Topic 10.7: Conservation of Electric Energy

What Topic 10.7 requires

Topic 10.7 is the shortest of the seven Unit 10 topics as the CED writes it: one learning objective, two essential knowledge statements, and no sub-statements at all. The next smallest, 10.2 and 10.4, carry three essential knowledge statements each.

• 10.7.A Describe changes in energy in a system due to a difference in electric potential between two locations.

The two statements under it are:

10.7.A.1 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 following equation.

10.7.A.2 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.

The suggested skills 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.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

There is no boundary statement on this topic. Topics 10.4, 10.5 and 10.6 each carry one, and 10.7 does not, so nothing here is fenced off by an explicit exclusion.

Short does not mean small. Two statements and one printed equation is the whole apparatus behind every accelerated-particle question on the exam, and Unit 10 is weighted at 15 to 18 percent of the multiple-choice section with about 14 to 21 class periods. Read 10.7 as the topic where the unit's three descriptive topics turn into a method: Topic 10.4 tells you the energy of a charge configuration, Topic 10.5 tells you the potential at each point, and 10.7 turns the difference between two of those points into a speed.

One equation, two independent signs

Everything on this page is $\Delta U_E = q \Delta V$, and the entire difficulty is that both factors on the right carry a sign, independently.

$\Delta V$ is always final minus initial: $\Delta V = V_f - V_i$. Compute it in that order and do not reach for the magnitude. $q$ is the signed charge of the moving object, $-1.60 \times 10^{-19}\ \text{C}$ for an electron, positive for a proton. Multiply the two signs and you get the four cases, which are worth learning as a block rather than rederiving under time pressure.

Read down the last two columns and the pattern is one sentence: every object released from rest moves so that $\Delta U_E$ is negative. Objects fall toward lower potential energy. What they do not all do is move toward lower potential, because for a negative charge lower potential energy sits at higher potential. The rule charges move from high voltage to low voltage is a positive-charge rule wearing a general disguise, and rows three and four of the table are where it costs marks.

Two details about the printed equation. The sheet writes the energy with subscript E, $U_E$, to keep it apart from $U_g$ and $U_s$, and the CED's phrasing in 10.7.A.1 assigns the energy to the object-field system rather than to the object alone. That is not pedantry: the energy lives in the arrangement, which is the same reason Topic 10.4 talks about the potential energy of a system of point charges. When you say a proton has 3 J of electric potential energy, you mean the proton together with whatever is producing the field has it.

The pattern that solves every problem on this topic

Statement 10.7.A.2 says the change in kinetic energy is consistent with the conservation of energy, and that is the licence to write one line and be done. With no friction, no other force, and nothing else in the system to absorb energy:

Released from rest, $\Delta K = \frac{1}{2}mv^2$, so

and the minus sign is not decoration. If the object speeds up, $q \Delta V$ is negative and $-q\Delta V$ is a positive number under the square root. If your radicand comes out negative, the object was not going to speed up on that trip, and the correct response is to reread the direction of travel rather than to drop the sign.

Deriving that expression symbolically before putting numbers in is suggested skill 2.A, and it is worth doing for its own sake because the structure tells you three scaling rules at a glance: $v \propto \sqrt{\Delta V}$, $v \propto \sqrt{q}$, and $v \propto 1/\sqrt{m}$. Quadruple the accelerating voltage and the exit speed doubles. Worked example 2 uses the second and third of those to compare two particles in one line.

A note on what is and is not printed. The AP Physics 2 equation sheet prints $\Delta U_E = q \Delta V$ in the Electricity group, and $K = \frac{1}{2}mv^2$, $\Delta K = \sum W_i$ and $\Delta U_g = mg\Delta y$ in the Mechanics and Fluids group, which the Physics 2 sheet reprints in full. What it does not print anywhere is a statement of energy conservation itself. There is no $K_i + U_i = K_f + U_f$ line to copy. You supply that, which is exactly what 10.7.A.2 is asking you to be able to do. The conservation of energy guide drills the general routine, and AP Physics 1 Topic 3.4 is the same principle before charge showed up.

The electron volt, and why it exists

Look at the size of the numbers. An electron carries $1.60 \times 10^{-19}\ \text{C}$, so pushing one through a few hundred volts produces an energy around $10^{-17}\ \text{J}$. Writing that repeatedly is how sign errors and exponent errors get made, so particle physics uses a unit sized for the job.

The Constants and Conversion Factors group of the AP Physics 2 sheet prints the elementary charge as $e = 1.60 \times 10^{-19}\ \text{C}$ and, on its own line, the conversion $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$. The unit-symbols box printed alongside the constants lists the electron volt with symbol eV. Those two lines carry the same digits for a reason: an electron volt is defined as the energy change when one elementary charge moves through a potential difference of one volt. Put $q = e$ and $\Delta V = 1\ \text{V}$ into $\Delta U_E = q\Delta V$ and the answer is $1.60 \times 10^{-19}\ \text{J}$, which is one electron volt by definition.

That gives you a shortcut worth having. A particle of charge magnitude $ne$ accelerated through $\Delta V$ volts gains $n \Delta V$ electron volts of kinetic energy. No arithmetic. An electron through 250 V picks up 250 eV. A proton through 250 V picks up 250 eV. An alpha particle, charge $2e$, through 250 V picks up 500 eV. Convert to joules only at the point where you need a speed, because $K = \frac{1}{2}mv^2$ wants SI units.

Two cautions. The electron volt is a unit of energy, not of voltage and not of charge, despite the name. And it is not an SI unit, so it never goes into a kinematics equation directly. Multiply by $1.60 \times 10^{-19}$ first, every time.

The same convention runs through the rest of the sheet, which is why Planck's constant is printed twice, once as $6.63 \times 10^{-34}\ \text{J} \cdot \text{s}$ and once as $4.14 \times 10^{-15}\ \text{eV} \cdot \text{s}$, and why the atomic mass unit is given a $\text{MeV}/c^2$ equivalent. Getting comfortable with eV in Unit 10 pays off later in the course.

Energy bar charts and the sketch you will be asked for

Suggested skill 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system, is the first skill listed for this topic, and the Unit 10 opener names the representation the CED has in mind. Describing the Translation Between Representations free-response question, the opener says the student might be asked to create energy bar charts to model the energy of a system containing a small point charge that is released from rest and the charged sphere, and then to explain how that representation and a sketch of equipotential lines are consistent with each other.

An energy bar chart for this topic has two bars, $K$ and $U_E$, drawn at an initial instant and again at a final instant. Four rules keep it honest.

1. The total height is the same at both instants, because nothing outside the system does work. That is conservation of energy drawn rather than written.
2. $U_E$ bars can point downward. Electric potential energy is negative whenever the charge sits closer than infinity to charge of the opposite sign, and negative energy is a bar below the axis, not a bar of length zero.
3. $K$ bars never point downward. Kinetic energy is $\frac{1}{2}mv^2$ and cannot be negative.
4. Label which system you drew. The object-field system holds the potential energy; the object alone holds only the kinetic energy, and for that smaller system the electric force is external and does work.

Skill 1.C also covers graph sketches, and the two shapes worth being able to draw freehand come straight from the unit's equations. For a charge moving radially away from a fixed point charge, $U_E \propto 1/r$, so the energy curve is a hyperbola approaching zero as $r$ grows, above the axis for like charges and below it for opposite charges. In the uniform field between parallel plates, the potential changes linearly with position, so $U_E$ against position is a straight line and $K$ against position is a straight line with the opposite slope. Those two sketches answer most qualitative parts on this topic.

The CED lists the Translation Between Representations question at 12 points with a suggested time of 25 to 30 minutes, and its skills are 1.A, 1.C, 2.A, 2.D, 3.B and 3.C. Three of Topic 10.7's four suggested skills, 1.C, 2.A and 3.C, are on that list.

When gravity and other forces join in

Most exam problems on this topic say ignore gravitational effects or use a particle whose weight is absurdly small next to the electric force, and then $\Delta K = -q\Delta V$ is the whole story. When a problem does not say that, the fix is to add a term rather than to change the method.

with $\Delta U_g = mg \Delta y$ from the Mechanics and Fluids group of the sheet and $g = 9.8\ \text{m}/\text{s}^2$ from the Table of Information. Take upward as positive for $\Delta y$ and keep that choice to the end of the problem.

Knowing when to bother is a matter of comparing magnitudes, and the comparison is stark for subatomic particles. A proton has a weight of about $1.6 \times 10^{-26}\ \text{N}$. In a field of $1000\ \text{V}/\text{m}$ the same proton feels an electric force of $1.6 \times 10^{-16}\ \text{N}$, ten orders of magnitude larger. For a dust grain, an oil drop or a pith ball the two can be comparable, which is precisely why those objects show up in problems that want you to include both. Worked example 3 is one of those.

If the problem introduces friction, a drag force, or a hand that pushes, then energy crosses the system boundary and $\Delta K + \Delta U_E$ no longer sums to zero. The general accounting is $\Delta K = \sum W_i$, which the sheet prints, with the work done by every force other than the conservative ones you have already written as potential energies. The routine is identical to the one in AP Physics 1 Topic 3.4; only the list of energy terms is longer.

The four traps on this topic

Using the magnitude of the charge out of habit. Coulomb's law and the point-charge field equation are printed with absolute value bars, so magnitudes are the right instinct there. $\Delta U_E = q \Delta V$ has no bars and the sign of $q$ is load-bearing. Electrons need their minus sign.

Confusing $\Delta V$ with $V$. A question that gives you $V_A = 150\ \text{V}$ and $V_B = 400\ \text{V}$ is not handing you $\Delta V$. Going from A to B, $\Delta V = +250\ \text{V}$; going from B to A it is $-250\ \text{V}$. Same two numbers, opposite energy change. Write the direction of travel down before you subtract.

Treating potential difference as energy. Volts are joules per coulomb, so a potential difference alone tells you nothing about energy until you multiply by a charge. Two particles crossing the same 500 V gap gain different amounts of energy if their charges differ, which is the whole point of worked example 2.

Forgetting that this topic is about a difference. Neither the absolute potential nor the absolute potential energy ever enters the calculation. Only $\Delta V$ does. A charge moving between two points at $-800\ \text{V}$ and $-500\ \text{V}$ is in exactly the same situation as one moving between $0\ \text{V}$ and $+300\ \text{V}$, because both trips have $\Delta V = +300\ \text{V}$. This is why the exam convention that fixes zero potential at infinity never actually changes an answer on this topic.

How the exam frames Topic 10.7

Three question shapes account for most of what this topic produces.

The accelerator. A particle of stated charge and mass starts from rest and crosses a stated potential difference. Find the final speed. This is one line of algebra and the only real work is the sign. Occasionally it is dressed as a pair of parallel plates, in which case Topic 10.6 supplies the potential difference from the field and the gap.

The comparison. Two particles, or one particle in two scenarios, with a question about the ratio of final speeds or energies. This is suggested skill 2.C and it never needs a calculator, because the ratio $v \propto \sqrt{q/m}$ collapses everything.

The justification. A claim about which way a charge moves, or whether its kinetic energy rises, supported by the sign of $q\Delta V$. This is suggested skill 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws, and the evidence the graders want is the sign argument written out rather than the number.

One structural point about the course, from the CED's Exam Information section: science practice 1 is not assessed in the multiple-choice section, while practices 1, 2 and 3 are all assessed in the free-response section, and required course content can be assessed with any skill. So the bar-chart sketching in skill 1.C belongs to the free-response half of the exam, and multiple-choice questions on 10.7 will come through skills 2.A, 2.C and 3.C.

Where this goes next. Unit 11 keeps the same energy accounting and adds a rate: an electric current carries charge through a potential difference continuously, and the sheet's $P = I \Delta V$ is $\Delta U_E = q\Delta V$ divided by time, since the current $I$ is itself $\Delta q / \Delta t$. Recognizing that Topic 11.4 is this topic per second saves learning it twice.