Electric Field vs Electric Potential: The Difference

Electric field is force per unit charge, a vector in newtons per coulomb. Electric potential is energy per unit charge, a scalar in volts. The field is the rate at which potential changes with distance, not a ratio, so a point can have zero field with a large potential, or the reverse.

AP Physics: Unit 10 (topics 10.3 Electric Fields, 10.5 Electric Potential). This pair spans AP Physics 2 Topic 10.3 (Electric Fields) and Topic 10.5 (Electric Potential), 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. EK 10.3.A.2 defines the electric field at a point as the ratio of the electric force exerted on a test charge there to the charge of the test charge, with the relevant equation E = F_E/q, and EK 10.3.A.3 records that the field is a vector quantity representable by vector field maps. EK 10.5.A.1 defines electric potential as the electric potential energy per unit charge at a point in space. Learning objective 10.5.B is devoted to the relationship between the two: EK 10.5.B.1 gives the average field as the potential difference divided by the distance, EK 10.5.B.2.iii states that an electric field vector points in the direction of decreasing potential, and EK 10.5.B.2.iv states that there is no component of an electric field along an isoline. Two boundary statements cap the arithmetic at four or fewer charged objects for the field and four or fewer particles for the potential, with more allowed in situations of high symmetry, and field analysis inside insulators is qualitative only. In AP Physics C: Electricity and Magnetism the same pair spans Unit 8 (Electric Charges, Fields, and Gauss's Law, 15 to 25 percent) and Unit 9 (Electric Potential, 10 to 20 percent), where the link is printed as both a derivative and an integral.

The distinction, stated once

Divide by the charge twice, in two different places, and you get these two quantities.

Divide the force on a charge by that charge and you get the electric field, $\vec{E} = \vec{F}_E / q$ . A force is a vector, so the field is a vector. Its unit is the newton per coulomb.

Divide the energy of a charge by that charge and you get the electric potential, $V = U_E / q$ . Energy is a scalar, so potential is a scalar. Its unit is the volt, which is a joule per coulomb.

That is the whole split, and everything below is a consequence of it. The AP Physics 2 CED states each half directly: EK 10.3.A.2 defines the electric field at a point as the ratio of the electric force exerted on a test charge there to the charge of that test charge, and EK 10.5.A.1 says electric potential describes the electric potential energy per unit charge at a point in space.

What trips people is the third fact, which is that the two are not a ratio of each other. There is no $E = V/q$ and no $V = Eq$ . The link between them is a rate of change with position, and for a point charge that produces $E = kq/r^2$ against $V = kq/r$ : one power of $r$ apart, which is exactly what differentiating does. Two quantities that differ by a derivative can be zero in completely different places, and that is what the rest of this page is about.

Side by side

	Electric field $\vec{E}$	Electric potential $V$
Defined as	Electric force per unit charge	Electric potential energy per unit charge
Symbol and unit	$\vec{E}$ , in N/C (same as V/m)	$V$ , in volts (same as J/C)
Scalar or vector	Vector	Scalar, signed
Point-charge formula	$\lvert \vec{E} \rvert = k \lvert q \rvert / r^2$	$V = kq/r$
Falls off as	$1/r^2$	$1/r$
Several charges add by	Vector superposition, with components	Signed scalar addition, no angles
What the sign of $q$ does	Sets the direction of the arrow	Sets the sign of the number
Gives you, once multiplied by $q$	The force, $\vec{F}_E = q\vec{E}$	The energy, $U_E = qV$
Zero reference	No reference needed; zero means no arrow	Zero taken at infinite distance, by AP exam convention
Represented by	Vector field maps and field line diagrams	Equipotential lines, also called isolines

Two rows carry most of the weight.

The superposition row is the one that changes your working. Fields from several charges must be resolved into components and added as vectors; potentials are added as plain signed numbers. That is why potential is usually the faster of the two to compute, and it is also why the two can cancel at completely different places, which the next section works out.

The falls off as row is the fastest self-check there is. Double your distance from a point charge and the field drops to a quarter while the potential only halves. If both of your answers changed by the same factor, one of them is wrong.

For the definitional detail on each side on its own, see electric field and electric potential.

The case that separates them, part one: zero field, large potential

Put two equal positive charges of $+4.0 \ \mathrm{nC}$ at $x = 0$ and $x = 0.40 \ \mathrm{m}$ , and stand at the midpoint.

Each charge is $0.20 \ \mathrm{m}$ away, so each produces a field of $900 \ \mathrm{N/C}$ . They point in opposite directions, one toward $+x$ and one toward $-x$ , and they cancel exactly. The net field at the midpoint is zero.

The potentials do not cancel, because they are not vectors and there is nothing to cancel with. Each charge contributes $+180 \ \mathrm{V}$ , and the total is $+360 \ \mathrm{V}$ .

So the midpoint has $E = 0$ and $V = 360 \ \mathrm{V}$ . A charge released from rest there feels no force and stays put, and yet it holds real potential energy: $U_E = qV$ is not zero. Nudge it off the midpoint along the line and it will be pushed away, converting that energy into kinetic energy.

Read that last paragraph again if the pair still feels interchangeable. Zero field means nothing is pushing. It does not mean nothing is stored.

The case that separates them, part two: zero potential, large field

Now flip one charge. Put $+5.0 \ \mathrm{nC}$ at $x = 0$ and $-5.0 \ \mathrm{nC}$ at $x = 0.80 \ \mathrm{m}$ , and stand on the perpendicular bisector, at the point $(0.40, 0.30)$ in metres.

That point is $0.50 \ \mathrm{m}$ from each charge, because $0.40$ , $0.30$ and $0.50$ form a right triangle. Equal distances and equal magnitudes mean the two potentials are $+90 \ \mathrm{V}$ and $-90 \ \mathrm{V}$ , and they sum to exactly zero. Every point on that perpendicular bisector has $V = 0$ , for the same reason.

The fields do not sum to zero. Each has magnitude $180 \ \mathrm{N/C}$ . The one from the positive charge points away from it, up and to the right; the one from the negative charge points toward it, down and to the right. Their vertical components are equal and opposite and cancel; their horizontal components are equal and both point in $+x$ . The net field is $288 \ \mathrm{N/C}$ , parallel to the line joining the charges, running from the positive charge toward the negative one.

So this point has $V = 0$ and $E = 288 \ \mathrm{N/C}$ . A charge released from rest there is shoved sideways immediately.

Put the two cases together and the pair is finished:

Situation	Net $E$	Net $V$
Midpoint of two equal positive charges	Zero	$+360 \ \mathrm{V}$
Perpendicular bisector of two opposite charges	$288 \ \mathrm{N/C}$	Zero

Neither zero implies the other. That table is the answer to the question this page is named after, and it is the shape of the multiple-choice item that tests it.

The relation is a derivative, not a ratio

Since the field is not the potential divided by anything you can point at, what is the actual link?

The field is the negative rate at which potential changes with position. AP Physics C: Electricity and Magnetism prints the derivative form:

E_x = -\frac{dV}{dx}

and the integral that runs the other way, $\Delta V = -\int_a^b \vec{E} \cdot d\vec{r}$ . AP Physics 2 prints the magnitude version of the same statement:

\lvert \vec{E} \rvert = \left\lvert \frac{\Delta V}{\Delta r} \right\rvert

EK 10.5.B.1 states it in words: the average electric field between two points in space equals the electric potential difference between the two points divided by the distance between them. The potential gradient entry carries the derivative relation in full.

Three consequences you can use directly.

Volts per metre is newtons per coulomb. The two units are the same unit written two ways, and that equality is the relation in disguise. If a question hands you a field in V/m, it is handing you a rate of change of potential.

The field points downhill in potential. EK 10.5.B.2.iii says an electric field vector points in the direction of decreasing potential. The minus sign in $E_x = -dV/dx$ is what encodes that.

The field has no component along an equipotential. EK 10.5.B.2.iv says there is no component of an electric field along an isoline, which is why field lines cross equipotential surfaces at right angles. Nothing changes along the surface, so the rate of change along it is zero, so the field component along it is zero.

The two cases above now read differently. At the midpoint of two equal positive charges the potential is at a local minimum along the line, and the slope of a graph at its minimum is zero, so the field is zero there. On the perpendicular bisector of a dipole the potential is zero and stays zero as you move along the bisector, but it changes steeply as you step across it, and the field is that cross-wise slope. A value of zero and a slope of zero are different claims about a graph.

What each equation sheet actually prints

The four AP sheets are not the same document, so check rather than recall.

The AP Physics 2 sheet Electricity block prints the field three ways and the potential two ways:

$\vec{E} = \dfrac{\vec{F}_E}{q}$ , the definition
$\lvert \vec{E} \rvert = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\lvert q \rvert}{r^2} = k\dfrac{\lvert q \rvert}{r^2}$ , the point-charge field
$\lvert \vec{E} \rvert = \left\lvert \dfrac{\Delta V}{\Delta r} \right\rvert$ , the field from the potential
$V = \dfrac{1}{4\pi\varepsilon_0} \sum_i \dfrac{q_i}{r_i}$ , the potential of a configuration
$\Delta U_E = q\Delta V$ , the bridge to energy

Read the bars. The point-charge field line is printed with $\lvert q \rvert$ , so it returns a magnitude and you assign the direction by inspection. The potential line has no bars and is printed as a sum, so signed numbers go straight in. Two lines of a sheet, two different habits.

The AP Physics C: E&M sheet is arranged for calculus and differs in a way worth knowing before the exam. It prints Coulomb's law and $\vec{E} = \vec{F}_E/q$ , then jumps straight to the integral forms $\vec{E} = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r^2}\hat{r}$ and $V = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r}$ . There is no separate $E = kq/r^2$ line on it and no summation form of $V$ . Both follow from what is printed, and you are expected to get there yourself.

One more thing the C sheet has and the Physics 2 sheet does not: the two-way link, $\Delta V = -\int \vec{E} \cdot d\vec{r}$ going one direction and $E_x = -dV/dx$ going the other.

When it costs a mark

Five errors this confusion produces, each of which yields an answer that looks respectable.

Answering a field question in volts, or a potential question in N/C. The units line alone is enough for a grader to mark it down without checking the arithmetic. Volts for $V$ , newtons per coulomb for $E$ .

Adding potentials as though they had directions. Resolving $V$ into $x$ and $y$ components, or drawing an arrow for it, is a defect even when the number that comes out is right. Potential is a signed scalar and adds like one.

Adding field magnitudes without checking directions. The reverse error. At the midpoint of two equal positive charges, adding $900 + 900$ gives $1800 \ \mathrm{N/C}$ where the true answer is zero, because the two arrows are opposite.

Concluding that zero potential means zero field, or the reverse. Both directions are wrong, and the two cases above are counterexamples to each. This appears constantly as a ranking task or as a where-is-it-zero item.

Using the wrong power of $r$ . Writing $V = kq/r^2$ or $E = kq/r$ is a single character each way, and both give a plausible-looking number with the wrong units. The check is dimensional: $kq/r$ carries volts, $kq/r^2$ carries newtons per coulomb, and V/m equals N/C only because there is a metre of difference between them.

One phrasing trap sits underneath all five. The word "potential" appears inside "potential energy", so an unqualified "the potential at that point" gets read as an energy. Potential is per coulomb. Energy is not, and that pair has its own page.

When they track each other, and why that lulls you

Around a single isolated point charge the two never disagree in any way you would notice. Both are largest near the charge, both fall off with distance, both change sign when the source charge changes sign, and both go to zero at infinity. Move outward and both decrease. Nothing in that picture forces you to treat them as different quantities, and a first course spends a long time in exactly that picture.

Inside a parallel-plate capacitor they also travel together in a tidy way, and this is the case worth memorising. The field is uniform, so $E$ is the same everywhere between the plates, while $V$ falls steadily from one plate to the other. Constant field, changing potential. That is a straight-line graph of $V$ against position whose slope is constant, and the constant slope is the field. It is the cleanest picture of the derivative relation available, and it is on the exam because a charged particle in that uniform field undergoes constant acceleration, which EK 10.6.A.3.ii points at projectile motion.

The pair only bites in three situations, which is why those three keep coming back:

A symmetric configuration where the vectors cancel but the scalars do not, or the scalars cancel but the vectors do not.
A ranking task across several points, where the field ordering and the potential ordering come out different because of the extra power of $r$ .
A question about a region rather than a point, where the answer depends on how the potential is changing rather than on its value.

Handle those three and the distinction is done.

How the exam frames it, and what to read next

In AP Physics 2 these are Topic 10.3 and Topic 10.5, inside Unit 10, which the CED weights at 15 to 18 percent of the multiple-choice section. Learning objective 10.5.B is literally named "Describe the relationship between electric potential and electric field", so the connection has its own objective rather than being left as an aside.

Two boundary statements fence the arithmetic, and both are worth knowing before you panic at a five-charge diagram. 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 fields inside insulators to qualitative analysis. Topic 10.5 sets the same four-or-fewer limit for calculating electric potential, again with more permitted under high symmetry.

In AP Physics C: Electricity and Magnetism the two are split across whole units: fields sit in Unit 8 (Electric Charges, Fields, and Gauss's Law), weighted at 15 to 25 percent, and potential gets Unit 9 to itself at 10 to 20 percent. The calculus course adds continuous charge distributions and the two-way integral and derivative link.

For the arithmetic routine on both quantities, worked end to end for a point charge, use the electric field and potential guide. For the energy side, which is a third quantity again, the electric potential vs electric potential energy page separates $V$ from $U_E$ . And if a diagram is what you are being asked to read rather than a number, electric field lines covers what a line diagram can and cannot tell you.

Two equal positive charges: the field cancels, the potential does not

Charges $q_1 = +4.0 \times 10^{-9} \ \mathrm{C}$ and $q_2 = +4.0 \times 10^{-9} \ \mathrm{C}$ are fixed at $x = 0$ and $x = 0.40 \ \mathrm{m}$ . Find the electric field and the electric potential at the midpoint, $x = 0.20 \ \mathrm{m}$ .

Each charge is $r = 0.20 \ \mathrm{m}$ from the midpoint. Compute $kq$ once, since both quantities need it: $kq = (9.0 \times 10^9)(4.0 \times 10^{-9}) = 36 \ \mathrm{N \cdot m^2/C}$ .
Field from $q_1$ : $\lvert \vec{E}_1 \rvert = kq/r^2 = 36/(0.20)^2 = 36/0.040 = 900 \ \mathrm{N/C}$ , pointing away from the positive charge, so in the $+x$ direction.
Field from $q_2$ : same magnitude, $900 \ \mathrm{N/C}$ , pointing away from $q_2$ , so in the $-x$ direction.
Add as vectors along $x$ : $E_{net} = 900 - 900 = 0$ . The field at the midpoint is exactly zero.
Potential from $q_1$ : $V_1 = kq/r = 36/0.20 = 180 \ \mathrm{V}$ . Same for $q_2$ : $V_2 = 180 \ \mathrm{V}$ .
Add as signed scalars: $V = 180 + 180 = 360 \ \mathrm{V}$ . There is no cancellation available, because there is no direction to cancel with.
Check the physics: a positive charge placed at the midpoint feels no net force, so it stays where it is, and yet it carries $U_E = qV$ joules of potential energy. Displace it slightly along the line and it accelerates away, which is where that energy goes.

$E_{net} = 0$ and $V = +360 \ \mathrm{V}$ at the midpoint. Zero field with a large potential, in the same place at the same time.

A dipole from the side: the potential cancels, the field does not

Charges $q_1 = +5.0 \times 10^{-9} \ \mathrm{C}$ at the origin and $q_2 = -5.0 \times 10^{-9} \ \mathrm{C}$ at $x = 0.80 \ \mathrm{m}$ lie along the $x$ -axis. Find the electric potential and the electric field at the point P with coordinates $(0.40 \ \mathrm{m}, 0.30 \ \mathrm{m})$ .

Distance from each charge to P: $r = \sqrt{(0.40)^2 + (0.30)^2} = \sqrt{0.16 + 0.09} = \sqrt{0.25} = 0.50 \ \mathrm{m}$ . Both distances are equal because P sits on the perpendicular bisector.
Potential from $q_1$ : $V_1 = kq_1/r = (9.0 \times 10^9)(5.0 \times 10^{-9})/0.50 = 45/0.50 = +90 \ \mathrm{V}$ .
Potential from $q_2$ : $V_2 = (9.0 \times 10^9)(-5.0 \times 10^{-9})/0.50 = -90 \ \mathrm{V}$ .
Total potential: $V = 90 + (-90) = 0 \ \mathrm{V}$ . Every point on the perpendicular bisector gives this, since the distances always match.
Field magnitude from each charge: $\lvert \vec{E} \rvert = k \lvert q \rvert /r^2 = 45/(0.50)^2 = 45/0.25 = 180 \ \mathrm{N/C}$ .
Directions. $\vec{E}_1$ points away from the positive charge, along the direction from the origin to P, whose components are $(0.40, 0.30)/0.50 = (0.80, 0.60)$ . So $\vec{E}_1 = 180(0.80, 0.60) = (144, 108) \ \mathrm{N/C}$ .
$\vec{E}_2$ points toward the negative charge, along the direction from P to $(0.80, 0)$ , whose components are $(0.40, -0.30)/0.50 = (0.80, -0.60)$ . So $\vec{E}_2 = 180(0.80, -0.60) = (144, -108) \ \mathrm{N/C}$ .
Add: the $y$ components cancel, $108 - 108 = 0$ , and the $x$ components reinforce, $144 + 144 = 288$ . So $E_{net} = 288 \ \mathrm{N/C}$ in the $+x$ direction, parallel to the line from the positive charge to the negative one.
Sanity check with the derivative relation: moving along the bisector keeps $V$ at zero, so there is no field component along the bisector, and indeed the net field came out perpendicular to it.

$V = 0$ and $E_{net} = 288 \ \mathrm{N/C}$ directed from the positive charge toward the negative one. Zero potential with a substantial field, the mirror image of the previous example.

Parallel plates: constant field, changing potential

Two parallel plates are separated by $d = 0.020 \ \mathrm{m}$ and held at a potential difference of $120 \ \mathrm{V}$ . Take the negative plate as the zero of potential. (a) Find the field between the plates. (b) Find the potential halfway across. (c) Find the field halfway across. (d) Say which quantity changes with position and which does not.

(a) The field between parallel plates is uniform, so the average field is the field, and the sheet relation applies exactly: $\lvert \vec{E} \rvert = \lvert \Delta V / \Delta r \rvert = 120/0.020 = 6000 \ \mathrm{V/m}$ .
That is also $6000 \ \mathrm{N/C}$ , because volts per metre and newtons per coulomb are the same unit. The direction is from the positive plate toward the negative plate, which is the direction of decreasing potential.
(b) Potential varies linearly across a uniform field, so at half the separation it has fallen through half the drop: $V = 120/2 = 60 \ \mathrm{V}$ measured from the negative plate.
(c) The field is unchanged: $6000 \ \mathrm{N/C}$ . Uniform means uniform, and the midpoint is no different from anywhere else away from the edges.
(d) Potential is a different number at every position between the plates, running from $0$ to $120 \ \mathrm{V}$ . The field takes one value everywhere. A graph of $V$ against position is a straight line, and the field is its slope, which is the same at every point on a straight line.
Check by working backwards from the midpoint: the potential difference between the midpoint and the negative plate is $60 \ \mathrm{V}$ across $0.010 \ \mathrm{m}$ , giving $60/0.010 = 6000 \ \mathrm{N/C}$ again, as it must.

$E = 6000 \ \mathrm{N/C}$ everywhere between the plates, while $V$ runs from $0$ to $120 \ \mathrm{V}$ and reads $60 \ \mathrm{V}$ at the midpoint. The field is the slope of the potential, not a value of it.

Frequently asked questions

What is the difference between electric field and electric potential?

Electric field is the electric force per unit charge at a point, a vector measured in newtons per coulomb. Electric potential is the electric potential energy per unit charge at that point, a scalar measured in volts, where one volt is one joule per coulomb. Both describe the point rather than anything placed there, but one has a direction and one does not. They are related by a rate of change with distance, not by a ratio: the field equals the negative rate at which the potential changes with position, which is why for a point charge the field goes as 1 over r squared and the potential goes as 1 over r.

Can the electric field be zero where the electric potential is not?

Yes. At the midpoint between two equal positive charges the two field vectors are equal in size and opposite in direction, so they cancel to zero, while the two potentials are both positive and add. For two charges of 4.0 nC separated by 0.40 m, the midpoint has a net field of zero and a potential of 360 V. A charge placed there feels no force but still has potential energy, and it will accelerate away as soon as it is nudged off the midpoint.

Can the electric potential be zero where the electric field is not?

Yes, and this is the more commonly tested version. Anywhere on the perpendicular bisector between two equal and opposite charges, the two distances are the same, so the potentials are equal and opposite and sum to zero. The fields do not cancel there: their components along the bisector cancel while their components across it add, leaving a field parallel to the line joining the charges. Zero potential is a statement about a number, not about whether anything is happening at the point.

What are the units of electric field and electric potential?

Electric field is measured in newtons per coulomb, which is exactly the same unit as volts per metre. Electric potential is measured in volts, and one volt is one joule per coulomb. The equivalence of N/C and V/m is not a coincidence: it is the field-potential relation written as units, since the field is a potential difference divided by a distance.

Is electric field the derivative of electric potential?

It is the negative of it. In one dimension the AP Physics C: Electricity and Magnetism sheet prints E_x = -dV/dx, and the AP Physics 2 sheet prints the magnitude version, the magnitude of E equals the magnitude of delta V over delta r. The minus sign makes the field point toward lower potential, which the AP Physics 2 CED states in words at EK 10.5.B.2.iii. Going the other way is an integral: delta V is minus the integral of E along the path.

Why does the electric field fall off as 1 over r squared while potential falls off as 1 over r?

Because one is the rate of change of the other with distance. Differentiating kq/r with respect to r gives -kq/r squared, so the potential of a point charge carries one power of r and its field carries two. This gives you a fast check on any answer: double the distance from a point charge and the field drops to a quarter of its value while the potential only halves.

Do I plug the sign of the charge into the electric field formula?

The AP Physics 2 sheet prints the point-charge field with absolute value bars around q, so the cleanest habit is to use the magnitude to get the strength and then assign the direction by inspection: away from positive charge, toward negative charge. The potential formula on the same sheet has no bars and is printed as a sum over the charges, so there you keep the sign, and negative charges genuinely produce negative potential.