AP Physics 2 · Topic 10.4

Topic 10.4: Electric Potential Energy

Unit 10: Electric Force, Field, and Potential15-18% of the multiple-choice section

Electric potential energy belongs to a pair of charges, not to either one alone. For two point charges it equals k times q1 times q2 divided by r, a scalar in joules. Keep both signs: like charges give a positive energy, opposite charges a negative one, and the zero sits at infinite separation.

AP Physics: Unit 10 (topics 10.4 Electric Potential Energy). AP Physics 2 Unit 10, Topic 10.4. One learning objective, 10.4.A, describe the electric potential energy of a system. Its three essential knowledge statements define the electric potential energy of two point charges as the work an external force must do to bring them from infinitely far away to their current positions, give the general form U_E = q1 q2 / (4 pi epsilon-nought r) = k q1 q2 / r, and require the total for a configuration to be summed over the individual interactions between each pair. The boundary statement limits calculations to configurations of four or fewer point charges, because the methods for extended charge distributions exceed the scope of the course. Unit 10 carries 15 to 18 percent of the multiple-choice section over roughly 14 to 21 class periods, and the suggested skills for this topic are 1.C, 2.A, 2.D, and 3.C.

What Topic 10.4 requires

Topic 10.4 is compact by the standards of Unit 10: one learning objective, three essential knowledge statements, no sub-statements at all.

10.4.A Describe the electric potential energy of a system.

10.4.A.1 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.
10.4.A.2 The general form for the electric potential energy of two charged objects is given by the equation $U_E = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r} = k\frac{q_1 q_2}{r}$ .
10.4.A.3 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.

The boundary statement reads, in full: "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." Both halves matter. The clause before the comma is the reason for the limit, and the limit itself is four or fewer point charges, which is a tighter phrase than the one Topic 10.3 uses for fields.

Suggested skills for this topic are 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Notice that 2.B, the calculate-a-number skill, is not among them, even though the boundary statement above does require students to calculate the energy of small configurations. Read the pairing as a hint about emphasis: sketch it, derive it, scale it, justify it. Unit 10 carries 15 to 18 percent of the multiple-choice section, over roughly 14 to 21 class periods.

The definition is work brought in from infinity (10.4.A.1)

The CED does not open with the formula. It opens with a construction: 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.

Read that as a recipe and three things fall out at once.

The energy belongs to the system. You cannot assemble one charge; assembling takes two, brought toward each other. So $U_E$ is a property of the pair and their separation, exactly as Topic 3.3 in AP Physics 1 insists that gravitational potential energy belongs to the object-Earth system and not to the object. Write the two charges store $-1.8 \ \mu\mathrm{J}$ , not this charge has $-1.8 \ \mu\mathrm{J}$ .

The zero is at infinite separation, and it is not your choice. Because the construction starts infinitely far away and counts work from there, $U_E = 0$ when $r$ grows without limit. This is the opposite of the near-surface gravitational case, where 3.3.A.3 says the observer picks the zero. Here the formula picks it, which is why absolute values of $U_E$ are meaningful rather than only differences.

Negative energy is built in, not a mistake. For two opposite charges the external agent does not have to push them together at all; they pull themselves in, and the external force must hold back, doing negative work. Negative work in gives a negative $U_E$ . For two like charges the external agent has to shove, doing positive work, so $U_E$ comes out positive. The sign is the answer to which way does this pair want to move, and it comes straight out of the definition rather than being a convention layered on top.

The equation keeps the signs, and the force law does not (10.4.A.2)

The AP Physics 2 equation sheet prints this as the fourth equation in its Electricity group:

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

Now compare it to Coulomb's law, the first equation in that same group:

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

The absolute-value bars are on the force line and not on the energy line. Everything about signs on this page follows from it. The force equation is written for a magnitude, so you strip the signs, compute a size, and decide attraction or repulsion separately. The energy equation is written for a signed scalar, so you keep both signs, multiply them, and the sign of the product is part of the answer.

Three more details in that line.

The denominator is $r$ , not $r^2$ . Energy falls off as one over the separation while force falls off as one over the separation squared.
$r$ is the center-to-center separation of the two charges, the same $r$ that appears in Coulomb's law.
$k = 9.0 \times 10^9 \ \mathrm{N \cdot m^2/C^2}$ and $\varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{C^2/(N \cdot m^2)}$ , both from the Constants and Conversion Factors group of the same sheet. Use the printed 9.0, not 8.99.

The unit is the joule, and it has to be. Newton metre squared per coulomb squared, times coulomb squared, divided by metre, leaves newton metres, which are joules. Running that check on the units is a fast way to catch an $r^2$ that should have been an $r$ .

Reading the sign of the potential energy

$U_E$ is a scalar, so it has no direction, but it does have a sign, and the sign carries physics. Suggested skill 3.C is justifying a claim from physical principles, and this is the claim it most often asks for.

Pair	Sign of $q_1 q_2$	Sign of $U_E$	What that means
Both positive	Positive	Positive	Repel; energy was stored pushing them together
Both negative	Positive	Positive	Repel; energy was stored pushing them together
One of each	Negative	Negative	Attract; the pair is bound, below the zero at infinity

Two habits keep this straight.

Positive means the pair would fly apart if released, and would give up energy doing it. Release two like charges and their potential energy falls toward zero while their kinetic energy climbs.

Negative means the pair is bound. Release two opposite charges from rest and they accelerate toward each other, so $r$ shrinks, so $U_E$ becomes more negative, and again kinetic energy climbs. Both cases move the system to lower potential energy; the sign of the starting value only tells you where the system sits relative to the infinitely separated zero.

The trap to avoid is importing a habit from Coulomb's law, where a negative product is a cue to write attraction and then drop the minus sign. Here the minus sign stays in the number. A configuration energy of $-5.4 \times 10^{-7} \ \mathrm{J}$ is not a mistake to be tidied into a positive value.

Scaling: one over r, not one over r squared (skill 2.D)

Suggested skill 2.D is predicting new values or factors of change using functional dependence between variables, and it is on this topic's list. Questions that hand you no numbers at all are testing exactly that.

Because $U_E \propto 1/r$ while $\left|\vec{F}_E\right| \propto 1/r^2$ , the same change in separation moves the two quantities by different factors:

Change to $r$	Factor on the energy $U_E$	Factor on the force magnitude
Double it	One half	One quarter
Triple it	One third	One ninth
Halve it	Double	Quadruple

Charge dependence is simpler and identical for both: double either charge and both the energy and the force double, because both are linear in $q_1$ and in $q_2$ .

One consequence worth carrying into Topic 10.7. Because energy falls off more slowly than force, two charges that are far apart can have a negligible force between them and still have a potential energy that matters to the energy accounting. Deciding that a distant charge is too far away to matter is safe for a free-body diagram and not safe for an energy bar chart.

Add up a configuration pair by pair (10.4.A.3)

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.

Because $U_E$ is a scalar, sum here means ordinary signed addition. No components, no angles, no diagrams of arrows. That is the whole advantage energy has over field: the geometry of the arrangement enters only through the separations $r$ , never through directions.

The bookkeeping rule is to count pairs, not charges. A system of $n$ charges has $n(n-1)/2$ pairs:

Charges	Pairs to add
2	1
3	3
4	6

The boundary statement caps you at four or fewer point charges, which means at most six terms. Count the pairs before you compute anything, so a term cannot go missing, and write each pair once. Adding both $U_{12}$ and $U_{21}$ double counts the same interaction.

Two errors show up repeatedly here. The first is using the distance from a charge to some chosen origin instead of the distance between the two charges in the pair. The second is forgetting that a pair of charges with opposite signs subtracts from the total, which can make a configuration containing large individual energies come out near zero. Worked example two below runs a three-charge configuration where exactly that happens.

The bridge to potential: delta U equals q delta V

The sheet's Electricity group also prints

\Delta U_E = q \Delta V

and this is the equation that connects Topic 10.4 to the rest of the unit. Two things about it are worth getting right.

Where the CED files it. This equation is not part of Topic 10.4's essential knowledge. It appears as essential knowledge 10.7.A.1 in Topic 10.7: when a charged object moves between two locations with different electric potentials, the resulting change in the electric potential energy of the object-field system is given by that equation. Topic 10.5 supplies the matching definition, essential knowledge 10.5.A.3: the electric potential difference between two points is the change in electric potential energy per unit charge when a test charge is moved between the two points, $\Delta V = \Delta U_E / q$ .

What each form needs. $U_E = kq_1q_2/r$ needs both charges and their separation, and gives you the energy of that pair. $\Delta U_E = q\Delta V$ needs one moving charge and a potential difference produced by everything else, and gives you the change in energy of that charge together with the field it moves through. The second form is what you reach for whenever the question hands you a voltage instead of a set of source charges.

Keep the two quantities apart in your head. $U_E$ is energy, a scalar in joules, and it belongs to the pair. $V$ is potential, a scalar in volts, energy per unit charge, and it belongs to a location. Multiplying a potential by the charge you place there converts one into the other, and dividing goes back. Potential itself, including how to compute it from a set of charges, belongs to Topic 10.5, and the electric field and potential guide works that procedure through with numbers.

Sketching the graph, and how the exam frames this

Suggested skill 1.C is creating qualitative sketches of graphs that represent features of a model or the behavior of a physical system, and it is the first skill listed for this topic. The graph to be able to draw from memory is $U_E$ against $r$ for a fixed pair of charges.

Like charges. The curve lies entirely above the horizontal axis, is very large at small $r$ , and falls toward zero as $r$ increases without limit. It never touches the axis and never crosses it.

Unlike charges. The mirror image below the axis: very large in magnitude and negative at small $r$ , rising toward zero from below as $r$ increases. Again it approaches the axis without crossing.

Both curves are $1/r$ shapes, not $1/r^2$ shapes, so they flatten more slowly than a force graph drawn on the same axes. A sketch that crosses the horizontal axis is wrong for either sign, because $q_1q_2/r$ cannot change sign while the charges stay put.

The CED's Unit 10 opener names one place this shows up. The second free-response question on the AP Physics 2 Exam, the Translation Between Representations question, asks students to create graphical and verbal models of a scenario and compare them to mathematical representations of the same situation. The example the CED gives runs through this topic: sketch equipotential lines around a small positively charged sphere, then create energy bar charts for a system containing that sphere and a small point charge released from rest, then explain how the two representations are consistent with each other. Energy bar charts are the representation this topic feeds. The CED also says content from any unit may appear in that question, so treat it as a pattern to practice rather than a prediction.

Where to go next. Topic 10.3 covers the same charges as a field instead of an energy, Topic 10.5 turns energy per unit charge into potential, and Topic 10.7 sets the energy in motion. The Coulomb's law guide and Coulomb's law calculator handle the force these charges also exert, the conservation of energy guide is the general method this feeds into, and every symbol used here is on the AP Physics 2 formula sheet.

Two charges, and the work needed to separate them

A charge $q_1 = +3.0 \times 10^{-9} \ \mathrm{C}$ and a charge $q_2 = -8.0 \times 10^{-9} \ \mathrm{C}$ are held $0.12 \ \mathrm{m}$ apart. Find (a) the electric potential energy of the pair, (b) the magnitude of the electric force between them, and (c) the work an external agent must do to move them to a separation of $0.36 \ \mathrm{m}$ .

(a) Use $U_E = kq_1q_2/r$ with the signs kept, as the sheet prints it. The product of charges is $(3.0 \times 10^{-9})(-8.0 \times 10^{-9}) = -2.4 \times 10^{-17} \ \mathrm{C^2}$ .
Multiply by $k$ : $(9.0 \times 10^9)(-2.4 \times 10^{-17}) = -2.16 \times 10^{-7}$ . Divide by $r = 0.12 \ \mathrm{m}$ : $U_E = -2.16 \times 10^{-7} / 0.12 = -1.8 \times 10^{-6} \ \mathrm{J}$ , that is $-1.8 \ \mu\mathrm{J}$ .
The sign is the check. The charges are opposite, so the pair is bound and $U_E$ must be below the zero at infinite separation. It is.
(b) The force equation carries absolute-value bars, so drop the signs: $\left|\vec{F}_E\right| = k|q_1q_2|/r^2 = 2.16 \times 10^{-7} / (0.12)^2 = 2.16 \times 10^{-7} / 0.0144 = 1.5 \times 10^{-5} \ \mathrm{N}$ , and it is attractive. Note that this magnitude is positive while the energy was negative: same two charges, two different equations, one with bars and one without.
(c) Recompute the energy at the new separation. Only $r$ changed, and it tripled, so $U_E$ scales by one third: $U_{E,f} = -2.16 \times 10^{-7} / 0.36 = -6.0 \times 10^{-7} \ \mathrm{J}$ .
The external work equals the change in the system's potential energy: $\Delta U_E = -6.0 \times 10^{-7} - (-1.8 \times 10^{-6}) = +1.2 \times 10^{-6} \ \mathrm{J}$ .
Sense check on the sign of the work. Opposite charges attract, so an external agent must pull against that attraction to separate them, doing positive work. Positive it is.

(a) $U_E = -1.8 \times 10^{-6} \ \mathrm{J}$ , negative because the charges are opposite. (b) $1.5 \times 10^{-5} \ \mathrm{N}$ , attractive, a positive magnitude from an equation written with absolute-value bars. (c) The external agent does $+1.2 \times 10^{-6} \ \mathrm{J}$ of work, raising the pair's energy toward the zero at infinite separation.

Three charges on a triangle: summing over pairs

Three point charges sit at the corners of an equilateral triangle of side $0.20 \ \mathrm{m}$ : $q_1 = +2.0 \times 10^{-9} \ \mathrm{C}$ , $q_2 = +2.0 \times 10^{-9} \ \mathrm{C}$ , and $q_3 = -4.0 \times 10^{-9} \ \mathrm{C}$ . Find the total electric potential energy of the configuration.

Count the pairs first, as 10.4.A.3 requires. Three charges give $3(3-1)/2 = 3$ pairs: $(1,2)$ , $(1,3)$ and $(2,3)$ . Three charges is inside the boundary statement's limit of four point charges. Every separation is $0.20 \ \mathrm{m}$ , so the geometry contributes nothing beyond that one number.
Pair $(1,2)$ , both positive: $U_{12} = (9.0 \times 10^9)(2.0 \times 10^{-9})(2.0 \times 10^{-9}) / 0.20$ . The charge product is $4.0 \times 10^{-18}$ , times $k$ is $3.6 \times 10^{-8}$ , divided by $0.20$ gives $U_{12} = +1.8 \times 10^{-7} \ \mathrm{J}$ .
Pair $(1,3)$ , opposite signs: the charge product is $(2.0 \times 10^{-9})(-4.0 \times 10^{-9}) = -8.0 \times 10^{-18}$ , times $k$ is $-7.2 \times 10^{-8}$ , divided by $0.20$ gives $U_{13} = -3.6 \times 10^{-7} \ \mathrm{J}$ .
Pair $(2,3)$ is identical to $(1,3)$ by symmetry, since $q_2 = q_1$ and the separation is the same: $U_{23} = -3.6 \times 10^{-7} \ \mathrm{J}$ .
Add as plain signed numbers, because $U_E$ is a scalar: $U_{\mathrm{total}} = 1.8 \times 10^{-7} - 3.6 \times 10^{-7} - 3.6 \times 10^{-7} = -5.4 \times 10^{-7} \ \mathrm{J}$ .
Sense check. Two of the three interactions are attractive and one is repulsive, and the attractive ones involve the largest charge, so a negative total is expected. Note also that the total, $-0.54 \ \mu\mathrm{J}$ , is smaller in magnitude than the sum of the individual magnitudes, $0.90 \ \mu\mathrm{J}$ : mixing signs partly cancels.

$U_{\mathrm{total}} = -5.4 \times 10^{-7} \ \mathrm{J}$ . The configuration is bound overall. An external agent would have to supply $5.4 \times 10^{-7} \ \mathrm{J}$ to pull all three charges infinitely far apart.

A proton and an electron through the same potential difference

A proton and an electron each move from a point at $250 \ \mathrm{V}$ to a point at $100 \ \mathrm{V}$ . 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 change in electric potential energy for each, in joules and in electron volts.

Compute the potential difference once, final minus initial: $\Delta V = 100 - 250 = -150 \ \mathrm{V}$ . Both particles make the same move, so both use this same $\Delta V$ .
Proton, $q = +1.60 \times 10^{-19} \ \mathrm{C}$ : $\Delta U_E = q\Delta V = (1.60 \times 10^{-19})(-150) = -2.40 \times 10^{-17} \ \mathrm{J}$ .
Convert with the sheet's electron volt: $-2.40 \times 10^{-17} / 1.60 \times 10^{-19} = -150$ , so $\Delta U_E = -150 \ \mathrm{eV}$ . The shortcut is worth noticing. A charge of one elementary charge moving through $\Delta V$ volts changes energy by exactly $\Delta V$ electron volts, which is the definition of the unit.
Electron, $q = -1.60 \times 10^{-19} \ \mathrm{C}$ : $\Delta U_E = (-1.60 \times 10^{-19})(-150) = +2.40 \times 10^{-17} \ \mathrm{J}$ , or $+150 \ \mathrm{eV}$ . Two negatives make a positive, and that is the whole of the physics here.
Interpret the signs. The proton lost potential energy, so a system left to itself would speed it up along that path. The electron gained potential energy, so something had to push it; released from rest at the $250 \ \mathrm{V}$ point, an electron would travel the other way. Turning those energy changes into speeds is Topic 10.7.
Note which quantity is not being used. Neither $q_1q_2$ nor $r$ appears anywhere, because $\Delta U_E = q\Delta V$ needs only the charge that moves and the potential difference it moves through.

Proton: $\Delta U_E = -2.40 \times 10^{-17} \ \mathrm{J}$ , or $-150 \ \mathrm{eV}$ . Electron: $\Delta U_E = +2.40 \times 10^{-17} \ \mathrm{J}$ , or $+150 \ \mathrm{eV}$ . Same path, same potential difference, opposite signs, entirely because of the sign of the charge.

Frequently asked questions

What is electric potential energy in AP Physics 2?

It is the energy stored in the arrangement of a set of charges. Essential knowledge 10.4.A.1 defines it for two point charges as the amount of work required for an external force to bring them to their current positions from infinitely far away. It is a scalar measured in joules, it belongs to the pair rather than to either charge, and for two point charges it equals k q1 q2 / r with both signs kept.

Why is electric potential energy negative for two opposite charges?

Because bringing them together from infinity takes negative work from an outside agent. Opposite charges attract, so they pull themselves inward and the external force has to hold them back rather than push them together. Since 10.4.A.1 defines the potential energy as that external work, and since the zero is fixed at infinite separation, a bound pair ends up below zero. Multiplying a positive by a negative charge in k q1 q2 / r produces the same minus sign.

Do you keep the negative signs in the electric potential energy formula?

Yes. The AP Physics 2 sheet prints the energy as k q1 q2 / r with no absolute value bars, unlike Coulomb's law, which is printed with bars around the charge product because it gives a magnitude. So substitute both charges with their signs, and the sign of the result tells you whether the pair repels, giving a positive energy, or attracts, giving a negative one.

Is electric potential energy the same as electric potential?

No. Electric potential energy is measured in joules and belongs to a pair of charges. Electric potential is measured in volts, is energy per unit charge, and belongs to a location in space. They are linked by the sheet equation delta U_E = q delta V, which the CED gives as essential knowledge 10.7.A.1, and by 10.5.A.3, which defines potential difference as the change in potential energy per unit charge. Multiply a potential by the charge you place there to get an energy.

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

Add the energy of every pair. Essential knowledge 10.4.A.3 says the total is the sum of the electric potential energies of the individual interactions between each pair of charged objects. Three charges make three pairs, so compute k q1 q2 / r for each pair using the distance between those two charges, then add the three results as ordinary signed numbers. There are no components to resolve, because potential energy is a scalar.

Where is the zero of electric potential energy?

At infinite separation, and it is not a choice you make. The definition in 10.4.A.1 counts work done bringing the charges in from infinitely far away, so the energy is zero when they are infinitely far apart. That differs from the near-surface gravitational case in AP Physics 1, where the CED states that the zero is a decision made by the observer. Here the equation k q1 q2 / r fixes it, which is why an absolute value of electric potential energy is meaningful.

How many charges can AP Physics 2 ask you to handle in an energy problem?

Four or fewer. The Topic 10.4 boundary statement says that because 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. Four charges make six pairs, so six terms is the largest sum you would be asked to build.