Electric field

Stand near a charged balloon and you feel nothing directly — no visible tether connects it to your hair standing on end. Yet something real fills the space around that balloon, something that will push or pull on any charge unlucky (or lucky) enough to wander into it. Physicists call that "something" the electric field, and it is one of the most productive ideas in all of physics: instead of asking "what force will charge A exert on charge B," we ask a subtler question first — "what has charge A already done to the space around it?" Once that question is answered, the force on any charge placed anywhere nearby follows in one line, without ever re-deriving Coulomb's law from scratch. This shift in viewpoint — from force-between-two-charges to field-filling-space — looks like bookkeeping at first, but it is the seed of an entire worldview. It is the same idea that later lets electromagnetic waves carry energy through empty space with no charges anywhere in sight, the idea Maxwell needed to unify electricity, magnetism and light. In this chapter you build that idea from the ground up: define the electric field precisely using an imaginary, vanishingly small test charge, work out exactly what field a single point charge produces, learn to read and draw field lines, and see how, when several charges act together, their fields simply add. By the end you should be able to compute the electric field of a point charge at any distance, judge its direction correctly for both positive and negative sources, and use $\vec F = q\vec E$ to find the force on any charge dropped into a known field — the working toolkit for everything electrostatic that follows in this chapter and the next.

Key takeaways

Electric field is force per unit test charge, $\vec E = \vec F/q_0$ , measured with an infinitesimally small positive test charge $q_0$ so it does not disturb the source.
A point charge $q$ produces a field $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ at distance $r$ , directed radially outward from a positive $q$ and radially inward toward a negative $q$ .
The field exists independently of any test charge — it is a property of space set up by the source charge, present whether or not anything is placed there to feel it.
Once $\vec E$ is known, the force on any charge $q$ dropped into it is simply $\vec F = q\vec E$ — no need to redo Coulomb's law for every new charge.
Field lines are a visual map: they start on positive charges, end on negative charges, never cross, and their spacing shows field strength — dense lines mean a strong field.
The SI unit of $\vec E$ is newton per coulomb (N/C), numerically equal to volt per metre (V/m), a link that becomes natural once potential is introduced in the next chapter.
Key idea: the electric field E = F/q0 at a point is what a source charge does to the space around it, so once E is known anywhere the force on any charge dropped there is just F = qE, and for a single point charge the field falls off as E = kq/r² directed radially away from a positive source and toward a negative one.

The core insight

The electric field at a point is the force per unit positive test charge placed there, so for a single point charge $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ directed radially outward from positive charge (inward toward negative charge), and once you know $\vec E$ anywhere, the force on any charge $q$ placed there follows at once from $\vec F = q\vec E$ .

Where it fits in the syllabus

Electric field sits squarely inside the CBSE Class 12 Physics chapter "Electric Charges and Fields" (Unit I: Electrostatics) and is fully part of the 2025-26 prescribed theory syllabus — nothing here has been rationalised out. It follows directly from Coulomb's law and the superposition principle covered just before it, and it feeds forward into everything electrostatic that comes after: the field of a system of charges and a dipole (the next node in this chapter), electric flux and Gauss's law (later in this same chapter), and electric potential (the following chapter). This section keeps to the field of a single point charge, its direction, and field-line pictures — the field due to multiple charges together, where the superposition principle is actually applied, is the dedicated next topic.

The physical picture

Before writing any formula, fix the picture. A source charge $q$ sits at some point in space. To "test" what it has done to a nearby point $P$ , imagine bringing in a tiny positive charge $q_0$ — so tiny that it cannot itself shove the source charge around — and measuring the force $\vec F$ it feels there. Dividing that force by the size of the test charge gives a number that no longer depends on how big the test charge happened to be; it depends only on the source and on the location of $P$ . That ratio is the electric field:

\vec E = \dfrac{\vec F}{q_0}

where: $\vec F$ is the electrostatic force on the test charge $q_0$ placed at $P$ , and $\vec E$ is the field at $P$ due to whatever source charge(s) are present. The physical reading is that $\vec E$ is a property of the point $P$ and the source charge — not of the test charge — because a bigger test charge would feel a proportionally bigger force, leaving the ratio unchanged. Strictly, the definition is a limit, $\vec E = \lim_{q_0\to 0}\vec F/q_0$ , taken as $q_0$ shrinks toward zero so it never disturbs the source. In practice, for board-level problems, any small, known test charge is treated as good enough. Now put a single point charge $q$ at the origin and ask what field it creates at distance $r$ . Coulomb's law gives the force on a test charge $q_0$ placed there as $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{qq_0}{r^2}$ ; dividing by $q_0$ removes the test charge entirely and leaves the field of the point charge on its own:

E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}

where: $q$ is the source charge, $r$ is the distance from it to the point where the field is being evaluated, and $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2/\text{C}^2$ is the same Coulomb constant used for the force law. The physical reading: the field magnitude falls as the inverse square of distance, exactly like the force, because it is the force with the test charge divided out — double the distance and the field drops to a quarter, treble it and the field drops to a ninth. Direction is fixed by the sign of the source: a positive $q$ pushes an imagined positive test charge away, so its field points radially outward; a negative $q$ pulls a positive test charge in, so its field points radially inward, toward the charge.

Figure

Field lines radiating outward from a positive point charge on the left and converging inward toward a negative point charge on the right, with arrowheads showing direction and closer spacing near each charge showing a stronger field there

Reading field-line pictures. Field lines are a drawing convention, not physical threads, but the rules for drawing them carry real information. They begin on positive charges (or come in from infinity) and end on negative charges (or go out to infinity); they never cross each other, because a crossing would mean the field pointed in two directions at once at that point; and their local spacing shows field strength — lines drawn close together mean a strong field, lines spread far apart mean a weak one, which is exactly why field lines around a point charge crowd close to it and thin out with distance, matching the inverse-square fall-off. A uniform field, produced later in this chapter by an infinite charged sheet, is drawn as equally spaced parallel lines, signalling equal strength everywhere.

Why bother with a field at all. The payoff of defining $\vec E$ is that it turns a two-body problem into a one-body problem asked twice: first find the field due to the source(s), once, then for any charge $q$ later placed anywhere in that field, the force follows immediately as $\vec F = q\vec E$ — no repeated Coulomb's-law bookkeeping for every new charge that wanders through. When several source charges are present, this same field can be built up charge by charge using the superposition principle: the total field at a point is the vector sum of the fields each source charge would produce alone. Working that superposition out in full for two or more charges together, however, is the job of the next topic in this chapter — here the goal is simply to nail down what the field of one charge looks like and means.

Derivation AElectric field of a single point charge, from Coulomb's law

The goal is to show, starting from nothing but Coulomb's law and the definition of the electric field as force per unit test charge, that the field of an isolated point charge $q$ at distance $r$ must be $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ , directed radially outward for positive $q$ and radially inward for negative $q$ — the single formula this entire chapter is built on, and the one every later field calculation (dipole, line charge, sheet, shell) ultimately reduces back to.

Step 1. Set up the source and a test point. Place the source charge $q$ at the origin, and pick any point $P$ at distance $r$ from it. Imagine bringing a small positive test charge $q_0$ to $P$ so its presence can be used, and then discarded, to probe the field there.

Step 2. Write the Coulomb force on the test charge. Since the two charges $q$ and $q_0$ are separated by distance $r$ , Coulomb's law gives the magnitude of the force on $q_0$ as $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q\,q_0}{r^2}$ , directed along the line joining the two charges — away from $q$ if both charges are positive (like charges repel), toward $q$ if $q$ is negative and $q_0$ positive (unlike charges attract).

Step 3. Apply the definition of the field. By definition, $E = F/q_0$ . Substituting the force from Step 2, $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q\,q_0}{r^2\,q_0}$ . The test charge $q_0$ cancels exactly, top and bottom, leaving a result that no longer contains $q_0$ at all: $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ .

Step 4. Fix the direction from the sign of $q$ , not from $q_0$ . Because $q_0$ was taken positive by convention, the direction found for the force in Step 2 is already the direction of $\vec E$ itself: radially outward from $q$ when $q$ is positive, radially inward (toward $q$ ) when $q$ is negative. This is the whole point of insisting the test charge be positive — it fixes a single, unambiguous sign convention for the direction of the field, so that $\vec E$ can be quoted once and reused for any actual charge placed there later, positive or negative, via $\vec F = q\vec E$ .

Step 5. Confirm the cancellation makes physical sense. That $q_0$ drops out entirely is not a coincidence of algebra — it is exactly what "field is independent of the test charge" requires. A test charge twice as large would feel twice the force in Step 2, but dividing by a test charge twice as large in Step 3 gives back the identical $E$ . The field $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ is therefore a property of $q$ and $r$ alone, exactly as the electric-field concept promised.

Worked examples

Board level - field of a point charge and the force it exerts

A point charge $q = +5\ \mu\text{C}$ is placed at the origin. Find the electric field at a point $P$ located $0.60\ \text{m}$ from the charge, and the force this field would exert on a test charge of $+2\ \text{nC}$ placed at $P$ .

Step 1. Write down the known values in SI units: $q = 5\times10^{-6}\ \text{C}$ , $r = 0.60\ \text{m}$ , so $r^2 = 0.36\ \text{m}^2$ , and use $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2/\text{C}^2$ .

Step 2. Substitute into $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ : $E = 9\times10^9 \times \dfrac{5\times10^{-6}}{0.36} = \dfrac{4.5\times10^4}{0.36} = 1.25\times10^5\ \text{N/C}$ .

Step 3. Fix the direction: since $q$ is positive, $\vec E$ at $P$ points radially away from the origin — outward, along the line joining the origin to $P$ .

Step 4. Find the force on the test charge using $\vec F = q_0\vec E$ , with $q_0 = 2\times10^{-9}\ \text{C}$ : $F = 2\times10^{-9}\times 1.25\times10^5 = 2.5\times10^{-4}\ \text{N}$ , in the same direction as $\vec E$ , since the test charge is positive.

Answer: $E = 1.25\times10^5\ \text{N/C}$ pointing radially outward from the source charge, and the force on the $+2\ \text{nC}$ test charge at $P$ is $2.5\times10^{-4}\ \text{N}$ in that same outward direction.

Competitive level - inverse-square scaling and reading a field backward for the source charge

At a point $0.30\ \text{m}$ from a point charge, the field is measured as $E_1 = 4\times10^4\ \text{N/C}$ . (a) Find the magnitude of the source charge. (b) Without recomputing from scratch, find the field $E_2$ at $r = 0.60\ \text{m}$ and $E_3$ at $r = 1.20\ \text{m}$ from the same charge, using inverse-square scaling alone.

Step 1. Find the charge from $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ , solved for $q$ : $q = \dfrac{E_1 r^2}{9\times10^9}$ , with $E_1 = 4\times10^4\ \text{N/C}$ and $r^2 = 0.09\ \text{m}^2$ . So $q = \dfrac{4\times10^4\times 0.09}{9\times10^9} = \dfrac{3600}{9\times10^9} = 4\times10^{-7}\ \text{C} = 0.4\ \mu\text{C}$ .

Step 2. For part (b), notice the distance doubles from $0.30\ \text{m}$ to $0.60\ \text{m}$ , and the field obeys $E \propto 1/r^2$ . Doubling $r$ divides $E$ by $2^2 = 4$ , without needing $q$ at all: $E_2 = E_1/4 = 4\times10^4/4 = 1\times10^4\ \text{N/C}$ .

Step 3. Going from $0.30\ \text{m}$ to $1.20\ \text{m}$ multiplies the distance by $4$ , so the field is divided by $4^2 = 16$ : $E_3 = E_1/16 = 4\times10^4/16 = 2.5\times10^3\ \text{N/C}$ . Both answers can be checked the long way through $q = 4\times10^{-7}\ \text{C}$ and $E = 9\times10^9\, q/r^2$ — they agree.

Answer: $q = 0.4\ \mu\text{C}$ ; $E_2 = 1\times10^4\ \text{N/C}$ at $0.60\ \text{m}$ ; $E_3 = 2.5\times10^3\ \text{N/C}$ at $1.20\ \text{m}$ . Timing note: under a minute once the pattern is seen. Whenever a question only changes the distance and asks for a new field value, resist the urge to recompute $q$ and redo the full formula — read off the distance ratio, square it, and scale. Doubling distance divides the field by $4$ ; tripling divides it by $9$ ; halving multiplies it by $4$ . This shortcut is worth marks on its own in objective/MCQ sections and saves real time in multi-part numericals.

Common traps

Confusing field with force. The electric field $\vec E$ is a property of space set up by the source charge — it exists at every point whether or not a test charge is actually there. The force $\vec F = q\vec E$ only appears once some particular charge $q$ is placed in that field. Losing marks here usually means writing $E$ where $F$ (or vice versa) was asked for.
Forgetting the field is a vector. Direction is as important as magnitude: radially outward for a positive source, radially inward for a negative source. A numerical answer with the wrong direction, or no stated direction at all, is an incomplete answer on the board exam.
Misapplying inverse-square scaling. $E \propto 1/r^2$ scales with the ratio of distances, not their difference — going from $r$ to $2r$ quarters the field, not halves it, and going from $r$ to $3r$ divides it by nine, not three.
Using a "large" test charge in reasoning. The formal definition needs $q_0 \to 0$ , small enough that it does not itself disturb the source charge's position or distribution; a genuinely large test charge would change the very thing being measured.
Mixing up N/C and V/m casually. The two units for $E$ are numerically equal, but a student should know why — that equivalence comes from the relation between field and potential, only introduced in the next chapter — rather than treating it as an arbitrary fact to memorise blindly.

Common trapElectric field is not the same thing as the force it can produce

The single most common source of lost marks in this topic is treating $\vec E$ and $\vec F$ as interchangeable. The electric field at a point is entirely determined by the source charge(s) and the location of that point — it is there, with a definite magnitude and direction, regardless of whether any other charge is nearby to feel it. The force is what happens only if a charge $q$ is actually placed at that point, and its value depends on $q$ as well as on $\vec E$ : double the charge placed there and the force doubles, but the field itself, set up by the source, is completely unchanged. When a question gives a field and a charge and asks for "the force," multiply; when it gives a charge and a force and asks for "the field," divide; and never quote a field value with force units, or a force value with field units — the two are related by $\vec F = q\vec E$ , not identical to each other.

You should be able to…

State the definition of electric field as force per unit positive test charge, $\vec E = \vec F/q_0$ , and give its SI unit.
Derive and apply the formula $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ for the field of a single point charge.
Correctly assign the direction of the field due to a positive or a negative point charge.
Use $\vec F = q\vec E$ to find the force on any charge placed in a known field, and use it in reverse to find an unknown field or charge.
Apply inverse-square scaling directly, without recomputing the source charge, when only the distance changes.
Read and interpret a field-line diagram: where lines start and end, why they never cross, and what their spacing means.

Ask the tutor: Want to work through how the field of two point charges combines by superposition, or see the field-line picture for a dipole before the next topic gets there? Ask the AI tutor to walk through it with a diagram.

Why do we even need the idea of a field — why not just use Coulomb's law directly every time?

Coulomb's law works fine for one pair of charges, but real problems usually involve one source charge (or a whole cloud of them) acting on many different test locations, or many different charges wandering through the same region one after another. Defining the field once, as a property of space, means you do that calculation a single time; after that, the force on any charge dropped anywhere in that field is just $F = qE$ , a one-line multiplication. It also generalises cleanly to continuous charge distributions and, much later in physics, to regions where a field exists with no charges present at all — something a force-between-two-charges picture cannot describe.

Does the electric field only exist when there is a test charge sitting there to feel it?

No — this is the trap the "vanishingly small test charge" language is meant to head off. The field is defined using a test charge, as a conceptual and mathematical device to measure it, but the field itself is a property of the source charge and the point in space, present continuously whether or not anything is actually placed there. Think of it like a contour map of altitude: the map exists and has a definite value at every point even if nobody ever climbs there to check.

Why must the test charge be positive, and does that make the field itself somehow "positive"?

The test charge is taken positive purely as a sign convention, so that direction can be assigned unambiguously and consistently: the field points the way the force on a positive test charge would point. It does not mean the field is a positive quantity in any deeper sense — a negative source charge produces a perfectly real field, it simply points inward, toward the charge, because a positive test charge placed there would be pulled in that direction. Once the convention is fixed, the force on any actual charge, positive or negative, is found from $\vec F = q\vec E$ , and the sign of $q$ then correctly flips the force direction for negative charges.

If two point charges are present, is the total field just the field of one plus the field of the other?

Yes, in principle — that is the superposition principle, and it is what makes the field concept so powerful for more complicated charge arrangements: work out the field due to each source charge alone, as if the others were not there, then add the results as vectors (not just their magnitudes) to get the net field. This section has focused on the field of a single isolated point charge to get that building block exactly right; combining two or more charges through superposition, including the dipole configuration, is worked out in detail in the very next topic.