Electric Potential vs Electric Potential Energy

Electric potential is energy per unit charge at a point in space, measured in volts. Electric potential energy is the energy a specific charge has because it sits there, measured in joules. The point has a potential whether or not anything is there. Put a charge there and you get energy: U = qV.

AP Physics: Unit 10 (topics 10.4 Electric Potential Energy, 10.5 Electric Potential). This pair spans two consecutive AP Physics 2 topics. Topic 10.4 (Electric Potential Energy) defines the potential energy of two point charges as the work an external force must do to bring them from infinitely far away, with the equation U_E = kq1q2/r, and requires configurations to be summed pair by pair. Topic 10.5 (Electric Potential) then defines electric potential as the electric potential energy per unit charge at a point in space (10.5.A.1) and potential difference as the change in electric potential energy per unit charge when a test charge is moved between two points (10.5.A.3). Both sit 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. Both topics carry a boundary statement limiting calculations to four or fewer point charges, with more allowed in situations of high symmetry. The pair returns in Topic 10.7 (Conservation of Electric Energy) and again in Unit 11, where energy per charge times charge per second gives electric power.

The distinction, stated once

Electric potential $V$ belongs to a point in space. It is set entirely by the source charges around that point, and it tells you how many joules each coulomb of charge would have if you put charge there. Its unit is the volt, and one volt is one joule per coulomb.

Electric potential energy $U_E$ belongs to a charge together with the field it sits in. It is set by the potential at the point and by how much charge you put there. Its unit is the joule.

The two are linked by one relationship:

U_E = qV

Read it as a conversion with a multiplier. The point supplies $V$ . You supply $q$ . The product is the energy.

That is also the entire reason they get confused. Every sentence about potential can be turned into a sentence about potential energy by multiplying by a charge, so the two ideas travel together through every problem and never separate on their own. They only separate when you change the charge and leave the point alone, or change the point and leave the charge alone, and the rest of this page is built around doing exactly that.

Side by side

	Electric potential $V$	Electric potential energy $U_E$
What it belongs to	A point in space	A charge, plus the field it sits in
Unit	Volt (V), which is J/C	Joule (J)
Scalar or vector	Scalar, signed	Scalar, signed
Exists with nothing there?	Yes	No, there must be a charge
Point charge source	$V = kq/r$	$U_E = kq_1q_2/r$
Depends on the test charge?	No	Yes, directly proportional to it
Sign is set by	The source charges	The product $q_1q_2$ , or by $qV$
Zero reference	Infinitely far from the sources	Infinite separation of the charges
What a battery labels	9 V is a potential difference	Never labelled on a battery

Two rows do most of the work. The exists with nothing there row is the definitional split: a region of space has a potential map the way a hill has a height map, and neither needs anything sitting on it. The depends on the test charge row is the one you can test numerically in ten seconds, and the worked examples below do it.

The rows they share matter too. Both are scalars, both carry a sign, and both add as ordinary signed numbers with no components. If you find yourself resolving either one into $x$ and $y$ parts, you have reached for the electric field instead.

The case that separates them: one point, four different charges

Fix a source charge of $+5.0 \ \mathrm{nC}$ at the origin and look at a point P a distance $0.30 \ \mathrm{m}$ away. The potential there is $V = kq/r = 150 \ \mathrm{V}$ , and that number is now fixed. Nothing you do at P changes it, because P's potential was set by the source.

Now put things at P, one at a time.

What is at P	Electric potential at P	Electric potential energy
Nothing	$150 \ \mathrm{V}$	Not defined, there is no charge
$+2.0 \ \mathrm{nC}$	$150 \ \mathrm{V}$	$+3.0 \times 10^{-7} \ \mathrm{J}$
$-2.0 \ \mathrm{nC}$	$150 \ \mathrm{V}$	$-3.0 \times 10^{-7} \ \mathrm{J}$
$+6.0 \ \mathrm{nC}$	$150 \ \mathrm{V}$	$+9.0 \times 10^{-7} \ \mathrm{J}$

One column never moves. The other takes four different values, including a negative one and an undefined one. If someone asks you "what is the potential energy at that point", the honest answer is a question back: the energy of what charge?

That is the difference made visible, and it is worth holding onto in exactly this form, because it is the shape of the multiple-choice question. Two identical points, two different charges, and one of the two quantities has changed.

The bridge: U = qV, and where the sign lives

The AP Physics 2 equation sheet prints the relationship in its difference form:

\Delta U_E = q \Delta V

Both factors on the right carry a sign, and the product decides. Four combinations, all of which appear on exams:

Charge moved	$\Delta V$ along the path	$\Delta U_E$	Speeds up if released?
Positive	Negative (toward lower $V$ )	Negative	Yes
Positive	Positive (toward higher $V$ )	Positive	No, must be pushed
Negative	Negative (toward lower $V$ )	Positive	No, must be pushed
Negative	Positive (toward higher $V$ )	Negative	Yes

The pattern in one sentence: a charge released from rest moves in whatever direction lowers its own potential energy, and for a negative charge that direction is toward higher potential. Positive charges roll downhill on the potential map; negative charges roll uphill on the same map.

Notice what this does to the language. "The electron moved to a point of higher potential" and "the electron lost potential energy" are the same sentence. If you translate one into the other and drop the sign of the charge on the way, you invert the physics while keeping every word true-sounding.

The electron volt exists precisely because of this bridge. One electron volt is the energy an elementary charge picks up across one volt of potential difference: $(1.60 \times 10^{-19} \ \mathrm{C})(1 \ \mathrm{V}) = 1.60 \times 10^{-19} \ \mathrm{J}$ , which is the sheet's printed conversion. The unit is a $qV$ product with the charge already substituted in, which is why an energy quoted in eV looks so much like a voltage. It is not one.

What the equation sheet actually prints

Check this rather than recall it, because the four AP sheets differ. On the AP Physics 2 sheet, the electricity block prints all of the following:

$U_E = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r} = k\frac{q_1q_2}{r}$ , the potential energy of a pair of point charges
$\Delta U_E = q\Delta V$ , the bridge, in difference form
$V = \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{r_i}$ , the potential of a configuration, as a sum
$\lvert \vec{E} \rvert = \lvert \frac{\Delta V}{\Delta r} \rvert$ , the field from the potential

Three details are worth reading off that list carefully.

The $U_E$ equation has no absolute value bars. Compare it with Coulomb's law two lines above it, which is printed as $k\lvert q_1q_2 \rvert / r^2$ and returns a magnitude only. The energy equation keeps the signed product, so opposite charges genuinely give a negative $U_E$ and you must carry that through. The Coulomb's law guide covers the force side of that contrast.

The potential equation is printed as a sum. $V$ for several charges is $\sum kq_i/r_i$ , signed numbers added directly, with no components and no angles.

There is no $U_E = qV$ on the sheet, only $\Delta U_E = q\Delta V$ . The non-difference form follows immediately once you take the zero of both at infinite separation, which is the standard reference, but the sheet's version is the one written in differences. Writing $\Delta$ in both places costs nothing and keeps you consistent with the printed relation.

When it costs a mark

Five specific errors, each of which produces a wrong answer that looks reasonable.

Answering in the wrong unit. "Find the electric potential at P" answered in joules, or "find the potential energy" answered in volts. A grader reading a units line does not need to check your arithmetic to mark this down. Volts for $V$ , joules for $U_E$ , every time.

Dropping the sign of the moving charge in $\Delta U_E = q\Delta V$ . This is the single most productive error on this pair, because it gives an answer of the right size with the wrong sign, and then the energy conservation step that follows turns a speeding-up into a slowing-down. Substitute $q$ with its sign attached, including the minus on an electron.

Assuming $V = 0$ means $U_E = 0$ means nothing is happening. At the midpoint between two equal and opposite charges the potential is zero and the field is large. A charge released there feels a force and accelerates. Zero potential is a statement about a number, not about the physics at the point.

Treating potential as if it needed a charge to exist. Questions phrased "a charge is removed from point P; what is the new potential at P?" are testing exactly this. If the removed charge was a test charge and not a source, the answer is that the potential is unchanged.

Reading eV as a voltage. "The electron gains 180 eV" is an energy statement. "The electron crosses 180 V" is a statement about two points. They happen to carry the same number for an elementary charge, which is convenient and a trap in equal measure.

When they coincide, and why that lulls you

They agree numerically when the charge you place is exactly $+1 \ \mathrm{C}$ . Then $U_E = qV$ reads $U_E = V$ as numbers, and the units differ while the digits do not. The definition of potential is often taught in exactly this form, as "the energy a unit charge would have", and that phrasing is where the conflation is born.

They also track each other perfectly in any problem where the charge never changes. Follow one proton through a whole multi-part question and the potential energy graph and the potential graph have identical shapes, differing only by the constant factor $e$ . Nothing in that problem forces you to notice that they are different quantities, and most single-particle problems are like this.

The distinction only bites in three situations, which is why those three keep appearing on exams.

Two different charges take the same path, so $\Delta V$ is shared and $\Delta U_E$ is not. A proton and an electron across the same potential difference gain equal and opposite energies.
A question changes the charge at a fixed point and asks what happened to the potential. The answer is nothing.
The charges have opposite signs, so $V$ and $U_E$ end up with opposite signs at the same location. A proton at a point of negative potential has negative energy; an electron at that same point has positive energy.

If you can handle those three, the pair is finished.

How the exam frames it, and what to read next

Potential energy is Topic 10.4 and potential is Topic 10.5, consecutive topics in Unit 10, which the CED weights at 15 to 18 percent of the multiple-choice section. The order is deliberate: the CED builds the energy of a pair of charges first, then divides out the charge to define the potential of a point. Reading them in that order is worth doing at least once, because it makes the volt look like what it is, an energy bookkeeping device rather than a new kind of quantity.

The pair then reappears in Topic 10.7, where a charge released in a field converts $\Delta U_E$ into kinetic energy, and in Unit 11, where the potential difference across a circuit element and the charge flowing through it multiply to give the energy delivered. That is the same $qV$ product, and it is why electrical power comes out as $P = I\Delta V$ : energy per charge times charge per second.

For the field side of the picture, which is a third quantity again and a vector, the electric field and potential guide sets $E = kq/r^2$ against $V = kq/r$ and shows why one falls off as an inverse square and the other does not.

One point, four different charges

A source charge $Q = +5.0 \times 10^{-9} \ \mathrm{C}$ is fixed at the origin. Point P lies $0.30 \ \mathrm{m}$ away. (a) Find the electric potential at P. (b) Find the electric potential energy when a charge $q = +2.0 \times 10^{-9} \ \mathrm{C}$ is placed at P, and check the answer a second way. (c) Repeat for $q = -2.0 \times 10^{-9} \ \mathrm{C}$ and for $q = +6.0 \times 10^{-9} \ \mathrm{C}$ . (d) State what changed and what did not.

(a) The potential of a point charge is $V = kQ/r$ with $k = 9.0 \times 10^9 \ \mathrm{N \cdot m^2/C^2}$ from the sheet.
$V = (9.0 \times 10^9)(5.0 \times 10^{-9}) / 0.30 = 45 / 0.30 = 150 \ \mathrm{V}$ . Note that the numerator $kQ$ is $45 \ \mathrm{V \cdot m}$ , and dividing by $0.30 \ \mathrm{m}$ leaves volts.
(b) Use the bridge: $U_E = qV = (2.0 \times 10^{-9} \ \mathrm{C})(150 \ \mathrm{V}) = 3.0 \times 10^{-7} \ \mathrm{J}$ .
Check with the pair equation from the sheet: $U_E = kQq/r = (9.0 \times 10^9)(5.0 \times 10^{-9})(2.0 \times 10^{-9}) / 0.30$ . The numerator is $(9.0 \times 10^9)(1.0 \times 10^{-17}) = 9.0 \times 10^{-8}$ , and dividing by $0.30$ gives $3.0 \times 10^{-7} \ \mathrm{J}$ . The two routes agree, as they must: the second is the first with $V = kQ/r$ substituted in.
(c) For $q = -2.0 \times 10^{-9} \ \mathrm{C}$ : $U_E = (-2.0 \times 10^{-9})(150) = -3.0 \times 10^{-7} \ \mathrm{J}$ . Same size, opposite sign, because the charges now attract.
For $q = +6.0 \times 10^{-9} \ \mathrm{C}$ : $U_E = (6.0 \times 10^{-9})(150) = 9.0 \times 10^{-7} \ \mathrm{J}$ , three times the first answer because the charge is three times as large.
(d) The potential at P was $150 \ \mathrm{V}$ in all four cases, including the case with nothing there. The potential energy took the values undefined, $+3.0 \times 10^{-7} \ \mathrm{J}$ , $-3.0 \times 10^{-7} \ \mathrm{J}$ and $+9.0 \times 10^{-7} \ \mathrm{J}$ .

$V = 150 \ \mathrm{V}$ at P regardless of what sits there. The potential energy is $+3.0 \times 10^{-7} \ \mathrm{J}$ , $-3.0 \times 10^{-7} \ \mathrm{J}$ and $+9.0 \times 10^{-7} \ \mathrm{J}$ for the three test charges. One number is a property of the point; the other is a property of what you put there.

Same two points, opposite charges: same delta V, opposite delta U

In some electric field, point A is at a potential of $+12.0 \ \mathrm{V}$ and point B is at $+4.0 \ \mathrm{V}$ . (a) Find $\Delta V$ for the trip from A to B. (b) Find $\Delta U_E$ for a charge of $+3.0 \ \mathrm{\mu C}$ making that trip. (c) Find $\Delta U_E$ for a charge of $-3.0 \ \mathrm{\mu C}$ making the same trip. (d) Say which charge speeds up if released from rest at A with only the electric force acting.

(a) $\Delta V = V_B - V_A = 4.0 \ \mathrm{V} - 12.0 \ \mathrm{V} = -8.0 \ \mathrm{V}$ . This belongs to the pair of points and does not know what will travel between them.
(b) $\Delta U_E = q\Delta V = (+3.0 \times 10^{-6} \ \mathrm{C})(-8.0 \ \mathrm{V}) = -2.4 \times 10^{-5} \ \mathrm{J}$ . The positive charge loses potential energy going from A to B.
(c) $\Delta U_E = (-3.0 \times 10^{-6} \ \mathrm{C})(-8.0 \ \mathrm{V}) = +2.4 \times 10^{-5} \ \mathrm{J}$ . Two negatives multiply to a positive: the negative charge gains potential energy over the identical path.
(d) With only the electric force acting, mechanical energy is conserved, so $\Delta K = -\Delta U_E$ . The positive charge gains $2.4 \times 10^{-5} \ \mathrm{J}$ of kinetic energy and speeds up on the way to B.
The negative charge would need $2.4 \times 10^{-5} \ \mathrm{J}$ supplied from outside to reach B. Released from rest at A it moves the other way, toward higher potential.

$\Delta V = -8.0 \ \mathrm{V}$ for both charges, because it belongs to the two points. $\Delta U_E = -2.4 \times 10^{-5} \ \mathrm{J}$ for the positive charge and $+2.4 \times 10^{-5} \ \mathrm{J}$ for the negative one. Only the positive charge speeds up going from A to B.

A negative potential, and why the sign of U does not have to match

A point Q in a field has an electric potential of $-180 \ \mathrm{V}$ . Using the sheet values $e = 1.60 \times 10^{-19} \ \mathrm{C}$ and $1 \ \mathrm{eV} = 1.60 \times 10^{-19} \ \mathrm{J}$ , find the electric potential energy of (a) a proton placed at Q and (b) an electron placed at Q, each in joules and in electron volts. (c) Explain why the potential is negative in both cases but the energy is not.

(a) A proton carries $q = +1.60 \times 10^{-19} \ \mathrm{C}$ . $U_E = qV = (+1.60 \times 10^{-19})(-180) = -2.88 \times 10^{-17} \ \mathrm{J}$ .
Convert: $(-2.88 \times 10^{-17} \ \mathrm{J}) / (1.60 \times 10^{-19} \ \mathrm{J/eV}) = -180 \ \mathrm{eV}$ . The digits repeat the potential because the charge is exactly one elementary charge, which is the whole point of the electron volt.
(b) An electron carries $q = -1.60 \times 10^{-19} \ \mathrm{C}$ . $U_E = (-1.60 \times 10^{-19})(-180) = +2.88 \times 10^{-17} \ \mathrm{J}$ , which is $+180 \ \mathrm{eV}$ .
(c) The potential is a property of Q and is the same negative number for both particles. The energy is the product $qV$ , and the sign of $q$ flips it. A negative potential means the point is surrounded on balance by negative source charge, which repels an electron and attracts a proton.
Sanity check on the direction: the electron sitting at a positive $180 \ \mathrm{eV}$ has energy available to give up. Released from rest at Q it moves toward higher potential, lowering its own energy, which matches the sign rule from the table above.

Proton: $U_E = -2.88 \times 10^{-17} \ \mathrm{J} = -180 \ \mathrm{eV}$ . Electron: $U_E = +2.88 \times 10^{-17} \ \mathrm{J} = +180 \ \mathrm{eV}$ . One potential, one sign; two energies, opposite signs.

Frequently asked questions

What is the difference between electric potential and electric potential energy?

Electric potential is the electric potential energy per unit charge at a point in space, measured in volts, where one volt is one joule per coulomb. Electric potential energy is the energy that a specific charge has because of where it sits, measured in joules. The point has a potential whether or not any charge is there; the energy only exists once a charge is placed at the point. They are connected by U = qV, so the potential energy is the potential multiplied by the charge you put in.

Is electric potential measured in volts or joules?

Volts. One volt is one joule per coulomb, so a volt is already an energy divided by a charge. Electric potential energy is measured in joules. If your answer to a find-the-potential question comes out in joules, you have found the energy of some particular charge instead of the property of the point.

Can electric potential be negative?

Yes, and so can electric potential energy, independently. The potential at a point is negative when the point is surrounded on balance by negative source charge, since V for a configuration is the signed sum of kq/r over the sources. The potential energy of a charge at that point is the product qV, so a positive charge at a negative potential has negative energy while a negative charge at the same point has positive energy. The two signs do not have to match.

Does electric potential energy depend on the test charge?

Yes, directly. Double the charge you place at a point and the potential energy doubles, because U = qV and V has not changed. This is the fastest way to tell the two quantities apart: change the charge at a fixed point and see which number moves. The potential stays put because it was set by the source charges, not by whatever you placed there.

What does the AP Physics 2 equation sheet print for potential and potential energy?

The electricity block prints the pair energy as U_E = kq1q2/r with no absolute value bars, so the sign of the product is kept. It prints the bridge in difference form as delta U_E = q delta V. It prints the potential of a configuration as V = the sum over i of kq_i/r_i, a signed scalar sum. It also prints the magnitude of the field as the magnitude of delta V over delta r. Coulomb's law on the same sheet does carry absolute value bars, which is the contrast to watch.

Why is one volt the same as one joule per coulomb?

Because potential is defined as energy per unit charge. Rearranging U = qV gives V = U/q, and joules divided by coulombs is the volt. This is also why an electron volt is an energy and not a voltage: it is the joules an elementary charge picks up across one volt, which comes to 1.60 times ten to the minus nineteen joules, the figure printed on the AP Physics 2 sheet.

If the potential at a point is zero, is the potential energy zero too?

Yes for the charge you place there, since U = qV gives zero for any q. But zero potential does not mean nothing is happening at the point. At the midpoint between two equal and opposite charges the potential is zero while the electric field is large, so a charge released there feels a force and accelerates. Zero potential is a statement about a number, not about whether the point is quiet.