Electric Field vs Electric Force: The Difference

An electric force is what one charge exerts on another, measured in newtons. An electric field is what a charge sets up in the space around it, measured in newtons per coulomb. The force needs two charges to exist. The field needs one, and it is still there when nothing is present to feel it.

AP Physics: Unit 10 (topics 10.1 Electric Charge and Electric Force, 10.3 Electric Fields). This pair spans AP Physics 2 Topic 10.1 (Electric Charge and Electric Force) and Topic 10.3 (Electric Fields), both in Unit 10, Electric Force, Field, and Potential, weighted at 15 to 18 percent of the multiple-choice section over a suggested 14 to 21 class periods. The definitional statement is EK 10.3.A.2: the electric field at a given point is the ratio of the electric force exerted on a test charge at that point to the charge of the test charge, with the relevant equation E = F_E/q. Three sub-points carry the rest of the distinction. EK 10.3.A.2.i defines a test charge as a point charge of small enough magnitude that its presence does not significantly affect an electric field in its vicinity. EK 10.3.A.2.ii fixes the field direction to the source: away from isolated positive charges and toward isolated negative charges. EK 10.3.A.2.iii fixes the force direction relative to the field: the electric force exerted on a positive test charge is in the same direction as the electric field, which is why a negative charge is pushed the other way. A boundary statement on Topic 10.3 limits electric field calculations to four or fewer charged objects or systems, with more allowed in situations of high symmetry, and restricts analysis of fields within insulators to qualitative treatment. AP Physics C: Electricity and Magnetism carries the same pair as Topic 8.1 and Topic 8.3 in Unit 8, weighted at 15 to 25 percent, with EK 8.3.A.2.i repeating the test-charge definition word for word.

The distinction, stated once

An electric force is an interaction. It takes two charged objects, it acts on one of them, and it is measured in newtons. Remove either object and there is no force to talk about.

An electric field is a property of a region of space. One charged object is enough to create it, it has a value at every point around that object, and it is measured in newtons per coulomb. Remove everything you might have placed in it and the field is unchanged, because the field was never about the thing you placed there.

The two are connected by one equation, which the AP Physics 2 sheet prints as a definition:

\vec{E} = \frac{\vec{F}_E}{q}

EK 10.3.A.2 states it in words: the electric field at a given point is the ratio of the electric force exerted on a test charge at that point to the charge of the test charge.

Read the direction of that equation carefully, because it is the source of the whole confusion. You do not compute the field by first finding a force. You use a force measurement to define what the field is, and the ratio comes out the same whichever probe you use, which is the only reason the ratio deserves a name of its own. After that the equation gets used backwards, as $\vec{F}_E = q\vec{E}$ : the field is known, you drop a charge into it, and the force follows.

Side by side

	Electric force $\vec{F}_E$	Electric field $\vec{E}$
What it is	An interaction between two charges	A condition of the space around one charge
Unit	Newton (N)	Newton per coulomb (N/C), the same as V/m
Scalar or vector	Vector	Vector
How many charges it needs	Two	One, the source
Exists with nothing there?	No	Yes
Point-source formula	$\lvert \vec{F}_E \rvert = k \lvert q_1 q_2 \rvert / r^2$	$\lvert \vec{E} \rvert = k \lvert q \rvert / r^2$
Doubling the placed charge	Doubles the force	Changes nothing
Doubling the source charge	Doubles the force	Doubles the field
Direction for a positive charge	Along the field	Set by the source: away from positive, toward negative
Direction for a negative charge	Opposite the field	Unchanged
Converted into the other by	Dividing by the placed charge	Multiplying by the placed charge

The exists with nothing there row is the definitional split and the reason the field concept was invented at all. A charge does not reach across empty space to grab another one; it changes the space, and the space then acts on whatever is in it. That is a physical claim rather than bookkeeping, and it becomes visible the moment you shake the source charge and the change propagates outward at the speed of light rather than arriving instantly.

The two direction rows are where the marks go, and the next-to-last section deals with them.

One row deserves care. The two point-source formulas differ by one factor of charge, so they look nearly identical and are easy to swap. The force equation carries $\lvert q_1 q_2 \rvert$ , the product of both charges. The field equation carries $\lvert q \rvert$ , the source charge only. If your field answer contains the charge you placed at the point, you have computed a force and divided by nothing.

The case that separates them: change the probe, then take it away

Fix a source charge $Q = +6.0 \ \mathrm{nC}$ at the origin and look at a point P, $0.30 \ \mathrm{m}$ away. The field there is $E = kQ/r^2 = 600 \ \mathrm{N/C}$ , pointing away from the source, and that number is now settled.

Now put things at P, one at a time.

What is at P	Field at P	Force on it
Nothing	$600 \ \mathrm{N/C}$ , away from $Q$	None. There is nothing to push
$+1.0 \ \mathrm{nC}$	$600 \ \mathrm{N/C}$ , away from $Q$	$6.0 \times 10^{-7} \ \mathrm{N}$ , away from $Q$
$+3.0 \ \mathrm{nC}$	$600 \ \mathrm{N/C}$ , away from $Q$	$1.8 \times 10^{-6} \ \mathrm{N}$ , away from $Q$
A proton	$600 \ \mathrm{N/C}$ , away from $Q$	$9.6 \times 10^{-17} \ \mathrm{N}$ , away from $Q$
An electron	$600 \ \mathrm{N/C}$ , away from $Q$	$9.6 \times 10^{-17} \ \mathrm{N}$ , toward $Q$

One column never moves. The other takes five different answers, one of which is "there is no force" and one of which points the opposite way.

The first row is the one that settles the argument. With nothing at P, the force is not zero, it is not defined: a force is an interaction, and there is nothing there to interact with. The field, meanwhile, has a perfectly definite value, and you could go and measure it a second later with a probe you had not yet brought. That asymmetry is why a field is a real object in the theory rather than a convenient shorthand for a force you have not computed yet.

Why the test charge has to be small

The definition $\vec{E} = \vec{F}_E/q$ has a condition attached, and AP states it. EK 10.3.A.2.i: a test charge is a point charge of small enough magnitude such that its presence does not significantly affect an electric field in its vicinity.

That clause is not decoration. Imagine measuring the field just outside a charged metal sphere with a probe carrying a full microcoulomb. Your probe has its own field, and that field pushes the mobile charges on the sphere's surface around. The source has been rearranged by the act of measuring it, so the force you read, divided by your probe charge, is not the field that was there before you arrived.

Shrink the probe and the disturbance shrinks with it. In the limit the ratio $F_E/q$ stops depending on the probe at all, and only then is it fair to call it a property of the point. The worked example below tests exactly that: three different probes at one location, three different forces, one field.

Two related idealisations get mixed up here, and the CED keeps them apart. A point charge is small in size, so that a single distance $r$ is unambiguous. A test charge is a point charge that is also small in effect. Every test charge is a point charge; the reverse is not true, and a question that says "point charge" has not told you the probe is negligible.

Direction: the same arrow for a positive charge, the opposite one for a negative charge

Here is the sign error students actually make, stated so you can check for it.

The field direction belongs to the source. EK 10.3.A.2.ii: an electric field points away from isolated positive charges and toward isolated negative charges. Nothing you put in the field changes that arrow.

The force direction depends on the sign of the charge you put in. EK 10.3.A.2.iii: the electric force exerted on a positive test charge by an electric field is in the same direction as the electric field. Flip the sign of the placed charge and $\vec{F}_E = q\vec{E}$ flips the force while leaving $\vec{E}$ alone.

So four combinations exist, and all four appear on exams:

Source charge	Charge placed in the field	Field at the point	Force on the placed charge
Positive	Positive	Away from the source	Away from the source (repelled)
Positive	Negative	Away from the source	Toward the source (attracted)
Negative	Positive	Toward the source	Toward the source (attracted)
Negative	Negative	Toward the source	Away from the source (repelled)

Column three depends only on column one. Column four depends on both. Reading the field arrow off a diagram and copying it as the force arrow works for a proton and fails for an electron, which is why an electron released in a field is the standard way of testing this.

There is a second sign convention that quietly rides along. On both the AP Physics 2 and the AP Physics C: E&M sheets, the point-source force and field equations are printed with absolute value bars: $\lvert \vec{F}_E \rvert = k \lvert q_1 q_2 \rvert / r^2$ and $\lvert \vec{E} \rvert = k \lvert q \rvert / r^2$ . Both return magnitudes. The signs are not meant to come out of the algebra; you put the numbers in as magnitudes and reason the direction separately, from the table above.

When it costs a mark

Answering a field question in newtons. Or a force question in newtons per coulomb. A grader scanning the units line marks this without checking the working.

Putting both charges into the field formula. $E = kq_1q_2/r^2$ is not an equation. The field of a source is set by the source, so only one charge belongs in it. This one is easy to catch by units: $kq_1q_2/r^2$ carries newtons, not newtons per coulomb.

Copying the field arrow onto a negative charge. The single most productive error in this pair. Write $\vec{F}_E = q\vec{E}$ with the sign of $q$ attached, then say in words which way that points, rather than reading it off the diagram.

Asking what the field is "on" a charge. Fields are not exerted on things; forces are. A question asking for "the field on the electron" is asking for the field at the electron's location, which the electron plays no part in creating and does not experience as a field. Fixing the language fixes the physics.

Treating a zero force as a zero field. Put a neutral object at a point and the electric force on it is zero while the field is whatever it was. The field only shows up as a force once there is charge to act on.

Forgetting that the source keeps its own field. A charge does not feel its own field. When two charges sit near each other, the force on charge 1 comes from the field that charge 2 creates at charge 1's location, and only that.

When they look like the same thing

For one fixed probe charge they are proportional, and that is what makes the confusion survive. Follow a single proton through a whole multi-part question and every force is exactly $1.60 \times 10^{-19}$ times the field at that spot. The graphs have the same shape, the zeros are in the same places, the ranking of points is identical. Nothing in that problem forces you to notice they are different quantities, and most single-particle problems are like this.

They also share the same geometry. Both fall off as $1/r^2$ from a point source, both obey superposition as vectors, and both point along the same line. The diagram you would draw for one is the diagram you would draw for the other with the arrows relabelled.

What breaks the proportionality is any question that changes the charge at the point, or removes it, or makes it negative. Those are the three question types that separate the pair, and they are the three that appear.

A fourth situation is worth flagging because it looks like a coincidence and is not. A field of $1 \ \mathrm{N/C}$ is the same thing as a field of $1 \ \mathrm{V/m}$ , so a field can be quoted in units that mention no force at all. That is the potential side of the field's personality, covered in electric field vs electric potential.

How the exam frames it, and what to read next

In AP Physics 2 the force comes first, in Topic 10.1, and the field follows in Topic 10.3. Both sit in Unit 10, weighted at 15 to 18 percent of the multiple-choice section. The order is the historical one and the pedagogical one: Coulomb's law is a statement about two objects, and the field is what you get when you divide one of them out.

A boundary statement on Topic 10.3 limits field calculations to four or fewer charged objects or systems, with more allowed in situations of high symmetry, and restricts fields inside insulators to qualitative analysis. In AP Physics C: Electricity and Magnetism the same two ideas are Topic 8.1 and Topic 8.3 inside Unit 8, weighted at 15 to 25 percent, and the course then extends the field to continuous charge distributions with an integral.

For the force procedure worked end to end, including attraction and repulsion and the inverse-square scaling, use the Coulomb's law guide, or the Coulomb's law calculator for the arithmetic. For the field procedure and its scalar partner, use the electric field and potential guide. If the thing in front of you is a diagram rather than a number, electric field lines covers what the picture is allowed to tell you.

One field, a proton and an electron

A source charge $Q = +6.0 \times 10^{-9} \ \mathrm{C}$ is fixed at the origin. Point P lies $0.30 \ \mathrm{m}$ away along the $+x$ axis. (a) Find the electric field at P. (b) Find the electric force on a proton placed at P. (c) Find the electric force on an electron placed at P. (d) State what changed between (b) and (c) and what did not.

(a) Use $\lvert \vec{E} \rvert = kQ/r^2$ with $k = 9.0 \times 10^9 \ \mathrm{N \cdot m^2/C^2}$ from the sheet. First $kQ = (9.0 \times 10^9)(6.0 \times 10^{-9}) = 54 \ \mathrm{N \cdot m^2/C}$ .
$\lvert \vec{E} \rvert = 54/(0.30)^2 = 54/0.090 = 600 \ \mathrm{N/C}$ . The source is positive, so the field points away from it: in the $+x$ direction at P.
(b) A proton carries $q = +1.60 \times 10^{-19} \ \mathrm{C}$ . Using $\vec{F}_E = q\vec{E}$ : $F_E = (1.60 \times 10^{-19})(600) = 9.6 \times 10^{-17} \ \mathrm{N}$ , in the $+x$ direction, since $q$ is positive.
(c) An electron carries $q = -1.60 \times 10^{-19} \ \mathrm{C}$ . The magnitude is identical, $9.6 \times 10^{-17} \ \mathrm{N}$ , but the sign of $q$ reverses the direction: the force is in the $-x$ direction, back toward the source charge.
(d) The field at P was $600 \ \mathrm{N/C}$ in the $+x$ direction in every case, including the case with nothing at P. The force reversed direction, because the field arrow belongs to the source and the force arrow belongs to $q\vec{E}$ .

$E = 600 \ \mathrm{N/C}$ away from $Q$ , whatever sits at P. The proton is pushed away with $9.6 \times 10^{-17} \ \mathrm{N}$ ; the electron is pulled toward $Q$ with the same $9.6 \times 10^{-17} \ \mathrm{N}$ .

Two sources, then a charge dropped in

Charges $q_1 = +8.0 \times 10^{-9} \ \mathrm{C}$ at $x = 0$ and $q_2 = -2.0 \times 10^{-9} \ \mathrm{C}$ at $x = 0.30 \ \mathrm{m}$ lie on the $x$ -axis. (a) Find the net electric field at $x = 0.10 \ \mathrm{m}$ . (b) Find the electric force on an electron placed there.

(a) Distances from the point of interest: $r_1 = 0.10 \ \mathrm{m}$ from $q_1$ , and $r_2 = 0.30 - 0.10 = 0.20 \ \mathrm{m}$ from $q_2$ .
Field from $q_1$ : $\lvert \vec{E}_1 \rvert = (9.0 \times 10^9)(8.0 \times 10^{-9})/(0.10)^2 = 72/0.010 = 7200 \ \mathrm{N/C}$ , pointing away from the positive charge, so in the $+x$ direction.
Field from $q_2$ : $\lvert \vec{E}_2 \rvert = (9.0 \times 10^9)(2.0 \times 10^{-9})/(0.20)^2 = 18/0.040 = 450 \ \mathrm{N/C}$ , pointing toward the negative charge, which also lies in the $+x$ direction from the point.
Both point the same way, so add the magnitudes: $E_{net} = 7200 + 450 = 7650 \ \mathrm{N/C}$ in the $+x$ direction. Note that only the two source charges appear anywhere in this calculation.
(b) Now place an electron there. $\lvert \vec{F}_E \rvert = \lvert q \rvert E = (1.60 \times 10^{-19})(7650) = 1.224 \times 10^{-15} \ \mathrm{N}$ , which rounds to $1.2 \times 10^{-15} \ \mathrm{N}$ at two significant figures.
Direction: the electron is negative, so the force is opposite the field, in the $-x$ direction, back toward the positive charge.
The two-step order matters on a free response. Work out the field from the sources alone, then multiply by the placed charge. Doing pairwise Coulomb forces instead gets the same answer and takes longer, and it fails immediately if the question asks what the field is with nothing there.

$E_{net} = 7650 \ \mathrm{N/C}$ in the $+x$ direction. The force on an electron placed at that point is $1.2 \times 10^{-15} \ \mathrm{N}$ in the $-x$ direction.

Three probes, one field: why the ratio is a definition

A source charge $Q = +12 \times 10^{-9} \ \mathrm{C}$ sits at the origin. At a point $0.20 \ \mathrm{m}$ away, place in turn charges of $1.0 \ \mathrm{nC}$ , $3.0 \ \mathrm{nC}$ and $5.0 \ \mathrm{nC}$ . Find the Coulomb force in each case, then divide each force by the charge that felt it.

Compute the shared factor once: $kQ/r^2 = (9.0 \times 10^9)(12 \times 10^{-9})/(0.20)^2 = 108/0.040 = 2700$ , in units of $\mathrm{N/C}$ .
With $q = 1.0 \times 10^{-9} \ \mathrm{C}$ : $\lvert \vec{F}_E \rvert = k \lvert Qq \rvert /r^2 = 2700 \times 1.0 \times 10^{-9} = 2.7 \times 10^{-6} \ \mathrm{N}$ . Dividing back: $2.7 \times 10^{-6} / 1.0 \times 10^{-9} = 2700 \ \mathrm{N/C}$ .
With $q = 3.0 \times 10^{-9} \ \mathrm{C}$ : $\lvert \vec{F}_E \rvert = 8.1 \times 10^{-6} \ \mathrm{N}$ , and $8.1 \times 10^{-6} / 3.0 \times 10^{-9} = 2700 \ \mathrm{N/C}$ .
With $q = 5.0 \times 10^{-9} \ \mathrm{C}$ : $\lvert \vec{F}_E \rvert = 1.35 \times 10^{-5} \ \mathrm{N}$ , and $1.35 \times 10^{-5} / 5.0 \times 10^{-9} = 2700 \ \mathrm{N/C}$ .
Three forces spanning a factor of five, one ratio. The ratio survived because $q$ appears once in the numerator of Coulomb's law and once in the denominator of the definition, and cancels.
This also reconciles the two formulas. $\lvert \vec{F}_E \rvert = k \lvert Qq \rvert /r^2$ and $\lvert \vec{F}_E \rvert = qE$ with $E = kQ/r^2$ are the same statement, grouped differently. The field version has the placed charge pulled out to one side, which is exactly what lets you answer questions about the point before deciding what goes there.

The forces are $2.7 \times 10^{-6} \ \mathrm{N}$ , $8.1 \times 10^{-6} \ \mathrm{N}$ and $1.35 \times 10^{-5} \ \mathrm{N}$ , all different. The ratio of force to charge is $2700 \ \mathrm{N/C}$ in all three cases, which is why that ratio, and not the force, is the property of the point.

Frequently asked questions

What is the difference between electric field and electric force?

Electric force is the push or pull one charge exerts on another, measured in newtons, and it needs two charges to exist. Electric field is the condition a single charge creates in the space around it, measured in newtons per coulomb, and it exists at every point whether or not anything is there to feel it. They are linked by E = F/q, so the field is the force per unit charge, and the force on a charge placed in a known field is F = qE.

Does an electric field exist if there is no charge in it?

Yes. The field is created by the source charge and fills the space around it. If nothing is placed at a point, the field at that point still has a definite magnitude and direction; what does not exist is a force, since a force is an interaction and there is nothing to interact with. This is the sharpest test of the difference: remove the object at the point and one quantity keeps its value while the other stops being defined.

Why is E = F/q a definition and not just a rearrangement?

Because the ratio comes out the same for every probe. Place a 1 nC charge, a 3 nC charge and a 5 nC charge at the same point near a source and you get three different forces, but dividing each force by the charge that felt it gives one number every time. That number depends only on the source and the location, which is what makes it a property of the point. The AP CED adds the condition that keeps this true: the test charge must be small enough that its presence does not significantly affect the field it is measuring.

Is the electric force always in the same direction as the electric field?

Only for a positive charge. EK 10.3.A.2.iii states that the electric force on a positive test charge is in the same direction as the electric field. For a negative charge, F = qE with a negative q reverses the force, so it points opposite the field. The field arrow itself never changes: it is set by the source charges, pointing away from positive and toward negative, and it does not care what you place in it.

What are the units of electric field and electric force?

Electric force is measured in newtons. Electric field is measured in newtons per coulomb, which is exactly the same unit as volts per metre. The units are a quick check on which quantity you have computed: if your answer to a find-the-field question came out in newtons, you found a force and forgot to divide by the charge that felt it.

Why does the test charge have to be small?

Because a large probe changes the thing it is measuring. Bring a big charge up to a charged conductor and its own field pushes the source charges around on the conductor's surface, so the force you then measure, divided by your probe charge, is not the field that was there before you arrived. AP Physics 2 EK 10.3.A.2.i defines a test charge as a point charge of small enough magnitude that its presence does not significantly affect the electric field in its vicinity, and the identical wording appears in the AP Physics C: E&M framework.

Do both charges go into the electric field formula?

No. Only the source charge does. The field of a point charge is the magnitude of E equals k times the magnitude of q over r squared, with one charge in it. Coulomb's law for the force carries the product of both charges. If a charge you placed at the point shows up in your expression for the field there, you have written down a force. Checking the units catches this instantly, since the force expression carries newtons rather than newtons per coulomb.