an electron, always an electron in exam problems — is pushed against it. Notice what equation (2) does not contain: the source charges' positions, sizes or count. All of that has been pre-digested into one local vector. This is the two-step dance: sources make the field (step one, sometimes hard), the field moves the charge (step two, always easy). For the simplest source — a single point charge $q$ — the field follows from Coulomb's law in one line (the derivation below does it honestly):

E = \frac{kq}{r^2} = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r^2} \qquad (3)

where: $r$ — distance from the source to the point; $k = 8.99\times10^9\ \mathrm{N\,m^2/C^2}$ ; direction radially away from a positive $q$ , radially toward a negative one. The inverse square is the same sphere-dilution geometry as Coulomb's law: double the distance, quarter the field. Feel the scale: a modest $5\ \mu$ C lab charge already produces $5\times10^5$ N/C at 30 cm, while the fair-weather atmospheric field is about $100$ N/C and air itself breaks down (sparks) near $3\times10^6$ N/C.

Figure

The two-step dance — sources make the field once; the field then bills every visitor locally

question	tool	status
force on a charge, field known	$\vec{F} = q\vec{E}$	this module
field of one point charge	$E = kq/r^2$ , radial	this module
field of several charges	vector-sum the individual fields	Superposition (previous module)
field of a rod, ring, sheet	chop into pieces and sum	Continuous Charge Distributions
picture of a field	draw its lines	Electric Field Lines (next)
symmetric sources, fast	flux + Gauss's law	later this chapter

Derivation AFrom Coulomb's law to the field of a point charge

We want equation (3) — the field of a point charge — extracted honestly from Coulomb's law, and we want to see exactly why the test charge drops out and why it had to be small. The strategy: write the Coulomb force a fixed source exerts on a probe, divide by the probe, and watch every trace of the probe cancel. The cancellation is the whole point — it is what makes the field a property of space rather than a property of the measurement.

Step 1. Fix the source charge $q$ at a point, and place a test charge $q_0$ at distance $r$ . Coulomb's law gives the force magnitude on the probe:

F = \frac{k\,q\,q_0}{r^2}

directed along the line joining them — away from $q$ if both are positive.

Step 2. Divide by the probe, per the definition in equation (1):

E = \frac{F}{q_0} = \frac{kq}{r^2}

Why does q0 vanish completely: Coulomb's force is directly proportional to q0, so dividing by q0 removes it exactly — a probe twice as big feels twice the force and reports the identical field.

Step 3. Attach the direction. The force on a positive probe points away from a positive $q$ , so $\vec{E}$ points radially outward; a negative source reverses every arrow inward. One rule, both signs.

Step 4. Honour the fine print: why must $q_0$ be small? Step 1 assumed the source stays put. A large probe would push or pull the source (or the charges on any nearby conductor) into new positions, and the measured $F/q_0$ would then describe a disturbed arrangement. Keeping $q_0$ tiny keeps the back-reaction negligible, so the quotient reports the undisturbed field — the "small test charge" clause in every board definition is this physics, not politeness. Verified: dimensions (N m²/C²)(C)/(m²) = N/C; with q = 4.0 μC and r = 0.20 m, E = (8.99×10⁹)(4.0×10⁻⁶)/(0.04) = 9.0×10⁵ N/C, recomputed numerically and reused in the worked example below.

One more consequence deserves its own sentence, because it decides exam signs: a charge never feels its own field. The field that pushes charge $q$ is made by all the other charges; a point charge's own field points symmetrically outward in every direction and gives it no net shove. In $\vec{F} = q\vec{E}$ , the $\vec{E}$ is always "everyone else's."

Limiting cases

case	result	meaning
double the distance from a point charge	$E \to E/4$	inverse-square dilution
double the source charge	$E \to 2E$ everywhere	field is proportional to its source
very far from any small charged object	$E \approx kQ_{\text{total}}/r^2$	everything looks like a point charge from far away
midpoint of two equal like charges	$E = 0$	superposition cancellation (previous module)
inside a conductor in equilibrium	$E = 0$	charges rearrange to cancel — Electrostatics of Conductors, later this chapter
test charge not small	measured $F/q_0$ depends on $q_0$	sources disturbed — the definition's fine print doing real work

Common trapThe field pushes electrons backwards

$\vec{E}$ points the way a positive charge would be pushed — that convention decides every sign in this subject, and forgetting it produces the most reliable wrong answer in board history: an electron released in a rightward field accelerates left, against the field, because $\vec{F} = q\vec{E}$ with $q = -e < 0$ . Three cousins of the same trap travel together: the field of a negative source points toward it (arrows inward); a charge feels only the field of other charges, never its own; and a point in empty space has a perfectly definite field even though nothing there feels a force — the field is the readiness, not the act. When any sign wobbles mid-problem, re-anchor on the one question that never lies: which way would a small positive charge go?

Worked examples

Board level — field first, then the force bill (CBSE/ISC pattern)

A point charge $q = +4.0\ \mu\mathrm{C}$ is fixed at the origin. (a) Find the electric field at point P, $0.20$ m away. (b) A charge $Q = -3.0$ nC is now placed at P; find the force on it. Take $k = 9\times10^9\ \mathrm{N\,m^2/C^2}$ if your board permits the rounded value.

Step 1. Field magnitude from equation (3): $E = kq/r^2 = (8.99\times10^9)(4.0\times10^{-6})/(0.20)^2 = 9.0\times10^5\ \mathrm{N/C}$ .

Step 2. Field direction: the source is positive, so $\vec{E}$ at P points away from the origin. State this in words — the direction earns its own mark.

Step 3. Force on $Q$ by equation (2): magnitude $F = |Q|E = (3.0\times10^{-9})(9.0\times10^5) = 2.7\times10^{-3}$ N.

Step 4. Force direction: $Q$ is negative, so the force is opposite to $\vec{E}$ — toward the source, as unlike charges demand. Answer: $E = 9.0\times10^5$ N/C outward; $F = 2.7$ mN toward the origin.

Step 5. The structure is the marks scheme: field computed once as a property of P (before $Q$ existed), then the force read off locally. Writing one combined Coulomb calculation gets the number but often forfeits the direction marks.

Competitive level — electron vs proton in the same field (NEET/JEE-Main pattern)

An electron and a proton are released from rest in a uniform field $E = 1.0\times10^4$ N/C. Compare their initial accelerations, and find how long the electron takes to cross $2.0$ cm. ( $m_e = 9.11\times10^{-31}$ kg, $m_p = 1.67\times10^{-27}$ kg, $e = 1.6\times10^{-19}$ C.)

Step 1. Same field, same magnitude of charge — so the same force magnitude $F = eE = (1.6\times10^{-19})(1.0\times10^4) = 1.6\times10^{-15}$ N on each, in opposite directions (proton along $\vec{E}$ , electron against it).

Step 2. Accelerations differ by mass alone: $a_e = eE/m_e = 1.76\times10^{15}\ \mathrm{m/s^2}$ , while $a_p = eE/m_p = 9.6\times10^{11}\ \mathrm{m/s^2}$ . The ratio is $m_p/m_e \approx 1836$ — the electron out-accelerates the proton nearly two-thousand-fold. (Both dwarf gravity: $a_e/g \sim 10^{14}$ , which is why $mg$ is deleted from such problems without apology.)

Step 3. Electron crossing $d = 2.0$ cm from rest: $d = \tfrac12 a_e t^2$ gives $t = \sqrt{2d/a_e} = \sqrt{2(0.020)/1.76\times10^{15}} = 4.8$ ns, exiting at $v = a_e t = 8.4\times10^6$ m/s — nearly 3% of light speed from a two-centimetre run.

Step 4. The MCQ pattern to cache: equal forces, accelerations in inverse ratio of masses, opposite directions. Any option claiming the proton accelerates faster "because it is positive" is bait. Timing note: 90 seconds. The only computation is one square root; the force-equal/acceleration-inverse logic should be instant. Full trajectory problems (deflection between plates, parabolic paths) are drilled in the archetype module Charged Particle in a Uniform Field.

Exam traps

Missing words in the definition. " $\vec{E}$ is the force on a charge" loses marks three ways: it is force per unit charge, the test charge is positive, and it is small (vanishingly small, so the sources are undisturbed). All three words are load-bearing; CBSE and ISC schemes list them.

Electron direction. Force on a negative charge is opposite $\vec{E}$ . Check the sign before the kinematics, not after.

Substituting signed charges into magnitudes. Compute magnitudes with $|q|$ ; assign directions by physics (away from $+$ , toward $-$ ; along $\vec{E}$ for $+$ , against for $-$ ). Feeding $-1.6\times10^{-19}$ into a magnitude formula and then "fixing" signs at the end is how directions get lost.

Inverse-square slips. $E$ at double distance is a quarter, not half. In ratio questions (JEE loves $E_1/E_2$ ), square the distance ratio first.

Units. N/C and V/m are the same unit — do not "convert" between them by any factor. And field is not force: an answer in newtons for a field question is an automatic zero for that line.

Scope notes by board. IGCSE (0625) needs only the qualitative story — field as a region where a charge feels a force, plus simple patterns; the $kq/r^2$ arithmetic is beyond its syllabus. IB-SL uses $E = F/q$ and the uniform field between plates; the point-charge formula is HL territory. CBSE/ISC/JEE/NEET/AP-2/A-Level all demand the full quantitative package on this page. Constants: $k = 9\times10^9$ is standard in Indian boards unless the paper says otherwise; JEE/NEET options are built from it.

Origins · 1 minMichael Faraday and how this was discovered

Michael Faraday

1791–1867 · Royal Institution, London

The field concept was born in the mind of a bookbinder's apprentice with almost no formal mathematics. Faraday, hired as Humphry Davy's laboratory assistant, became the greatest experimentalist of his century — and because he did not think in equations, he thought in pictures: space around charges and magnets filled with lines of force, real as stretched rigging, transmitting push and pull through the space between. His contemporaries mostly indulged the idea as a crutch while serious people computed action-at-a-distance. It was Maxwell who insisted Faraday had it exactly right — "Faraday, in his mind's eye, saw lines of force traversing all space where the mathematicians saw centres of force attracting at a distance" — and built the equations of electromagnetism on Faraday's pictures. Every $\vec{E}$ you write is Faraday's intuition wearing mathematical clothes; the field lines you will draw in the next module are his original invention, almost unchanged.

History

1785

— Coulomb quantifies the force between charges; the interaction is read as instantaneous action at a distance

1837

— Faraday, experimenting with insulators, argues electric action is transmitted through the intervening space by lines of force

1855

— Maxwell's first paper "On Faraday's Lines of Force" begins turning the picture into mathematics

1865

— Maxwell's field theory makes the field a physical system carrying energy — and identifies light as its waves

1887

— Hertz generates and detects electromagnetic waves: fields verifiably detach from their sources and travel

Going further

Texts. NCERT Physics Class 12, Part 1, Ch. 1 (§1.7–1.8, Electric Field) — the CBSE/NEET canonical statement · HC Verma, Concepts of Physics Vol 2, Ch. 29 · Halliday, Resnick & Walker, Ch. 22. The full undergraduate treatment — including the deflection-plate derivation, the field's energy and momentum, and the local field equations — is The Electric Field (UG deep dive), which this module distils.

Provenance. All numbers recomputed: $E = (8.99\times10^9)(4.0\times10^{-6})/(0.20)^2 = 9.0\times10^5$ N/C and $F = (3.0\times10^{-9})(9.0\times10^5) = 2.7\times10^{-3}$ N; $a_e = (1.6\times10^{-19})(1.0\times10^4)/(9.11\times10^{-31}) = 1.76\times10^{15}\ \mathrm{m/s^2}$ ; $a_p = 9.6\times10^{11}\ \mathrm{m/s^2}$ ; ratio $1833 \approx m_p/m_e$ ; $t = \sqrt{2(0.020)/1.76\times10^{15}} = 4.8\times10^{-9}$ s; $v = 8.4\times10^6$ m/s.

Prerequisites. Coulomb's Law · Superposition: Forces & Fields of Many Charges. Leads to. Electric Field Lines · Charged Particle in a Uniform Field (archetype) · The Electric Field (UG deep dive).

Is the electric field real, or just a calculation trick for Coulomb's law?

In statics you honestly cannot tell the difference — every static prediction of field language matches action-at-a-distance exactly. The tiebreaker is motion. Wiggle a charge, and the news reaches distant charges only after a delay (at light speed); during the transit, the energy and momentum must be somewhere, and a mere bookkeeping trick cannot carry momentum across a vacuum. The field can, and does — light pressing on a surface is measured routinely. Verdict of physics: the field is a real physical system; statics is just the special case where its reality hides.

How is E different from F? They seem to point the same way.

$\vec{F}$ is what a particular charge feels; $\vec{E}$ is what the place offers. The field exists before and after your charge visits, and is the same for every visitor — while the force depends on who visits ( $F = qE$ scales with $q$ , and flips direction for negative charges, so $\vec{F}$ and $\vec{E}$ point the same way only for positive charges). Units betray the difference too: newtons versus newtons per coulomb. Exam questions exploit the confusion by asking for a field and accepting only N/C.

If a big test charge distorts the field, how does anyone measure fields in real life?

Three practical routes. Use a probe so small that its back-reaction is below your error bars — a nanocoulomb near lab sources moves nothing measurably. Or measure the potential difference between two nearby points with a voltmeter and divide by the separation, exploiting V/m = N/C — no test charge needed. Or use a field mill, a spinning-shutter instrument that reads the atmospheric field (about 100 N/C on a clear day) from the charge it induces. The "vanishingly small" clause is conceptual hygiene in the definition; laboratories have workarounds.

Why doesn't a charge feel its own field? It's strongest right where the charge is.

Symmetry. A point charge's own field points radially outward in every direction at once, so the self-pushes cancel pairwise — there is no direction the charge could pick to push itself. A static charge cannot bootstrap itself into motion (energy conservation would collapse if it could). So in $\vec{F} = q\vec{E}$ , the $\vec{E}$ always means the field of everything else. The refined question — what happens when an accelerating charge interacts with its own field — is real physics, but it lives far beyond this course.

Does the field at a point exist even when there's no charge there to feel it?

Yes — that is exactly what makes it a field: a definite vector at every point of space, like temperature on a weather map existing at places with no thermometer. "The field here is 500 N/C" is a certified conditional: if a positive charge were placed here, it would feel 500 N per coulomb. And the conditional has physical teeth — space containing a field stores energy (you will meet the energy density when capacitors arrive), so a vacuum with a field genuinely differs from a vacuum without one.

Why away from positive? Could the convention have gone the other way?

It could have — the direction of $\vec{E}$ encodes a single human choice: the test charge is positive. Had physics standardised on negative probes, every field arrow would flip and every law would survive with adjusted signs. What is not conventional is the relational fact that unlike charges attract. The historical wrinkle is Franklin's: his "positive" turned out to label the charge that mostly doesn't move in metals, so conventional current runs opposite to electron drift — an inversion the entire electrical industry lives with comfortably, and so will you, provided you re-anchor on "which way would a positive charge go?"

N/C or V/m — which unit should I write?

Either; they are identical. $\mathrm{V} = \mathrm{J/C}$ , so $\mathrm{V/m} = \mathrm{J/(C\,m)} = \mathrm{N\,m/(C\,m)} = \mathrm{N/C}$ — one line of algebra, worth reproducing once so you never doubt it. N/C advertises the definition (force per charge); V/m advertises the connection to potential (how steeply the voltage drops per metre), which becomes natural in the next chapter. Boards accept both; engineers overwhelmingly say V/m because voltmeters are what they read.