AP Physics C: E&M · Topic 9.3

Topic 9.3: Conservation of Electric Energy

Q: What is conservation of electric energy in AP Physics C?

It is the statement that a charged object moving between two locations at different electric potentials changes the electric potential energy of the object-field system by q times the potential difference, and that the object's kinetic energy changes to match, consistent with the conservation of energy. Those are CED statements 9.3.A.1 and 9.3.A.2. In practice, if no other force does work, the change in kinetic energy is the negative of q delta V.

Q: Is Topic 9.3 different in AP Physics C from AP Physics 2?

Barely. AP Physics C Topic 9.3 and AP Physics 2 Topic 10.7 have identical essential knowledge, statement for statement, with the same equation, and neither prints a boundary statement. The learning objective differs by two words: Physics 2 says describe changes in energy in a system, Physics C says describe changes in a system. What actually differs is what feeds the equation. In the calculus-based course the potential difference can come from integrating a charge distribution or from a line integral of a non-uniform field, so the same relation is applied to problems the algebra-based course cannot pose.

Q: How fast does a charge move after crossing a potential difference?

If it starts from rest and nothing else does work, its kinetic energy is the magnitude of q delta V, so the speed is the square root of two times that divided by the mass. The kinetic energy depends only on the charge and the potential difference, not on the geometry that produced them, so a proton and an electron pushed through the same potential difference gain equal energies but very different speeds, in the ratio of the square root of the inverse masses.

Q: Why does a negative charge speed up when it moves to higher potential?

Because the change in potential energy is q times delta V, and q is negative. A negative charge moving to higher potential has a positive delta V multiplied by a negative charge, so the potential energy falls, and by conservation of energy the kinetic energy rises. The general rule behind all four sign cases is that a free charge accelerates toward wherever its own potential energy is lower: positive charges toward low potential, negative charges toward high potential.

Q: What is an electron volt?

An electron volt is the kinetic energy gained by an object carrying one elementary charge when it crosses a potential difference of one volt. The AP Physics C: Electricity and Magnetism constants table prints the conversion, 1 eV = 1.60 x 10^-19 J, which is numerically the elementary charge because that is how the unit is defined. It is a convenient energy unit for this topic: a proton through 500 V gains 500 eV, and an alpha particle carrying two elementary charges gains 1000 eV through the same 500 V.

Q: When can you not use the constant-acceleration equations in an electric field?

Whenever the field is not uniform, which is most of the time outside parallel plates. Near a point charge the force goes as one over r squared, along a ring's axis it varies with position, and near a charged wire it goes as one over r. In all of those the acceleration changes as the object moves, so the kinematic equations do not apply. Energy methods still do, because the potential difference already contains the integral of the force along the path.

Q: Why does the CED say the object-field system rather than the object?

Because potential energy belongs to an interaction, not to a single object. Statement 9.3.A.1 says the change is in the electric potential energy of the object-field system, which keeps the accounting honest: the object on its own has kinetic energy, and the potential energy is shared with the field, or equivalently with the charges that produced the field. On a free-response question, drawing a system boundary and naming what is inside it is skill 1.A, which the CED lists for this topic.

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

When a charged object moves between two locations at different electric potentials, the electric potential energy of the object and field system changes by the charge times the potential difference. Energy is conserved, so that change shows up as kinetic energy with the opposite sign.

AP Physics: Unit 9 (topics 9.3 Conservation of Electric Energy). Topic 9.3 of the current AP Physics C: Electricity and Magnetism course and exam description, inside Unit 9, weighted 10 to 20% of the multiple-choice section at about 10 to 20 class periods. One learning objective, 9.3.A, using the task verb describe, with two essential-knowledge statements: 9.3.A.1 gives the change in the electric potential energy of the object-field system as delta U_E = q delta V, and 9.3.A.2 states that movement between two points at different potentials changes the object's kinetic energy consistent with the conservation of energy. Topic 9.3 prints no boundary statement. Suggested skills are 1.A, 2.C, 2.D, 3.A and 3.C, five in all, more than either other Unit 9 topic, and 3.A appears nowhere else in the unit. The equation delta U_E = q delta V is printed on both the AP Physics C: E&M and the AP Physics 2 equation sheets. The corresponding AP Physics 2 topic, 10.7, has identical essential knowledge word for word and also prints no boundary statement; its learning objective reads describe changes in energy in a system, where the Physics C objective reads describe changes in a system, and its suggested skills are 1.C, 2.A, 2.C and 3.C. The calculus-based difference is not in the relation itself but in how the potential difference is obtained, whether by integrating a charge distribution or by taking a line integral of a non-uniform field.

What Topic 9.3 requires

Topic 9.3 is the shortest topic in Unit 9: one learning objective, two essential-knowledge statements, one equation.

9.3.A, describe changes in a system due to a difference in electric potential between two locations.

9.3.A.1 states that 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 the equation $\Delta U_E = q\Delta V$ .
9.3.A.2 states that the movement of a charged object between two points with different electric potentials results in a change in kinetic energy of the object consistent with the conservation of energy.

Topic 9.3 prints no boundary statement. Neither does Topic 9.1. Unit 9's only boundary statement sits under Topic 9.2 and restricts which distributions you integrate for potential.

The CED lists five suggested skills, more than either other topic in the unit: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Skill 3.A appears in Unit 9 only here, which is a signal about how the topic is examined. Energy conservation is the part of this unit you can measure on a bench, so it is the part the experimental-design question can be built on.

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

This is the same physics as AP Physics 2, and the CED says so almost word for word

Say this plainly rather than manufacture a difference. Compare Topic 9.3 with Topic 10.7 in AP Physics 2:

9.3.A.1 and 10.7.A.1 are identical, sentence for sentence, and both give the same relevant equation $\Delta U_E = q\Delta V$ .
9.3.A.2 and 10.7.A.2 are identical, sentence for sentence.
Neither course prints a boundary statement here.
Both sheets print $\Delta U_E = q\Delta V$ , the AP Physics 2 sheet and the AP Physics C: E&M sheet alike.

There is exactly one wording difference in the whole topic, and it is in the learning objective. AP Physics 2 says "Describe changes in energy in a system due to a difference in electric potential between two locations." AP Physics C drops those two words: "Describe changes in a system due to a difference in electric potential between two locations."

That is a widening, not a narrowing. In the calculus-based course the question can be about anything that changes, not just an energy: a speed, an acceleration, a time of flight, the shape of a trajectory. Which is exactly how the C exam poses it, and it is why the derivative and integral relations from Topic 9.2 are in the room.

The suggested skills also differ. AP Physics 2 lists 1.C, 2.A, 2.C and 3.C. AP Physics C lists 1.A, 2.C, 2.D, 3.A and 3.C, swapping the graph-sketching skill for diagram creation and adding experimental design.

If you want the algebra-based treatment, read the AP Physics 2 topic. Nothing on it is wrong for a Physics C student. The rest of this page is about the difference calculus makes to what you can apply it to, because that is where the two courses actually part.

What calculus changes: the fields you can apply this to

$\Delta U_E = q\Delta V$ is not a calculus statement. What changes in AP Physics C is where the $\Delta V$ comes from.

In the algebra-based course, potential differences arrive one of three ways: from a scalar sum over a handful of point charges, from a stated battery voltage, or from a uniform field via $\lvert \vec{E} \rvert = \lvert \Delta V/\Delta r \rvert$ . All three give you a number to put into $q\Delta V$ without any integration.

In AP Physics C the same equation is fed by:

A potential computed by integrating a continuous distribution. $V = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r}$ for a ring, an arc, or a finite line, then $\Delta V$ between two points on that object's axis.
A potential difference computed from a non-uniform field. $\Delta V = -\int_a^b \vec{E} \cdot d\vec{r}$ , which is how you handle a charge moving radially away from a point charge, or outward from a charged wire where the field goes as $1/r$ .
A potential given to you as a function, from which you can also recover the force at any point through $E_x = -dV/dx$ and then $F_x = qE_x$ .

The last of these is the connection that makes Physics C's version genuinely different. In the algebra-based course, energy methods and force methods are two separate toolkits. Here they are linked by a derivative: $F_x = qE_x = -q\,\dfrac{dV}{dx} = -\dfrac{dU_E}{dx}$ , which is the same relation the mechanics table on the same sheet prints as $F_x = -\dfrac{dU(x)}{dx}$ .

That gives you a fast reading of any potential-energy curve, exactly as in mechanics: the force is minus the slope, equilibrium points are where the slope is zero, and a minimum is stable while a maximum is not. Skill 1.C in Topic 9.1 asks you to sketch such graphs, and skill 2.D here asks you to predict factors of change from them.

One consequence to be careful with. Because the field is often non-uniform, $\Delta V = -Ed$ usually does not apply and neither does constant acceleration. A charge released near a point charge speeds up at a decreasing rate, so kinematics is out and only energy will do.

Getting the signs right, which is the whole difficulty

Nothing in this topic is conceptually hard. Marks are lost on sign bookkeeping, and there is a reliable way through it.

Write the two statements as one line. If gravity and other forces are negligible and nothing external does work,

\Delta K + \Delta U_E = 0 \quad \text{so} \quad \Delta K = -q\Delta V

Now work through the four sign combinations, because all four appear on exams.

Charge	Moves to	$\Delta V$	$\Delta U_E = q\Delta V$	$\Delta K$
Positive	Lower potential	Negative	Negative	Positive, it speeds up
Positive	Higher potential	Positive	Positive	Negative, it slows
Negative	Higher potential	Positive	Negative	Positive, it speeds up
Negative	Lower potential	Negative	Positive	Negative, it slows

The pattern worth remembering is not the table but the sentence behind it: a free charge accelerates toward wherever its own potential energy is lower. Positive charges fall toward low potential; negative charges fall toward high potential. Both statements say $\Delta U_E < 0$ .

Two habits keep this straight.

Compute $\Delta V$ as final minus initial, and carry the sign into $q$ as well. Substituting an electron as $q = -1.60 \times 10^{-19}$ C and letting the algebra decide beats deciding first and checking later.

Sanity-check the direction against the field. Statement 9.2.B.3.iii says the field points toward decreasing potential, and a positive charge feels a force along the field. Those two together reproduce the first row of the table without any arithmetic.

One more piece of vocabulary. Statement 9.3.A.1 says the change belongs to the object-field system, not to the object. That phrasing is deliberate and it matters when a question asks what the energy of the object is: on its own, a charge has kinetic energy and nothing else, and the potential energy is shared with the field or, equivalently, with the other charges that created it. Draw a system boundary on your diagram, which is skill 1.A.

Electron volts, and the accelerating-potential shortcut

The most common single application of this topic is a charged particle starting from rest and crossing a potential difference. It is worth having as a reflex.

Start from $\Delta K = -q\Delta V$ with $K_i = 0$ :

\tfrac{1}{2}mv^2 = \lvert q\Delta V \rvert \quad \Rightarrow \quad v = \sqrt{\frac{2\lvert q\Delta V \rvert}{m}}

Two features are worth noticing. The speed depends on the charge-to-mass ratio, so an electron and a proton accelerated through the same potential difference reach the same kinetic energy but very different speeds. And the kinetic energy depends on the potential difference alone, not on the geometry that produced it: the same $500$ V does the same job whether it comes from a battery, a pair of plates or a point charge.

That second fact is what the electron volt is built on. One electron volt is the kinetic energy an object carrying one elementary charge gains crossing a potential difference of one volt, and the constants table on the equation sheet prints the conversion $1\ \mathrm{eV} = 1.60 \times 10^{-19}$ J alongside the elementary charge $e = 1.60 \times 10^{-19}$ C. They are numerically the same because that is the definition.

So a proton pushed through $500$ V gains $500$ eV, an alpha particle carrying $2e$ gains $1000$ eV through the same $500$ V, and you convert to joules only when you need a speed in metres per second.

Be careful about one thing the shortcut hides. $v = \sqrt{2\lvert q\Delta V\rvert/m}$ assumes the particle starts from rest and that no other force does work. If gravity matters, or if a second charged object also moves, add those terms. The second case is the one that catches people: when two free charges repel each other apart, both gain kinetic energy, and momentum conservation, not the potential alone, decides how the energy splits. That is the second worked example below, and it is where AP Physics 1's conservation of linear momentum comes back.

Traps

Using $q\Delta V$ with the magnitude of the charge. Put the sign in. An electron moving to higher potential loses potential energy, which looks backwards until you substitute $q = -e$ .

Confusing $\Delta V$ with $V$ . A question that gives you the potential at two points wants the difference; a question that gives a battery voltage has already handed you the difference. Both appear.

Assuming constant acceleration. Only a uniform field gives that. Between parallel plates, yes, and then the projectile analogy of statement 10.3.A.3.iii applies. Near a point charge or along a ring's axis, no.

Forgetting that both objects move. If the second charge is not held fixed, energy alone underdetermines the answer. Use momentum conservation as well, and state that the total momentum starts at zero if both start at rest.

Dropping gravity when it matters. Most electrostatics problems have a charge-to-mass ratio large enough that gravity is negligible, and the exam usually says so. A millimetre-scale oil drop or a suspended pith ball is the case where it does not.

Quoting a speed without checking the energy is positive. If $\Delta K$ comes out negative and the object started from rest, the motion you assumed cannot happen. That is a real answer to a "can it reach point B?" question, not an arithmetic error.

Treating potential difference as if it had a direction. $\Delta V$ is a signed scalar, not a vector. A charge that moves perpendicular to the field crosses no potential difference at all, whatever distance it covers, because it stays on an isoline.

How Topic 9.3 is assessed

The AP Physics C: Electricity and Magnetism exam is 3 hours long: 42 multiple-choice questions worth 50% in 85 minutes, and 4 free-response questions worth 50% in 95 minutes. The four free-response questions always appear in the same order: Mathematical Routines (10 points, 20 to 25 minutes), Translation Between Representations (12 points, 25 to 30 minutes), Experimental Design and Analysis (10 points, 25 to 30 minutes), and Qualitative/Quantitative Translation (8 points, 15 to 20 minutes).

Topic 9.3 is the Unit 9 topic that carries skill 3.A, create experimental procedures that are appropriate for a given scientific question, which is the science practice the third free-response question is built on. The CED describes that question as expecting a scientifically sound method: vary a single parameter, measure how that change affects a single characteristic, using equipment realistically available in a high school laboratory. A question in this topic might ask you to design a way to determine an unknown charge, or an unknown accelerating potential, from measurements of speed or deflection.

Skill 2.D, predict new values or factors of change using functional dependence, is the other distinctive one here, and it is what the fourth free-response question rewards. If the potential difference doubles, the kinetic energy doubles and the speed goes up by $\sqrt{2}$ . If the charge doubles at fixed potential difference, the energy doubles. Being able to say that without recomputing is the skill being tested.

The Unit 9 Progress Check in AP Classroom is about 18 multiple-choice questions and 4 free-response questions.

For the general energy-conservation routine that this topic specialises, see the conservation of energy guide. For the potential differences that feed it, Topic 9.2; for the energy stored when charge is separated rather than moved, Topic 10.3.

A proton accelerated through 500 volts

A proton starts from rest and moves from a location at $+500$ V to a location at $0$ V. Find the change in the electric potential energy of the proton-field system, the proton's kinetic energy in both electron volts and joules, and its final speed. Take $e = 1.60 \times 10^{-19}$ C and $m_p = 1.67 \times 10^{-27}$ kg, and treat gravity as negligible.

Write the potential difference as final minus initial. $\Delta V = 0 - 500 = -500$ V. The proton moves to lower potential.
Apply 9.3.A.1 with the signed charge. $\Delta U_E = q\Delta V = (+1.60 \times 10^{-19})(-500) = -8.0 \times 10^{-17}$ J. Negative, so the system energy falls.
Apply 9.3.A.2. Nothing else does work, so $\Delta K = -\Delta U_E = +8.0 \times 10^{-17}$ J. Starting from rest, that is the final kinetic energy.
In electron volts. The proton carries one elementary charge and crossed $500$ V, so $K = 500$ eV by definition. Check: $(500)(1.60 \times 10^{-19}) = 8.0 \times 10^{-17}$ J, matching.
Solve for speed. $v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2(8.0 \times 10^{-17})}{1.67 \times 10^{-27}}} = \sqrt{9.58 \times 10^{10}} = 3.10 \times 10^5$ m/s.
Check the direction makes sense. A positive charge accelerates toward lower potential, which is the direction of the field. Consistent with statement 9.2.B.3.iii.
Predict without recomputing, which is skill 2.D. An electron through the same $500$ V would also gain $500$ eV, but with $m_e = 9.11 \times 10^{-31}$ kg its speed would be larger by $\sqrt{m_p/m_e} = \sqrt{1833} = 42.8$ times.

$\Delta U_E = -8.0 \times 10^{-17}$ J, $K = 8.0 \times 10^{-17}$ J, which is $500$ eV, and $v = 3.10 \times 10^5$ m/s. The kinetic energy in electron volts is just the charge in elementary charges times the potential difference in volts, which is why the unit exists.

Two charged spheres released from rest: energy is not enough on its own

Two small spheres are held $0.10$ m apart on a frictionless horizontal surface. Sphere 1 has mass $2.0 \times 10^{-6}$ kg and charge $+30$ nC; sphere 2 has mass $6.0 \times 10^{-6}$ kg and the same charge $+30$ nC. They are released from rest. Find the speed of each when they are very far apart.

Find the initial potential energy of the pair. From Topic 9.1, $U_i = k\dfrac{q_1 q_2}{r} = (9.0 \times 10^9)\dfrac{(30 \times 10^{-9})^2}{0.10} = \dfrac{8.1 \times 10^{-6}}{0.10} = 8.1 \times 10^{-5}$ J.
At large separation, $U_f = 0$ , by the printed convention that the potential of an isolated point charge is zero at infinity. So the total kinetic energy at the end is $8.1 \times 10^{-5}$ J.
Energy alone is one equation and there are two unknowns. This is the step that catches people. You cannot split the energy in the ratio of anything without a second condition.
Use momentum conservation. The system starts at rest and the electric forces are internal and equal and opposite, so total momentum stays zero: $m_1 v_1 = m_2 v_2$ , giving $v_1 = \dfrac{m_2}{m_1}v_2 = 3v_2$ . The lighter sphere ends up faster.
Substitute into the energy equation. $\tfrac{1}{2}m_1(3v_2)^2 + \tfrac{1}{2}m_2 v_2^2 = \tfrac{1}{2}(2.0 \times 10^{-6})(9v_2^2) + \tfrac{1}{2}(6.0 \times 10^{-6})v_2^2 = (9.0 \times 10^{-6} + 3.0 \times 10^{-6})v_2^2 = (1.2 \times 10^{-5})v_2^2$ .
Solve. $v_2^2 = \dfrac{8.1 \times 10^{-5}}{1.2 \times 10^{-5}} = 6.75$ , so $v_2 = 2.60$ m/s and $v_1 = 3v_2 = 7.79$ m/s.
Check both conditions. Kinetic energy: $\tfrac{1}{2}(2.0 \times 10^{-6})(7.79)^2 + \tfrac{1}{2}(6.0 \times 10^{-6})(2.60)^2 = 6.07 \times 10^{-5} + 2.03 \times 10^{-5} = 8.1 \times 10^{-5}$ J. Momentum: $(2.0 \times 10^{-6})(7.79) = (6.0 \times 10^{-6})(2.60) = 1.56 \times 10^{-5}$ kg m/s. Both hold.

$v_1 = 7.79$ m/s for the lighter sphere and $v_2 = 2.60$ m/s for the heavier one. The shortcut $v = \sqrt{2\lvert q\Delta V\rvert/m}$ does not apply here, because it assumes the source of the potential is held fixed. When both objects are free, energy conservation and momentum conservation are needed together.

A bead falling toward a fixed charge in a field that is not uniform

A point charge $Q = +8.0$ nC is fixed in place. A small bead of mass $5.0 \times 10^{-6}$ kg and charge $q = -4.0$ nC is released from rest at $r_i = 0.30$ m from it and slides on a frictionless insulating rod straight toward it. Find the bead's speed as it passes $r_f = 0.10$ m. Explain why the constant-acceleration equations cannot be used.

Write the potentials at the two locations. The fixed charge produces $V = kQ/r$ , so $V_i = \dfrac{(9.0 \times 10^9)(8.0 \times 10^{-9})}{0.30} = 240$ V and $V_f = \dfrac{72}{0.10} = 720$ V.
Potential difference, final minus initial. $\Delta V = 720 - 240 = +480$ V. The bead moves to higher potential, as it must when approaching a positive charge.
Apply 9.3.A.1 with the signed charge. $\Delta U_E = q\Delta V = (-4.0 \times 10^{-9})(+480) = -1.92 \times 10^{-6}$ J. Negative, because a negative charge moving to higher potential loses potential energy. This is the third row of the sign table.
Apply 9.3.A.2. $\Delta K = +1.92 \times 10^{-6}$ J, and the bead started from rest, so that is its kinetic energy at $r_f$ .
Solve for speed. $v = \sqrt{\dfrac{2(1.92 \times 10^{-6})}{5.0 \times 10^{-6}}} = \sqrt{0.768} = 0.876$ m/s.
Why kinematics fails. The force is $\lvert F \rvert = k\lvert qQ \rvert/r^2$ , which grows from $3.2 \times 10^{-6}$ N at $0.30$ m to $2.88 \times 10^{-5}$ N at $0.10$ m, a factor of nine. The acceleration is not constant, so no equation with $\frac{1}{2}at^2$ in it applies. Energy does not care, because the potential difference already contains the whole integral of the force along the path.
Check with the line integral. $\Delta V = -\displaystyle\int_{0.30}^{0.10} \dfrac{kQ}{r^2}dr = kQ\left[\dfrac{1}{r}\right]_{0.30}^{0.10} = 72\left(10 - 3.33\right) = 480$ V, matching.

$v = 0.876$ m/s. The energy route works because $\Delta V$ already carries the integral of the non-uniform field along the path; the kinematic equations do not apply, since the electric force rises by a factor of nine over this displacement.

Frequently asked questions

What is conservation of electric energy in AP Physics C?

It is the statement that a charged object moving between two locations at different electric potentials changes the electric potential energy of the object-field system by q times the potential difference, and that the object's kinetic energy changes to match, consistent with the conservation of energy. Those are CED statements 9.3.A.1 and 9.3.A.2. In practice, if no other force does work, the change in kinetic energy is the negative of q delta V.

Is Topic 9.3 different in AP Physics C from AP Physics 2?

Barely. AP Physics C Topic 9.3 and AP Physics 2 Topic 10.7 have identical essential knowledge, statement for statement, with the same equation, and neither prints a boundary statement. The learning objective differs by two words: Physics 2 says describe changes in energy in a system, Physics C says describe changes in a system. What actually differs is what feeds the equation. In the calculus-based course the potential difference can come from integrating a charge distribution or from a line integral of a non-uniform field, so the same relation is applied to problems the algebra-based course cannot pose.

How fast does a charge move after crossing a potential difference?

If it starts from rest and nothing else does work, its kinetic energy is the magnitude of q delta V, so the speed is the square root of two times that divided by the mass. The kinetic energy depends only on the charge and the potential difference, not on the geometry that produced them, so a proton and an electron pushed through the same potential difference gain equal energies but very different speeds, in the ratio of the square root of the inverse masses.

Why does a negative charge speed up when it moves to higher potential?

Because the change in potential energy is q times delta V, and q is negative. A negative charge moving to higher potential has a positive delta V multiplied by a negative charge, so the potential energy falls, and by conservation of energy the kinetic energy rises. The general rule behind all four sign cases is that a free charge accelerates toward wherever its own potential energy is lower: positive charges toward low potential, negative charges toward high potential.

What is an electron volt?

An electron volt is the kinetic energy gained by an object carrying one elementary charge when it crosses a potential difference of one volt. The AP Physics C: Electricity and Magnetism constants table prints the conversion, 1 eV = 1.60 x 10^-19 J, which is numerically the elementary charge because that is how the unit is defined. It is a convenient energy unit for this topic: a proton through 500 V gains 500 eV, and an alpha particle carrying two elementary charges gains 1000 eV through the same 500 V.

When can you not use the constant-acceleration equations in an electric field?

Whenever the field is not uniform, which is most of the time outside parallel plates. Near a point charge the force goes as one over r squared, along a ring's axis it varies with position, and near a charged wire it goes as one over r. In all of those the acceleration changes as the object moves, so the kinematic equations do not apply. Energy methods still do, because the potential difference already contains the integral of the force along the path.

Why does the CED say the object-field system rather than the object?

Because potential energy belongs to an interaction, not to a single object. Statement 9.3.A.1 says the change is in the electric potential energy of the object-field system, which keeps the accounting honest: the object on its own has kinetic energy, and the potential energy is shared with the field, or equivalently with the charges that produced the field. On a free-response question, drawing a system boundary and naming what is inside it is skill 1.A, which the CED lists for this topic.