AP Physics C: E&M · Topic 9.1

Topic 9.1: Electric Potential Energy

Unit 9: Electric Potential10-20% of the multiple-choice section

The electric potential energy of two point charges is the work an external force must do to bring them from infinitely far apart to where they are. It is k times q1 q2 over r, a scalar that carries the signs of both charges, and for more than two charges you add one term for every pair.

AP Physics: Unit 9 (topics 9.1 Electric Potential Energy). Topic 9.1 of the current AP Physics C: Electricity and Magnetism course and exam description, inside Unit 9, which is weighted 10 to 20% of the multiple-choice section at about 10 to 20 class periods. One learning objective, 9.1.A, using the task verb describe, with three essential-knowledge statements: 9.1.A.1 defines the electric potential energy of a system of two point charges as the work required for an external force to bring them to their current positions from infinitely far away; 9.1.A.2 gives the general form as U_E = (1/(4 pi epsilon-zero)) q1 q2 / r = k q1 q2 / r; 9.1.A.3 gives the total as the sum over the individual interactions between each pair of charged objects. Topic 9.1 prints no boundary statement, so nothing caps the number of charges. Suggested skills are 1.C, 2.C, 3.B and 3.C. The equation sheet prints U_E in the permittivity form only, without the k form that statement 9.1.A.2 also shows, and prints no pair-sum formula. The corresponding AP Physics 2 topic, 10.4, has an identical learning objective and essential knowledge that matches except for one word in 10.4.A.2, which reads potential energy of two charged objects where 9.1.A.2 reads between two charged objects, and it adds a boundary statement capping calculations at four or fewer point charges; AP Physics 2 also lists different suggested skills, 1.C, 2.A, 2.D and 3.C.

What Topic 9.1 requires

Topic 9.1 has one learning objective and three essential-knowledge statements under it. That is the whole of the required content.

9.1.A, describe the electric potential energy of a system.

9.1.A.1 states that the electric potential energy of a system of two point charges equals the amount of work required for an external force to bring the point charges to their current positions from infinitely far away.
9.1.A.2 states that the general form for the electric potential energy between two charged objects is given by the equation $U_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r} = k\dfrac{q_1 q_2}{r}$ .
9.1.A.3 states that the total electric potential energy of a system can be determined by finding the sum of the electric potential energies of the individual interactions between each pair of charged objects in the system.

Topic 9.1 prints no boundary statement. Unit 9 carries exactly one, and it sits under Topic 9.2, restricting which charge distributions you integrate for potential. Nothing in the framework caps how many charges a 9.1 question may hand you.

The CED lists four suggested skills for the topic: 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Skill 1.C is the one people skip. A graph of $U_E$ against separation for a pair of charges is a sketchable object with a sign, an asymptote and a shape, and the CED asks for it first.

Unit 9 is weighted 10 to 20% of the multiple-choice section over a suggested 10 to 20 class periods.

The AP Physics 2 version of this topic is all but identical

This is worth saying plainly rather than dressing up. Compare Topic 9.1 here with Topic 10.4 in AP Physics 2. The learning objective is identical. Statements 9.1.A.1 and 9.1.A.3 are identical to Physics 2's 10.4.A.1 and 10.4.A.3, sentence for sentence. Statement 9.1.A.2 differs from 10.4.A.2 by a single word, "the electric potential energy between two charged objects" here against "of two charged objects" there, and prints the same equation in the same two forms.

Beyond that one word, two things change.

AP Physics 2 prints a boundary statement here and AP Physics C does not. The Physics 2 statement reads: "As the methods to calculate the electric potential energy due to extended charge distributions exceed the scope of the course, AP Physics 2 only requires that students calculate the electric potential energy of configurations of four or fewer point charges." Physics C prints nothing of the kind. The cap is lifted, and lifting it is the entire calculus consequence of this topic: extended distributions are back on the table because you can integrate.

The suggested skills differ. Physics 2 lists four, 1.C, 2.A, 2.D and 3.C. Physics C lists four, 1.C, 2.C, 3.B and 3.C. Physics C drops the derive-a-symbolic-expression skill from this topic and adds apply-a-law-to-make-a-claim, which matches how 9.1 actually shows up: as a claim about whether the energy of a configuration went up or down, justified, rather than as a derivation.

If you want the algebra-based treatment, the AP Physics 2 page is the one to read: same physics, four-charge ceiling, no integrals anywhere. This page is for the calculus-based course, where the same three statements have to survive contact with continuous charge and with the Topic 9.2 integral.

The definition is a statement about work, not a formula

Read 9.1.A.1 again and notice that it never mentions $1/r$ . It says the energy equals the work required for an external force to bring the charges from infinitely far away.

That is a definition you can act on, and it is where the formula comes from. The C: E&M equation sheet reprints the whole mechanics table, so you already have the work integral:

W = \int_a^b \vec{F} \cdot d\vec{r}

Hold charge $q_1$ fixed and walk $q_2$ in from infinity along a radial line, moving it slowly so its kinetic energy never changes. The external force has to balance the Coulomb force at every point, so the work you do is

W_{\text{ext}} = \int_{\infty}^{r} \left(-k\frac{q_1 q_2}{r'^2}\right) dr' = k\frac{q_1 q_2}{r}

and that integral is where the $r^2$ in Coulomb's law turns into the $r$ in the energy. It is also on the sheet in the other direction, as the mechanics relation $\Delta U = -\int_a^b \vec{F}_c(r) \cdot d\vec{r}$ for a conservative force, and as $F_x = -\dfrac{dU(x)}{dx}$ going back the other way.

Two consequences follow immediately.

The zero of potential energy is at infinite separation, and that is a convention the exam commits to in writing. The conventions box printed with the AP Physics C: E&M equation sheet says: "The electric potential is zero at an infinite distance from an isolated point charge." You do not have to argue for it on an exam, and you do not get to move it.

The path does not matter. The Coulomb force is conservative, so the integral above gives the same answer whatever route you walk the charge in along. That is what licenses talking about a potential energy at all, and it is the same argument that makes the line integral in Topic 9.2 path independent. See conservative vs nonconservative force if that distinction is shaky.

Signs, and the two places they go missing

$U_E$ is a scalar, and it carries the signs of both charges. This is the single most common place to lose a mark in Unit 9, and the equation sheet is quietly complicit.

Look at the two lines as the sheet actually prints them:

\left| \vec{F}_E \right| = \frac{1}{4\pi\varepsilon_0} \frac{\lvert q_1 q_2 \rvert}{r^2}

U_E = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}

The force line has absolute-value bars on both sides. The energy line has none. That is not decoration. The force expression gives a magnitude and you supply the direction from a diagram; the energy expression gives a signed number and there is no diagram to fix it afterwards. Substitute $-3$ nC as $-3 \times 10^{-9}$ C, not as $3 \times 10^{-9}$ C with a note to yourself.

Opposite charges have negative potential energy. Bring a positive and a negative charge together from infinity and they pull themselves in: the external force points outward while the displacement points inward, so the external work is negative. Negative $U_E$ means you would have to supply energy to separate them again.

Like charges have positive potential energy. You push them together against their own repulsion, so you do positive work and it is stored.

The second place the sign goes missing is in reading the sentence "the potential energy increased." For two positive charges, moving them closer increases $U_E$ . For a positive and a negative charge, moving them closer makes $U_E$ more negative, which is a decrease. Both statements come from the same formula. The trap works because "closer means more energy" feels like a rule and is not one.

Configuration	Sign of $U_E$	Moving them closer	Released from rest, they
Two positive charges	Positive	$U_E$ increases	Fly apart
Two negative charges	Positive	$U_E$ increases	Fly apart
One positive, one negative	Negative	$U_E$ decreases	Pull together

The last column is the physical content of the sign, and it is the version to reason with: a system released from rest moves so as to lower its potential energy, exactly as in mechanics. Topic 9.3 makes that bookkeeping explicit.

Adding a system up pair by pair

Statement 9.1.A.3 is short and it is doing real work: the total energy is the sum over pairs, not over charges.

U_{\text{total}} = \sum_{\text{pairs } i<j} k\frac{q_i q_j}{r_{ij}}

The number of terms is the number of ways to choose two objects from $N$ , which is $\dfrac{N(N-1)}{2}$ .

Two charges: 1 term.
Three charges: 3 terms.
Four charges: 6 terms.
Five charges: 10 terms.

Counting three terms for three charges is easy. Counting six for four is where the marks go, because the two diagonals of a square are easy to forget and they sit at a different separation from the four sides.

There is a good reason the sum is over pairs and not over charges, and it is the assembly picture again. Bring the charges in one at a time from infinity. The first one is free: there is nothing there yet. The second costs its interaction with the first. The third costs its interactions with the first two. Every interaction gets paid for exactly once, and a sum over pairs is precisely a list of interactions counted once each.

That also answers a question students ask and rarely get a straight answer to: no, you do not divide by two. The $\frac{1}{2}$ appears only if you sum over all ordered pairs, $i \neq j$ , which counts each interaction twice. The CED's phrasing, "between each pair of charged objects," is the unordered version, so there is no factor to insert.

One more thing this statement quietly licenses. Nothing in it says the objects have to be point charges, and nothing in Topic 9.1 caps how many there may be. Push the sum to a continuum and it becomes an integral, which is the machinery Topic 9.2 gives you for potential and which Topic 10.3 reuses for the energy stored on a capacitor.

What the equation sheet prints, and what it leaves to you

Check this against the sheet rather than against memory, because there is a real gap between what the CED prints in Topic 9.1 and what you get in the exam booklet.

The AP Physics C: E&M equation sheet prints the potential energy of two point charges in one form only:

U_E = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}

The $k$ form is not on the sheet. Statement 9.1.A.2 in the CED prints both, " $U_E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r} = k\dfrac{q_1 q_2}{r}$ ", but the exam booklet's version stops at the permittivity form. Contrast the Coulomb force line directly above it, which does print both forms. This costs nothing if you know that the constants table defines $k = \dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9\ \mathrm{N \cdot m^2/C^2}$ , which it does, on the same page.

Three related lines are printed and are worth knowing sit there:

$\Delta U_E = q\Delta V$ , which is Topic 9.3's equation and the fastest route to most 9.1 numbers once a potential is known.
$U_C = \frac{1}{2}Q\Delta V$ , the energy stored on a capacitor, which belongs to Topic 10.3.
The mechanics work and potential-energy integrals, because the C: E&M sheet reprints the entire C: Mechanics table.

And two useful constants: the elementary charge $e = 1.60 \times 10^{-19}$ C, and the conversion $1\ \mathrm{eV} = 1.60 \times 10^{-19}$ J. The electron volt is a unit of energy, defined as the work done moving one elementary charge through one volt, and it turns awkward exponents into readable numbers.

What is not printed anywhere is a pair-sum formula. Statement 9.1.A.3 is prose in the framework and stays prose. You write the sum yourself, and on a free-response question you show the terms.

Where potential energy stops being about point charges

Topic 9.1's three statements are all about point charges and pairs of them. The course does not stop there, and knowing where the idea reappears saves you from treating Unit 9 as an island.

Potential energy per unit charge is potential. Divide $U_E$ by the charge you brought in and you get $V$ , which is Topic 9.2. Statement 9.2.A.1 defines electric potential as exactly that, "the electric potential energy per unit charge at a point in space." The practical consequence is that once you have $V$ at a location, the work to bring a charge $q$ there from infinity is $W_{\text{ext}} = qV$ with no pair sums at all. That is usually the fast route. See electric potential vs electric potential energy if the two keep sliding into each other.

Energy stored on a capacitor is the same assembly idea. Statement 10.3.A.4 says the electric potential energy stored in a capacitor is equal to the work done by an external force to separate that amount of charge on the capacitor. Same sentence shape as 9.1.A.1, same physics, different geometry.

A charged conductor's energy lives on its surface. Once you reach Topic 10.1, all the excess charge sits on the surface and the whole conductor is one equipotential, which is why a conductor's stored energy can be written from $Q$ and $V$ alone.

Gravitation is the same algebra with one sign flipped. The mechanics table on the same sheet prints $U_G = -G\dfrac{m_1 m_2}{r}$ . The minus sign is explicit there because masses always attract, so gravitational potential energy is always negative; in the electric case the sign rides on the charges instead. If you are comfortable with orbital energy diagrams, you already know the shape of a $U_E$ against $r$ curve.

Finally, watch the frame. $U_E$ belongs to a system, not to one charge. Statement 9.3.A.1 is careful about this, calling it "the electric potential energy of the object-field system." A single isolated charge in empty space has no electric potential energy, because there is nothing for it to interact with.

Traps, and how Topic 9.1 is assessed

Using $\lvert q_1 q_2 \rvert$ in the energy. The bars belong on the force line. Strip them from the energy and let the signs through.

Counting charges instead of pairs. Four charges give six terms. Write them out; do not multiply anything by four.

Squaring the separation. $U_E$ goes as $1/r$ , the force as $1/r^2$ . If your energy has an $r^2$ in it, you have written the force expression and multiplied by a length out of habit.

Assuming $U_E$ and $V$ vanish in the same places. They do not. On the perpendicular bisector of a dipole the potential is zero everywhere, but a charge placed there still has a well-defined potential energy of zero relative to infinity, while the two-charge system it joined has a large negative energy of its own. Ask which system you are being asked about.

Reporting a negative "amount of work" as a magnitude. If the external work comes out negative, the answer is negative: the field did the work and you resisted.

Forgetting that $r$ is centre-to-centre. For spheres treated as point charges, $r$ runs between centres, not between surfaces.

On assessment: the AP Physics C: Electricity and Magnetism exam is 3 hours long, with 42 multiple-choice questions worth 50% in 85 minutes and 4 free-response questions worth 50% in 95 minutes. The CED's Progress Check for Unit 9 is about 18 multiple-choice questions and 4 free-response questions.

The Unit 9 opener names skills 1.A, 1.C, 2.A and 3.C as the ones this unit builds, and points at the second free-response question, Translation Between Representations, worth 12 points with a suggested time of 25 to 30 minutes. Its example is telling: a student might be asked to create an energy diagram for a point charge moving inside a region with an electric field. That is 9.1 content in graphical form, and it is skill 1.C again.

For the numerical work, Coulomb's law covers the force side and the Coulomb's law calculator will check your arithmetic on the constant.

Three charges on a right triangle: the total energy of the system

Three point charges are fixed at the vertices of a right triangle. Charge $q_A = +5.0$ nC sits at the right angle, $q_B = -2.0$ nC sits $0.040$ m away along one leg, and $q_C = +3.0$ nC sits $0.030$ m away along the other. Find the total electric potential energy of the system, and state what it means physically. Use $k = 9.0 \times 10^9\ \mathrm{N \cdot m^2/C^2}$ .

Count the pairs first. Three charges give $\dfrac{3 \times 2}{2} = 3$ terms: AB, AC and BC. Not three charges times something.
Get the third separation. The legs are $0.030$ m and $0.040$ m, so the hypotenuse is $r_{BC} = \sqrt{0.030^2 + 0.040^2} = 0.050$ m.
Pair AB. $U_{AB} = k\dfrac{q_A q_B}{r_{AB}} = (9.0 \times 10^9)\dfrac{(5.0 \times 10^{-9})(-2.0 \times 10^{-9})}{0.040}$ . The numerator is $-1.0 \times 10^{-17}$ , so $U_{AB} = -2.25 \times 10^{-6}$ J. Negative, because the charges are opposite.
Pair AC. $U_{AC} = (9.0 \times 10^9)\dfrac{(5.0 \times 10^{-9})(3.0 \times 10^{-9})}{0.030} = +4.50 \times 10^{-6}$ J. Positive, both charges positive.
Pair BC. $U_{BC} = (9.0 \times 10^9)\dfrac{(-2.0 \times 10^{-9})(3.0 \times 10^{-9})}{0.050} = -1.08 \times 10^{-6}$ J.
Add them as signed scalars. $U_{\text{total}} = -2.25 + 4.50 - 1.08 = +1.17\ \mu$ J. No components, no angles: potential energy is a scalar and the geometry entered only through the three separations.
Read the sign. The total is positive, so assembling this configuration from infinity took net positive external work. Release all three and they will fly apart, converting $1.17\ \mu$ J into kinetic energy by the time they are far apart, which is the Topic 9.3 statement.

$U_{\text{total}} = U_{AB} + U_{AC} + U_{BC} = -2.25\ \mu\mathrm{J} + 4.50\ \mu\mathrm{J} - 1.08\ \mu\mathrm{J} = +1.17\ \mu\mathrm{J}$ . Positive total energy means an external agent did $1.17\ \mu$ J of net work to assemble the configuration, and that much kinetic energy is released if all three are freed.

Four equal charges on a square: six terms, two lengths

Four identical charges $q = +2.0$ nC are held at the corners of a square of side $a = 0.10$ m. Find the work an external agent must do to assemble this configuration by bringing the charges in one at a time from infinity.

The work equals the total potential energy, by statement 9.1.A.1. So compute $U_{\text{total}}$ .
Count the pairs. Four charges give $\dfrac{4 \times 3}{2} = 6$ terms. Four of them are sides of the square at separation $a$ ; two are diagonals at separation $a\sqrt{2}$ . Missing the diagonals is the standard error here.
One side term. $k\dfrac{q^2}{a} = (9.0 \times 10^9)\dfrac{(2.0 \times 10^{-9})^2}{0.10} = (9.0 \times 10^9)\dfrac{4.0 \times 10^{-18}}{0.10} = 3.60 \times 10^{-7}$ J, or $360$ nJ.
One diagonal term. Same numerator, separation $a\sqrt{2} = 0.1414$ m, so it is the side term divided by $\sqrt{2}$ : $360/\sqrt{2} = 254.6$ nJ.
Sum with the right multiplicities. $U_{\text{total}} = 4(360) + 2(254.6) = 1440 + 509.1 = 1949\ \mathrm{nJ}$ .
Or factor it symbolically first, which is what a free-response answer should show: $U_{\text{total}} = \dfrac{kq^2}{a}\left(4 + \dfrac{2}{\sqrt{2}}\right) = \dfrac{kq^2}{a}\left(4 + \sqrt{2}\right)$ , and $4 + \sqrt{2} = 5.414$ , so $U_{\text{total}} = (360\ \mathrm{nJ})(5.414) = 1.95\ \mu$ J.
Sanity check the assembly order. Bring them in one at a time and the running costs are $0$ , then $1$ term, then $2$ terms, then $3$ terms: $0 + 1 + 2 + 3 = 6$ terms in total, which matches. The order you choose changes the intermediate numbers and not the total.

$W_{\text{ext}} = U_{\text{total}} = \dfrac{kq^2}{a}(4 + \sqrt{2}) = 1.95 \times 10^{-6}$ J, or $1.95\ \mu$ J. Four side terms at $360$ nJ each and two diagonal terms at $255$ nJ each. All six terms are positive because every pair is a pair of like charges.

Bringing a charge in from infinity near a dipole

Charges $q_1 = +6.0$ nC and $q_2 = -6.0$ nC are fixed on the $x$ -axis at $x = -0.050$ m and $x = +0.050$ m. Find (a) the potential energy of that pair alone, (b) the work an external force must do to bring $q_3 = +2.0$ nC in from infinity to the point $x = +0.150$ m on the axis, and (c) the total potential energy of the three-charge system afterwards.

(a) One pair, one term. The separation is $0.100$ m, so $U_{12} = (9.0 \times 10^9)\dfrac{(6.0 \times 10^{-9})(-6.0 \times 10^{-9})}{0.100} = -3.24 \times 10^{-6}$ J, that is $-3240$ nJ. Negative, as it must be for opposite charges.
(b) Use the potential rather than two pair terms. The work to bring $q_3$ from infinity is $W_{\text{ext}} = q_3 V$ , where $V$ is the potential at the destination due to $q_1$ and $q_2$ . This is faster and it is the Topic 9.2 shortcut.
Distances from the destination. From $x = +0.150$ m, the charge at $x = -0.050$ m is $0.200$ m away and the charge at $x = +0.050$ m is $0.100$ m away.
Potential there, a scalar sum. $V = k\left(\dfrac{6.0 \times 10^{-9}}{0.200} + \dfrac{-6.0 \times 10^{-9}}{0.100}\right) = (9.0 \times 10^9)(3.0 \times 10^{-8} - 6.0 \times 10^{-8}) = -270$ V.
The work. $W_{\text{ext}} = q_3 V = (2.0 \times 10^{-9})(-270) = -5.4 \times 10^{-7}$ J, that is $-540$ nJ. Negative: the pair pulls the positive test charge in, so you have to hold it back, and the field does the work.
Check against pair terms. $U_{13} = k(6.0 \times 10^{-9})(2.0 \times 10^{-9})/0.200 = +540$ nJ and $U_{23} = k(-6.0 \times 10^{-9})(2.0 \times 10^{-9})/0.100 = -1080$ nJ. Their sum is $-540$ nJ, matching.
(c) Add the new terms to the old total. $U_{\text{total}} = -3240 + 540 - 1080 = -3780$ nJ. The pair term is unchanged, because moving $q_3$ in did nothing to $q_1$ and $q_2$ .

(a) $U_{12} = -3.24\ \mu$ J. (b) $W_{\text{ext}} = q_3 V = (2.0\ \mathrm{nC})(-270\ \mathrm{V}) = -540$ nJ, negative because the destination is at negative potential. (c) $U_{\text{total}} = -3.78\ \mu$ J. Note that the potential at $x = +0.150$ m is negative even though the nearer charge is negative and the further one is positive: the $1/r$ weighting decides it.

Frequently asked questions

What is electric potential energy in AP Physics C?

It is the work an external force must do to bring a set of charges from infinitely far apart to their present positions. For two point charges the CED gives it as U_E = (1/(4 pi epsilon-zero)) q1 q2 / r, equivalently k q1 q2 / r. It is a scalar, it belongs to the system rather than to any one charge, and it carries the signs of both charges: like charges give a positive value, opposite charges a negative one. The zero is at infinite separation, which the exam's printed conventions fix for you.

Is electric potential energy the same in AP Physics 2 and AP Physics C?

Almost entirely. AP Physics C Topic 9.1 and AP Physics 2 Topic 10.4 share the same learning objective, the same equation, and the same three essential-knowledge statements, two of them word for word and the third differing by a single word, potential energy between two charged objects in Physics C against of two charged objects in Physics 2. The substantive difference is a boundary statement: AP Physics 2 caps calculations at configurations of four or fewer point charges because extended distributions exceed that course's scope, while AP Physics C prints no boundary statement at all for this topic. In the calculus-based course, extended charge distributions are in scope, handled through the potential integral in Topic 9.2.

Why is electric potential energy negative for opposite charges?

Because the external work needed to assemble them is negative. Opposite charges attract, so as you bring one in from infinity the electric force already pulls it inward; the external force must point outward while the displacement points inward, so the external work is negative. The physical meaning is that you would have to put energy in to pull them back apart. Like charges give the opposite result: you push them together against their repulsion, doing positive work, and the stored energy is positive.

How do you find the total electric potential energy of several point charges?

Add one term k q_i q_j / r_ij for every pair, using the signed charges. The number of terms is N(N-1)/2, so three charges give three terms, four give six and five give ten. Do not divide by two: the pairwise sum already counts each interaction once, and the factor of one half only appears if you sum over ordered pairs. This is CED statement 9.1.A.3, which is prose in the framework and is not printed on the equation sheet, so you write the sum out yourself.

Is U = k q1 q2 / r on the AP Physics C equation sheet?

The sheet prints the potential energy of two point charges in the permittivity form only, U_E = (1/(4 pi epsilon-zero)) q1 q2 / r. The k form is printed in the CED's own statement 9.1.A.2 but not in the exam booklet, unlike the Coulomb force line just above it, which does show both forms. This costs nothing, because the same page's constants table defines k = 1/(4 pi epsilon-zero) = 9.0 x 10^9 N m^2 / C^2.

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

Electric potential energy is a property of a configuration of charges and is measured in joules. Electric potential is energy per unit charge at a point in space and is measured in volts, one volt being one joule per coulomb. A point in space has a potential whether or not any charge is sitting there; potential energy only exists once a charge occupies the point. The link is U_E = qV for a charge q at a location of potential V, and its change form, delta U_E = q delta V, is printed on the equation sheet.

Where is electric potential energy zero?

At infinite separation, by convention, and the AP Physics C exam commits to this in print: the conventions box on the equation sheet states that the electric potential is zero at an infinite distance from an isolated point charge. That choice is what makes the formula k q1 q2 / r come out with no additive constant. It also means a value of U_E is always relative to the charges being infinitely far apart, so a negative value signals a bound configuration rather than a physically negative amount of energy.