AP Physics 2 · Topic 10.5

Topic 10.5: Electric Potential

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

Electric potential is the electric potential energy per unit charge at a point in space, measured in volts, and one volt is one joule per coulomb. Potential is a scalar, so several charges add as signed numbers with no components. The electric field points toward decreasing potential.

AP Physics: Unit 10 (topics 10.5 Electric Potential). AP Physics 2 Unit 10, Topic 10.5. Two learning objectives: 10.5.A, describe the electric potential due to a configuration of charged objects, and 10.5.B, describe the relationship between electric potential and electric field. The load-bearing statements are 10.5.A.1 (electric potential describes the electric potential energy per unit charge at a point in space), 10.5.A.2 (scalar superposition of the potential due to each point charge), 10.5.A.3 (potential difference is the change in electric potential energy per unit charge when a test charge is moved between two points), 10.5.A.4 (conductors in electrical contact end at the same surface potential), 10.5.B.1 (the average field between two points is the potential difference divided by the distance), and the four sub-statements of 10.5.B.2 on isolines. The boundary statement limits students to calculating the electric potential of configurations of four or fewer particles, or more in situations of high symmetry. Unit 10 carries 15 to 18 percent of the multiple-choice section and about 14 to 21 class periods, and the suggested skills for this topic are 1.A, 2.A, 2.C, and 3.B.

What Topic 10.5 requires

Topic 10.5 carries two learning objectives, and they cut the topic cleanly in half.

10.5.A Describe the electric potential due to a configuration of charged objects.
10.5.B Describe the relationship between electric potential and electric field.

Four essential knowledge statements sit under 10.5.A (10.5.A.1 through 10.5.A.4, with one sub-statement, 10.5.A.3.i), and two sit under 10.5.B (10.5.B.1 and 10.5.B.2, with four sub-statements, 10.5.B.2.i through 10.5.B.2.iv). The CED's suggested skills for this topic are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 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.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

The boundary statement tells you exactly where the arithmetic stops, and its parenthetical is the half people drop:

As the methods to calculate the electric potential due to extended charges exceed the scope of the course, AP Physics 2 only expects that students calculate the electric potential of configurations of four or fewer particles (or more in situations of high symmetry).

Four particles is the working limit for a brute-force sum. A symmetric arrangement can hand you more than four and still be fair game, because the symmetry does the summing for you.

Unit 10 is weighted at 15 to 18 percent of the multiple-choice section of the AP Physics 2 exam, and the CED allots it about 14 to 21 class periods. Within the unit, 10.5 is the hinge. Topic 10.3 gave you a vector field at every point in space. Topic 10.4 gave you the energy of a charge configuration. Topic 10.5 divides that energy by the charge and gets back a scalar you can draw as a contour map.

Potential is energy per coulomb (10.5.A.1 and 10.5.A.3)

Statement 10.5.A.1 is one sentence long: electric potential describes the electric potential energy per unit charge at a point in space.

Read per unit charge as per coulomb and the definition stops being abstract. Park a test charge at a point, ask how much electric potential energy the charge and the field have because it is there, then divide by the size of the test charge. What is left does not depend on the test charge at all. It is a property of the point.

Statement 10.5.A.3 writes the same idea as a difference, which is the form you actually use: 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. The relevant equation printed alongside it is

\Delta V = \frac{\Delta U_E}{q}

and the AP Physics 2 equation sheet prints the rearranged version, $\Delta U_E = q \Delta V$ , in the Electricity group. They are the same statement solved for different letters, and Topic 10.7 is built entirely on the second one.

The unit comes straight out of the definition. Divide joules by coulombs and you get the volt, so one volt is one joule per coulomb. A 9 V battery is a device that moves one coulomb between its terminals with a 9 J change in electric potential energy. That is the whole meaning of the number on the label.

One more piece of bookkeeping. Only differences in potential are physical, so somebody has to choose where zero sits. The AP Physics 2 exam makes the choice for you: the conventions list printed with the Table of Information says the electric potential is zero at an infinite distance from an isolated point charge. Every potential you calculate below is therefore measured against infinity, which is why the potential near a lone negative charge comes out negative rather than merely small.

Scalar superposition, and why the sign goes in (10.5.A.2)

Statement 10.5.A.2 says the electric potential due to multiple point charges can be determined by the principle of scalar superposition of the electric potential due to each of the point charges. The equation printed with it, and on the equation sheet, is

V = \frac{1}{4\pi\varepsilon_0} \sum_i \frac{q_i}{r_i}

Three rules run the sum, and every one of them is a place where students who are fluent with fields go wrong.

The sign of each charge goes in. No absolute value bars. A negative charge contributes a negative term, and the sum can come out positive, negative, or exactly zero.
Each distance appears to the first power. Potential falls off as $1/r$ while the field of a point charge falls off as $1/r^2$ . Squaring $r$ here is the single most common arithmetic slip on this topic.
There are no components and no angles. Add the numbers in any order. Where the charges sit relative to each other affects only the distances.

Compare the workload. Three charges give a potential in three divisions and two additions. The same three charges give a field in three magnitudes, six components, two component sums, a Pythagorean combination and an inverse tangent. That asymmetry is the practical reason potential exists as a quantity.

A detail if you plan to work off the printed sheet. Three point-charge lines in the Electricity group give the constant both ways, as $1/(4\pi\varepsilon_0)$ and as $k$ : Coulomb's law, the magnitude of a point charge's field, and the two-charge potential energy $U_E$ . The superposition line above prints only the $1/(4\pi\varepsilon_0)$ form. The Constants group does give the Coulomb constant as $k = 1/(4\pi\varepsilon_0) = 9.0 \times 10^9$ in SI units, so writing $V = k \sum_i q_i / r_i$ is legitimate, but the substitution is yours to make.

The electric field and potential guide works the single-charge arithmetic step by step. This page stays with what the CED asks you to describe.

Getting the field from the potential (10.5.B.1)

Learning objective 10.5.B asks you to describe the relationship between electric potential and electric field, and 10.5.B.1 states it: the average electric field between two points in space is equal to the electric potential difference between the two points divided by the distance between the two points. The sheet prints it as

\left| \vec{E} \right| = \left| \frac{\Delta V}{\Delta r} \right|

Four things about that line repay attention.

The word average is load-bearing. Between two points in a non-uniform field this returns a mean value along the path, not the field at either endpoint. In a uniform field, the field between two charged parallel plates being the standard case, average and actual are the same number and the equation becomes exact. That is the setting where Topic 10.6 uses it constantly.

Both sides wear absolute value bars. The printed equation returns a magnitude and nothing else. Direction is not in it; direction comes from 10.5.B.2.iii, which says an electric field vector points in the direction of decreasing potential.

Volts per meter and newtons per coulomb are the same unit. A volt is a joule per coulomb, so a volt per meter is a joule per coulomb-meter, and a joule is a newton-meter, so the meters cancel and you are left with newtons per coulomb. Either label is correct on a field answer. Volts per meter is the more useful of the two here because it says out loud what the field is: a rate at which potential drops off with distance.

$\Delta r$ is the distance between the two points, not any distance you like. If your two points are not displaced along the field direction, $\Delta V / \Delta r$ returns the component of the field along the line joining them, which is smaller than the full magnitude. Worked example 2 puts a number on that.

Run the relationship the other way and you get the reading skill the exam tests. A field of 500 V/m means the potential drops by 500 V for every meter you travel along the field, so 5 V across a centimeter. Steep potential means strong field. Flat potential means weak field. That sentence is most of what an equipotential map is telling you.

Reading an equipotential map (10.5.B.2)

Statement 10.5.B.2 is the representation half of the topic: electric field vector maps and equipotential lines are tools to describe the field produced by a charge or configuration of charges and can be used to predict the motion of charged objects in the field. Its four sub-statements are short enough to learn verbatim, and they are the rules you apply to every map you meet.

10.5.B.2.i Equipotential lines represent lines of equal electric potential in space. These lines are also referred to as isolines of electric potential.
10.5.B.2.ii Isolines are perpendicular to electric field vectors. An isoline map of electric potential can be constructed from an electric field vector map, and an electric field map may be constructed from an isoline map.
10.5.B.2.iii An electric field vector points in the direction of decreasing potential.
10.5.B.2.iv There is no component of an electric field along an isoline.

The word isoline is the CED's own vocabulary, borrowed from contour maps, and the analogy is exact. Contour lines join points of equal height, they crowd together where the slope is steep, water runs straight downhill across them, and walking along one costs no climbing. Swap height for potential and gravity for the electric field and every one of those sentences is 10.5.B.2.

The same content as a lookup table:

What you see on the map	What it tells you
Closely spaced isolines	Large field magnitude, since the same $\Delta V$ falls across a smaller $\Delta r$
Widely spaced isolines	Small field magnitude
Evenly spaced parallel isolines	Uniform field, constant in magnitude and direction
The direction that crosses isolines from high $V$ to low $V$	The direction of $\vec{E}$ , at right angles to the lines
A path that follows one isoline	$\Delta V = 0$ , so $\Delta U_E = 0$ for any charge carried along it
Two isolines that cross	An impossible map, since one point cannot hold two potentials

Statement 10.5.B.2.ii promises the translation runs in both directions, and suggested skill 1.A asks you to create the representation rather than only read it. To build isolines from a field vector map, walk at right angles to the local field vector and keep turning so you stay perpendicular. To build field vectors from an isoline map, draw an arrow at each point perpendicular to the local line, pointing toward the lower-potential neighbor, with a length set by how tightly packed the lines are.

Statement 10.5.B.2.iv is the one that turns into free marks. No component of the field lies along an isoline, so the electric force on a charge carried along one is perpendicular to its motion, so that force does no work, so the electric potential energy is unchanged. If a question moves a charge between two points on the same equipotential and asks for the change in electric potential energy, the answer is zero and the path is irrelevant.

Batteries and conductors in contact (10.5.A.3.i and 10.5.A.4)

Two statements under 10.5.A look like footnotes and are not.

10.5.A.3.i Electric potential difference may also result from chemical processes that cause positive and negative charges to separate, such as in a battery.

That single sentence is the bridge from electrostatics to Unit 11. A potential difference does not require a visible charge configuration you can sum over. A chemical reaction that pushes positive and negative charge to opposite terminals produces one just as well, and the volt on the battery label is measured in exactly the same joules per coulomb as the volt you computed from $V = k \sum q_i / r_i$ . When Topic 11.2 puts a battery in a loop, this is the statement that licenses it.

10.5.A.4 When conductors are in electrical contact, electrons will be redistributed such that the surfaces of the conductors are at the same electric potential.

Note what is equalized and what is not. Potential is equalized. Charge is not, field is not, and charge density is not. Connect a large conductor to a small one and charge flows until both surfaces sit at one potential, which in general means the two carry different amounts of charge and have different surface fields. Worked example 3 runs the numbers, and the result explains the unit opener's essential question about where the safest place to be during a lightning storm is: sharp, small-radius conductors reach large surface fields at modest potentials.

The same statement is why every point of an ideal wire is at one potential, and why two points joined by a wire can be treated as one node. Skill 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim, often tests exactly this: given two connected conductors, which quantity are you entitled to set equal?

The four confusions that cost marks

Potential is not potential energy. $V$ is a property of a location and exists whether or not anything is sitting there. $U_E$ is a property of a charge together with everything producing the field, and it needs both. They are related by one factor of $q$ , and the units differ: volts against joules. Reserve $V$ for the empty point and $U_E$ for the charge that arrives at it, and half of this topic's error rate disappears. Topic 10.4 owns the energy side in full.

Positive charges and negative charges do not agree about which way is downhill. Released from rest, a positive charge accelerates toward lower potential and a negative charge accelerates toward higher potential. Both accelerate toward lower potential energy, which is the actual rule, because $\Delta U_E = q \Delta V$ flips sign with $q$ . Saying that charges move from high potential to low potential is true only for positive charges, and it is a claim you will be marked on under skill 3.B.

A large potential does not mean a large field, and zero potential does not mean zero field. Potential is the accumulated value at a point; field is the rate at which that value changes with position. Worked example 1 below builds a point where the potential is exactly zero and the field is 1920 N/C.

A negative potential is a real answer, not a mistake. Because the exam pins zero at infinity, any point closer to net negative charge than to net positive charge has a negative potential. A charge that moves from $-50$ V to $-20$ V has moved to higher potential, and $\Delta V = +30$ V. Subtract in the order final minus initial, keep the signs, and do not silently reach for the magnitude.

How the exam frames Topic 10.5

The Unit 10 opener singles out this material for the second free-response question. It says the second free-response question on the AP Physics 2 exam is the Translation Between Representations question, the TBR, which requires students to create graphical and verbal models of scenarios, as well as compare these models to mathematical representations of the same situation. The opener then gives an example: In the TBR, a student might be asked to sketch equipotential lines around a small, positively charged sphere. It adds that the student might then be asked to create energy bar charts for a small point charge released from rest near that sphere, and finally to explain how the two representations are consistent with each other. The opener closes that paragraph by noting that while Unit 10 content provides especially good practice for the TBR, content from any unit may be included in this free-response question.

The CED's Exam Information section lists the TBR at 12 points with a suggested time of 25 to 30 minutes and skills 1.A, 1.C, 2.A, 2.D, 3.B and 3.C. Three of Topic 10.5's four suggested skills, 1.A, 2.A and 3.B, are on that list, which is a fair signal about the shape of the practice worth doing: sketch the map, derive the relationship, then justify a claim from it.

The same Exam Information section states that science practices 2 and 3 are assessed in the multiple-choice section, that science practice 1 will not be assessed there, and that practices 1, 2 and 3 are all assessed in the free-response section, while noting that required course content can be assessed with any skill. So the drawing itself belongs to the free-response section, and multiple-choice questions on 10.5 lean on skills 2.A, 2.C and 3.B instead.

Practical consequences. Learn to answer what happens to $V$ if questions without computing anything, since $V \propto q$ and $V \propto 1/r$ make most of them one line. Expect an isoline map with a field magnitude, a direction, or an energy change attached, all three of which come from the four sub-statements of 10.5.B.2. And keep the scalar sum ready for four or fewer particles. From here, Topic 10.6 builds the device that stores a potential difference, and Topic 10.7 turns potential differences into speeds.

Two point charges: the potential, and the point where it vanishes

A charge $q_1 = +5.0\ \text{nC}$ sits at $x = 0$ and a charge $q_2 = -3.0\ \text{nC}$ sits at $x = 0.40\ \text{m}$ . (a) Find the electric potential at $x = 0.10\ \text{m}$ . (b) Find the point between the charges where the potential is zero. (c) Find the magnitude and direction of the electric field at that point, and say what the two answers together show.

Use $k = 9.0 \times 10^9\ \text{N} \cdot \text{m}^2/\text{C}^2$ , the Coulomb-constant value printed in the Constants group of the AP Physics 2 sheet, and take rightward as the positive direction throughout.
(a) The point at $x = 0.10\ \text{m}$ is $r_1 = 0.10\ \text{m}$ from $q_1$ and $r_2 = 0.30\ \text{m}$ from $q_2$ . Both distances are first powers.
$V_1 = \dfrac{(9.0 \times 10^9)(+5.0 \times 10^{-9})}{0.10} = \dfrac{45}{0.10} = +450\ \text{V}$ and $V_2 = \dfrac{(9.0 \times 10^9)(-3.0 \times 10^{-9})}{0.30} = \dfrac{-27}{0.30} = -90\ \text{V}$ .
Scalar superposition, per 10.5.A.2: $V = V_1 + V_2 = 450 + (-90) = +360\ \text{V}$ . No components, no angles, and the minus sign on $q_2$ carried straight through.
(b) Let the zero sit a distance $x$ from $q_1$ , so it is $0.40 - x$ from $q_2$ . Setting the sum to zero: $\dfrac{k(5.0 \times 10^{-9})}{x} + \dfrac{k(-3.0 \times 10^{-9})}{0.40 - x} = 0$ .
Cancel $k$ and $10^{-9}$ , cross-multiply: $5.0(0.40 - x) = 3.0x$ , so $2.0 = 8.0x$ and $x = 0.25\ \text{m}$ . Check it: $45/0.25 = +180\ \text{V}$ and $-27/0.15 = -180\ \text{V}$ , which sum to zero.
(c) At $x = 0.25\ \text{m}$ the field contributions are magnitudes with $r$ squared. From $q_1$ : $E_1 = \dfrac{45}{(0.25)^2} = \dfrac{45}{0.0625} = 720\ \text{N}/\text{C}$ , pointing away from the positive charge, so in the $+x$ direction.
From $q_2$ : $E_2 = \dfrac{27}{(0.15)^2} = \dfrac{27}{0.0225} = 1200\ \text{N}/\text{C}$ , pointing toward the negative charge, which also means the $+x$ direction. The two contributions are vectors and they point the same way, so $E = 720 + 1200 = 1920\ \text{N}/\text{C}$ in the $+x$ direction.

(a) $V = +360\ \text{V}$ . (b) The potential is zero at $x = 0.25\ \text{m}$ . (c) At that same point the field is 1920 N/C in the $+x$ direction, nowhere near zero. Potential is a signed scalar and its two terms cancelled; field is a vector and its two terms reinforced. Zero potential is a statement about energy per coulomb, not about force per coulomb, and a charge released at $x = 0.25\ \text{m}$ would accelerate hard.

Reading a field magnitude off an isoline map

A region of space is mapped with parallel, evenly spaced equipotential lines labelled 60 V, 40 V, 20 V and 0 V. Consecutive lines are 2.0 cm apart. (a) Find the magnitude and direction of the electric field. (b) Point A lies on the 60 V line and point B lies on the 20 V line, offset 3.0 cm sideways along the lines, so that A and B are 5.0 cm apart in a straight line. Compute $\Delta V / \Delta r$ for A to B and explain why it does not equal the answer to part (a). (c) A charge is carried from A to another point on the 60 V line. How much does its electric potential energy change?

(a) Evenly spaced parallel isolines mean a uniform field, so the average field from 10.5.B.1 is also the actual field everywhere in the region.
Take any adjacent pair: $\left| \Delta V \right| = 20\ \text{V}$ across $\Delta r = 2.0\ \text{cm} = 0.020\ \text{m}$ , measured perpendicular to the lines. $\left| \vec{E} \right| = \dfrac{20}{0.020} = 1000\ \text{V}/\text{m}$ , the same as $1000\ \text{N}/\text{C}$ .
Direction: 10.5.B.2.ii puts the field perpendicular to the isolines and 10.5.B.2.iii points it toward decreasing potential, so the field runs at right angles to the lines, from the 60 V line toward the 0 V line.
(b) From A to B the potential change is $20 - 60 = -40\ \text{V}$ , and the straight-line distance is $0.050\ \text{m}$ . So $\left| \Delta V / \Delta r \right| = 40 / 0.050 = 800\ \text{V}/\text{m}$ .
That is smaller than 1000 V/m because AB is not along the field. AB has a 4.0 cm component across the isolines and a 3.0 cm component along them, giving the 5.0 cm total. Only the crossing component changes the potential, and $1000 \times (4.0/5.0) = 800\ \text{V}/\text{m}$ , which matches exactly. What $\Delta V / \Delta r$ returns for an arbitrary pair of points is the component of the field along the line joining them.
(c) Both endpoints are on the 60 V isoline, so $\Delta V = 0$ and $\Delta U_E = q \Delta V = 0$ for any charge and any path. This is 10.5.B.2.iv: with no component of the field along the isoline, the electric force is perpendicular to the motion and does no work.

(a) 1000 V/m, perpendicular to the isolines and directed from the 60 V line toward the 0 V line. (b) $\Delta V / \Delta r$ from A to B is 800 V/m, which is the component of the field along AB rather than the field magnitude, because AB runs partly along the isolines. (c) Zero, whatever the charge and whatever the path.

Two conducting spheres joined by a wire

A solid conducting sphere of radius 0.10 m and a solid conducting sphere of radius 0.20 m sit 3.0 m apart, far enough that each one's charge spreads out as though the other were absent. A total charge of 12 nC is placed on the pair and they are joined by a long thin wire. (a) Find the charge on each sphere. (b) Find the common potential. (c) Compare the electric field magnitudes just outside each surface.

Set up with 10.5.A.4: when conductors are in electrical contact, electrons are redistributed so that the surfaces are at the same electric potential. The wire makes the two spheres one conductor, so $V_1 = V_2$ . Note that this equalizes potential, not charge.
Because the spheres are far apart and each carries a spherically symmetric charge, the potential at the surface of a sphere of radius $R$ carrying charge $Q$ is $V = kQ/R$ , the same value a point charge $Q$ at its centre would produce at distance $R$ . This is the high-symmetry case the Topic 10.5 boundary statement allows alongside four-or-fewer-particle sums.
(a) Set the potentials equal: $\dfrac{kQ_1}{0.10} = \dfrac{kQ_2}{0.20}$ , which gives $Q_2 = 2Q_1$ . Charge is conserved, so $Q_1 + Q_2 = 12\ \text{nC}$ , giving $3Q_1 = 12\ \text{nC}$ .
So $Q_1 = 4.0\ \text{nC}$ on the small sphere and $Q_2 = 8.0\ \text{nC}$ on the large one. The charges split in the ratio of the radii, not equally.
(b) $V = \dfrac{(9.0 \times 10^9)(4.0 \times 10^{-9})}{0.10} = \dfrac{36}{0.10} = 360\ \text{V}$ . Check against the other sphere: $\dfrac{(9.0 \times 10^9)(8.0 \times 10^{-9})}{0.20} = \dfrac{72}{0.20} = 360\ \text{V}$ . They agree, as they must.
(c) Surface fields use $r$ squared. Small sphere: $E_1 = \dfrac{36}{(0.10)^2} = 3600\ \text{N}/\text{C}$ . Large sphere: $E_2 = \dfrac{72}{(0.20)^2} = 1800\ \text{N}/\text{C}$ . The smaller sphere has twice the surface field at identical potential.

(a) 4.0 nC on the 0.10 m sphere and 8.0 nC on the 0.20 m sphere. (b) Both sit at 360 V. (c) The surface field is 3600 N/C on the small sphere against 1800 N/C on the large one, a factor of 2, matching the inverse ratio of the radii. Equal potential does not mean equal charge and it does not mean equal field. Small radius plus shared potential means a concentrated field, which is why charge leaks from points and spikes rather than from broad smooth surfaces.

Frequently asked questions

What is electric potential in simple terms?

Electric potential is the electric potential energy per unit charge at a point in space, which is essential knowledge statement 10.5.A.1 in the AP Physics 2 CED. Park one coulomb of charge at a point, ask how much electric potential energy that costs, and the answer in joules is the potential in volts, because one volt is one joule per coulomb. Potential belongs to the point and exists whether or not a charge is sitting there. On the AP Physics 2 exam it is measured against a zero at infinite distance from an isolated point charge, which is stated in the conventions list printed with the Table of Information.

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

Electric potential, V, measured in volts, is a property of a location in a field. Electric potential energy, U_E, measured in joules, is a property of a charge together with whatever produces the field, and it takes both to exist. One factor of charge separates them: the change in potential energy equals the charge times the change in potential, which the AP Physics 2 equation sheet prints as a change in U_E equal to q times a change in V. So a point can have a potential of 500 V with nothing there at all, but it has no potential energy until a charge arrives.

Is electric potential a vector or a scalar?

A scalar. Statement 10.5.A.2 calls the combination rule scalar superposition: to find the potential from several point charges you divide each charge by its distance and add the results as signed numbers, with no components and no angles. The sign of each charge goes into the sum, so the total can be positive, negative or zero. This is what makes potential cheaper to work with than electric field, which needs a full vector sum for the same configuration.

Why are equipotential lines perpendicular to the electric field?

Because moving along an equipotential produces no change in potential, and statement 10.5.B.2.iv states there is no component of an electric field along an isoline. If any part of the field pointed along the line, then walking along it would change the potential and the line would not be an equipotential. So the field can only point across the lines, which is statement 10.5.B.2.ii: isolines are perpendicular to electric field vectors. A practical consequence is that carrying a charge along an equipotential involves no work by the electric force and no change in electric potential energy.

Does the electric field point from high potential to low potential?

Yes. Statement 10.5.B.2.iii says an electric field vector points in the direction of decreasing potential. Be careful about extending that to charges. A positive charge released from rest does accelerate toward lower potential, but a negative charge released from rest accelerates toward higher potential, because the change in electric potential energy equals the charge times the change in potential and that flips sign with a negative charge. Both charges move toward lower potential energy; only the positive one moves toward lower potential.

How do you find the electric field from a potential difference?

Divide the potential difference by the distance between the two points. The AP Physics 2 equation sheet prints the magnitude of the electric field as the absolute value of a change in V divided by a change in r, and statement 10.5.B.1 defines that as the average electric field between the two points. In a uniform field, such as the region between two charged parallel plates, the average equals the actual field, so 12 V across a 1.5 mm gap gives 8000 V/m. Volts per meter and newtons per coulomb are the same unit.

How many charges can AP Physics 2 ask you to find the electric potential of?

The Topic 10.5 boundary statement says that because the methods to calculate the electric potential due to extended charges exceed the scope of the course, AP Physics 2 only expects students to calculate the electric potential of configurations of four or fewer particles, or more in situations of high symmetry. So a direct sum will never run past four terms, but a symmetric arrangement, such as charges spaced evenly around a circle or a uniformly charged sphere, can involve more because the symmetry collapses the sum.