AP Physics 2 · Topic 10.3

Topic 10.3: Electric Fields

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

An electric field is the electric force per unit charge at a point in space: put a small test charge there, measure the force on it, and divide by that charge. The field is a vector in newtons per coulomb. It points away from isolated positive charges and toward isolated negative charges.

AP Physics: Unit 10 (topics 10.3 Electric Fields). AP Physics 2 Unit 10, Topic 10.3. Two learning objectives: 10.3.A, describe the electric field produced by a charged object or configuration of point charges, and 10.3.B, describe the electric field generated by charged conductors or insulators. Their essential knowledge statements define the field as force per unit charge, fix its direction away from isolated positive and toward isolated negative charges, make it a vector that superposes, distinguish vector field maps from the simplified field line diagrams that give only relative magnitude, and set out what happens in conductors and insulators at electrostatic equilibrium. The boundary statement limits calculations to four or fewer charged objects or systems, allows more in situations of high symmetry, and restricts insulator work to qualitative analysis. 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.A, 2.A, 2.B, and 3.B.

What Topic 10.3 requires

Topic 10.3 carries two learning objectives. The first is about point charges; the second is about conductors and insulators, and it is easy to miss that it is part of this topic at all.

10.3.A Describe the electric field produced by a charged object or configuration of point charges.
10.3.B Describe the electric field generated by charged conductors or insulators.

Five essential knowledge statements sit under them, with eight sub-statements attached. Under 10.3.A:

10.3.A.1 Electric fields may originate from charged objects.
10.3.A.2 The electric field at a given point is the ratio of the electric force exerted on a test charge at that point to the charge of the test charge. The CED prints $\vec{E} = \vec{F}_E / q$ beside it as the relevant equation.
10.3.A.2.i A test charge is a point charge of small enough magnitude such that its presence does not significantly affect an electric field in its vicinity.
10.3.A.2.ii An electric field points away from isolated positive charges and toward isolated negative charges.
10.3.A.2.iii The electric force exerted on a positive test charge by an electric field is in the same direction as the electric field.
10.3.A.3 The electric field is a vector quantity and can be represented in space using vector field maps.
10.3.A.3.i The net electric field at a given location is the vector sum of individual electric fields created by nearby charged objects.
10.3.A.3.ii Electric field maps use vectors to depict the magnitude and direction of the electric field at many locations within a given region.
10.3.A.3.iii Electric field line diagrams are simplified models of electric field maps and can be used to determine the relative magnitude and direction of the electric field at any position in the diagram.

Under 10.3.B:

10.3.B.1 While in electrostatic equilibrium, the excess charge of a solid conductor is distributed on the surface of the conductor, and the electric field within the conductor is zero.
10.3.B.1.i At the surface of a charged conductor, the electric field is perpendicular to the surface.
10.3.B.1.ii The electric field outside an isolated sphere with spherically symmetric charge distribution is the same as the electric field due to a point charge with the same net charge as the sphere located at the center of the sphere.
10.3.B.2 While in electrostatic equilibrium, the excess charge of an insulator is distributed throughout the interior of the insulator as well as at the surface, and the electric field within the insulator may have a nonzero value.

The suggested skills 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.B, calculate or estimate an unknown quantity with units from known quantities; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Unit 10 is weighted at 15 to 18 percent of the multiple-choice section, over roughly 14 to 21 class periods.

Field is force per unit charge (10.3.A.2)

The definition is a ratio, and it is the second equation in the Electricity group of the AP Physics 2 equation sheet:

\vec{E} = \frac{\vec{F}_E}{q}

Read it as an instruction. Put a test charge $q$ at the point you care about, measure the electric force $\vec{F}_E$ on it, divide. The test charge divides out, and what is left describes the point rather than the probe used to find it. That is what per unit charge means, and the units follow from the division: newtons per coulomb, N/C.

Three things follow, and the CED states all three.

Direction comes from the sign of the source (10.3.A.2.ii). The field points away from isolated positive charges and toward isolated negative charges. Nothing about the test charge enters that rule.

Force follows the field only for a positive charge (10.3.A.2.iii). The statement says the force on a positive test charge is in the same direction as the field. Rearranged, $\vec{F}_E = q\vec{E}$ , so a negative $q$ reverses the arrow: an electron in a field accelerates opposite to the field vector.

The test charge has to be small (10.3.A.2.i). It is a point charge of small enough magnitude such that its presence does not significantly affect an electric field in its vicinity. A large probe would push the source charges around, most obviously on a conductor where charge is free to move, and you would be measuring a field you had changed. Once the field is known the probe is discarded: the field is still there when nothing sits in it.

Four quantities in this unit are easy to mix up. Here is the whole distinction on one line each:

Quantity	Symbol	Type	SI unit	What it needs
Electric force	$\vec{F}_E$	Vector	newton, N	Two charges
Electric field	$\vec{E}$	Vector	newton per coulomb, N/C	A source charge and a location
Electric potential energy	$U_E$	Scalar	joule, J	Two charges
Electric potential	$V$	Scalar	volt, V	A source charge and a location

Read the last column. The rows that need a second charge describe an interaction; the rows that need only a location describe the space itself. Topic 10.4 owns the energy row and Topic 10.5 owns the potential row.

The field of a single point charge

The third equation in that same group gives the magnitude for a point charge:

\left|\vec{E}\right| = \frac{1}{4\pi\varepsilon_0}\frac{|q|}{r^2} = k\frac{|q|}{r^2}

Both the field and the charge are wrapped in absolute-value bars, and that is deliberate: the equation returns a size and nothing else. The direction is not hidden in the algebra. You supply it from 10.3.A.2.ii by asking whether the source charge is positive or negative. Feeding a minus sign in and reporting a negative field is a category error, because a vector's magnitude cannot be negative.

Both constants come off the sheet's Constants and Conversion Factors group, which prints $k = 1/(4\pi\varepsilon_0) = 9.0 \times 10^9 \ \mathrm{N \cdot m^2 / C^2}$ on one line and $\varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{C^2/(N \cdot m^2)}$ on another. Use $9.0 \times 10^9$ , not the more precise 8.99: the sheet value is the one the exam's answer choices were built from. The two forms of the equation above are the same equation, so use whichever the question hands you.

Where the formula comes from is one line of suggested skill 2.A. Topic 10.1 supplies Coulomb's law, printed on the same sheet as $\left|\vec{F}_E\right| = k|q_1 q_2| / r^2$ . Put a test charge $q_2$ at distance $r$ from a source $q_1$ , divide the force by $|q_2|$ as 10.3.A.2 tells you to, and $q_2$ cancels, leaving $k|q_1|/r^2$ .

Treat the inverse square as a ratio rather than a calculation. Double the distance and the field drops to a quarter; triple it and the field drops to a ninth; halve it and the field quadruples. Questions that hand you no numbers at all are testing exactly this, and the Coulomb's law guide runs the same scaling on the force side.

Superposition, and the four-charge ceiling (10.3.A.3.i)

The net electric field at a given location is the vector sum of individual electric fields created by nearby charged objects. Vector sum, not arithmetic sum, and that is the single largest difference between this topic and Topic 10.5, where electric potential adds as plain signed numbers because it is a scalar.

The routine is four moves: find each magnitude with $k|q|/r^2$ , draw each arrow using the away-from-positive rule, resolve into components, add the components. The electric field and potential guide runs that procedure end to end with a full set of numbers, so this page will not repeat it.

How many charges you can be asked about is capped, and the cap has an exception. The boundary statement printed with Topic 10.3 reads, in full: "AP Physics 2 only expects students to make calculations of the electric field resulting from four or fewer charged objects or systems. Analysis of the electric field resulting from more charges is allowed in situations of high symmetry. Students will only be expected to perform qualitative analysis of electric fields within insulators."

All three sentences matter. The first caps calculation at four. The second lifts that cap for symmetric arrangements, so an evenly charged sphere is fair game even though it holds far more than four charges, because the symmetry does the summing. The third restricts insulator work to description rather than computation, which is why the conductor material below is quantitative and the insulator material is not.

Superposition also produces null points. Two positive charges on a line have a location between them where the fields cancel exactly, and it sits nearer the smaller charge, which needs the shorter distance to keep up. Two charges of opposite sign have no such point between them, because both fields point the same way there; the cancellation lies outside the pair, beyond the smaller-magnitude charge.

Field maps and field line diagrams are not the same thing (10.3.A.3.ii and 10.3.A.3.iii)

The CED names two representations and keeps them apart.

An electric field map uses vectors to depict the magnitude and direction of the electric field at many locations within a given region. Every arrow is anchored at the point it describes, longer arrows mean a stronger field, and a value can be read off one.

An electric field line diagram is a simplified model of a field map, and can be used to determine the relative magnitude and direction of the electric field at any position in the diagram. The word doing the work is relative. Field lines tell you where the field is stronger and which way it points, not how many newtons per coulomb it is. For a number, go back to the map or the equation.

Given those two statements plus 10.3.A.2.ii and 10.3.A.3.i, the reading rules follow rather than needing to be memorised:

The field at a point is tangent to the line through it, with the arrowhead giving the direction.
Lines leave positive charge and arrive at negative charge, which is 10.3.A.2.ii drawn instead of written.
Closer spacing means a stronger field.
Lines cannot cross. Statement 10.3.A.3.i gives one vector sum at each location, and a crossing would need two directions at one point.
By the usual drawing convention the number of lines on a charge is proportional to its magnitude, so $-2q$ absorbs twice as many lines as $+q$ emits.

Suggested skill 1.A for this topic is creating diagrams, tables, charts, or schematics, and the CED's own sample instructional activity for Topic 10.3 is exactly this exercise: given an arrangement of two or three charges of various signs and either the same or different magnitude, students sketch the electric field vector diagram and the potential isolines. Isolines belong to Topic 10.5, which states in 10.5.B.2.ii that they are perpendicular to the electric field vectors, so the two sketches carry the same information and either can be built from the other.

Conductors and insulators in electrostatic equilibrium (10.3.B)

Learning objective 10.3.B has nothing to do with point charges, and it is where the phrase electrostatic equilibrium earns its place. Equilibrium here means the charges have finished moving.

Inside a solid conductor the field is zero (10.3.B.1). While in electrostatic equilibrium, the excess charge of a solid conductor is distributed on the surface of the conductor, and the electric field within the conductor is zero. The reason is built into the definition: a conductor has charges free to move, so any leftover interior field would keep pushing them, and while they are still being pushed the conductor is not yet at equilibrium. This is behind the CED's Unit 10 essential question about the safest place to be during a lightning storm.

At the surface the field is perpendicular (10.3.B.1.i). A parallel component would drag surface charge sideways, which again would mean equilibrium had not been reached.

Outside a spherical conductor, treat it as a point charge (10.3.B.1.ii). The electric field outside an isolated sphere with spherically symmetric charge distribution is the same as the electric field due to a point charge with the same net charge as the sphere located at the center of the sphere. So $r$ in $k|q|/r^2$ is measured from the center, not the surface, and the result is exact as long as the distribution really is spherically symmetric. The Gauss's law guide explains why it holds, though Gauss's law itself appears nowhere in the AP Physics 2 CED and nowhere on its equation sheet, so that is background rather than required content.

Insulators behave differently (10.3.B.2). While in electrostatic equilibrium, the excess charge of an insulator is distributed throughout the interior of the insulator as well as at the surface, and the electric field within the insulator may have a nonzero value. Charge cannot migrate freely through an insulator, so it stays where it was put and the interior field is not forced to zero. The boundary statement caps what you do with that: only qualitative analysis.

That contrast suits skill 3.B. Same excess charge on a metal sphere and a plastic sphere: the metal puts it all on the surface and has zero field inside, the plastic keeps it through the bulk and can have a field inside.

Uniform fields, and where the CED files parallel plates

A uniform electric field has the same magnitude and direction everywhere in a region, so its field line diagram is a set of straight, evenly spaced, parallel lines. Note where the CED puts it: not in Topic 10.3, but in Topic 10.6, under the parallel-plate capacitor.

Essential knowledge 10.6.A.3 states that the electric field between two charged parallel plates with uniformly distributed electric charge, such as in a parallel-plate capacitor, is constant in both magnitude and direction, except near the edges of the plates. That exception clause is part of the statement, and Topic 10.6's boundary statement then adds that edge effects will be ignored unless explicitly stated otherwise.

Two results hang off it. Statement 10.6.A.3.i gives the magnitude, where the plate separation is much smaller than the dimensions of the plates:

E_C = \frac{Q}{\kappa \varepsilon_0 A}

Statement 10.6.A.3.ii gives the payoff: a charged particle between two oppositely charged parallel plates undergoes constant acceleration, and therefore its motion shares characteristics with the projectile motion of an object with mass in the gravitational field near Earth's surface. Constant field means constant force through $\vec{F}_E = q\vec{E}$ , so every habit from projectile motion transfers, with $qE/m$ in the role of $g$ .

The other route to a uniform field runs through potential rather than charge. The sheet prints $\left|\vec{E}\right| = \left|\Delta V / \Delta r\right|$ , which the CED attaches to essential knowledge 10.5.B.1: 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. In a uniform field the average field is the field, which is why plate problems reduce to dividing plate voltage by plate separation. Potential itself belongs to Topic 10.5; this page stops at the field.

How the exam frames electric fields, and what to read next

The four suggested skills map onto four question types: 1.A draw the field, 2.A derive an expression symbolically with no numbers, 2.B calculate a field or force with units, and 3.B make a claim and name the law behind it, which is where the 10.3.B statements pay off.

The CED's Unit 10 opener flags this unit as especially good practice for the second free-response question on the AP Physics 2 Exam, the Translation Between Representations question, which requires students to create graphical and verbal models of scenarios and compare them to mathematical representations of the same situation. The example the CED gives sits in this corner of the course: sketch equipotential lines around a small positively charged sphere, 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. The CED also says content from any unit may appear in that question, so this is a pattern to practice rather than a prediction. The classroom Progress Check for Unit 10 carries about 24 multiple-choice and 4 free-response questions.

Where to go next. The electric field and potential guide is the procedure page, working through finding a field and a potential step by step with numbers. The Coulomb's law guide and Coulomb's law calculator cover the force this field was defined from. Topic 10.4 takes the same charges and asks about energy instead of field, and Topic 10.7 turns that energy into motion. Every symbol here is on the AP Physics 2 formula sheet, and the Unit 10 overview shows how the seven topics fit together.

Field from a point charge, then the force on two different charges

A small sphere carrying charge $Q = -5.0 \times 10^{-9} \ \mathrm{C}$ is fixed at the origin. Point P lies $0.15 \ \mathrm{m}$ from it. Find (a) the magnitude and direction of the electric field at P, (b) the electric force on a charge $q = +2.0 \times 10^{-9} \ \mathrm{C}$ placed at P, and (c) the electric force on a charge $q = -2.0 \times 10^{-9} \ \mathrm{C}$ placed at P.

Take the direction from the origin toward P as positive along the line joining them, and hold it to the end.
(a) Magnitude first, with absolute values only as the sheet equation requires: $\left|\vec{E}\right| = k|Q|/r^2 = (9.0 \times 10^9)(5.0 \times 10^{-9}) / (0.15)^2$ . The numerator is $45$ and $(0.15)^2 = 0.0225$ , so $\left|\vec{E}\right| = 45 / 0.0225 = 2000 \ \mathrm{N/C}$ .
Direction from 10.3.A.2.ii: the source is negative, so the field points toward it. At P that is back toward the origin, the negative direction on the axis just chosen.
(b) Use $\vec{F}_E = q\vec{E}$ with $q$ positive: $(2.0 \times 10^{-9})(2000) = 4.0 \times 10^{-6} \ \mathrm{N}$ . Statement 10.3.A.2.iii puts the force on a positive charge along the field, so it points toward the origin. That is attraction between opposite charges, which is the check.
(c) Same magnitude, $4.0 \times 10^{-6} \ \mathrm{N}$ , because only $|q|$ sets the size. But $q$ is negative, so $\vec{F}_E = q\vec{E}$ points opposite the field: away from the origin. Two negative charges repel, which is again the check.
Note what did not change: the field at P is a property of $Q$ and the location, so it stayed $2000 \ \mathrm{N/C}$ toward the origin in both parts. Only the force flipped.

(a) $2000 \ \mathrm{N/C}$ , directed from P toward the sphere. (b) $4.0 \times 10^{-6} \ \mathrm{N}$ toward the sphere. (c) $4.0 \times 10^{-6} \ \mathrm{N}$ away from the sphere. Same field, same magnitude of force, opposite directions, decided entirely by the sign of the charge placed at P.

Two positive charges: the net field, and where it vanishes

Charge $q_1 = +4.0 \times 10^{-9} \ \mathrm{C}$ sits at $x = 0$ and charge $q_2 = +9.0 \times 10^{-9} \ \mathrm{C}$ sits at $x = 0.50 \ \mathrm{m}$ . Find (a) the net electric field at the midpoint $x = 0.25 \ \mathrm{m}$ , and (b) the point between them where the net electric field is zero.

Set $+x$ to the right. Both charges are positive, so between them the field from $q_1$ points right and the field from $q_2$ points left. Two charges is inside the boundary statement's limit of four.
(a) From $q_1$ at the midpoint, $r = 0.25 \ \mathrm{m}$ and $r^2 = 0.0625$ : $E_1 = (9.0 \times 10^9)(4.0 \times 10^{-9}) / 0.0625 = 36 / 0.0625 = 576 \ \mathrm{N/C}$ in $+x$ . From $q_2$ , same $r$ : $E_2 = (9.0 \times 10^9)(9.0 \times 10^{-9}) / 0.0625 = 81 / 0.0625 = 1296 \ \mathrm{N/C}$ in $-x$ .
Add as vectors along the one axis, which here is a subtraction: $E_{\mathrm{net}} = 576 - 1296 = -720 \ \mathrm{N/C}$ , meaning $720 \ \mathrm{N/C}$ in $-x$ , back toward the smaller charge. That direction is the sense check, since the larger charge pushes harder from the same distance.
(b) Let the null point sit at $x$ , so it is $x$ from $q_1$ and $0.50 - x$ from $q_2$ . Setting the magnitudes equal, $k$ divides out: $q_1 / x^2 = q_2 / (0.50 - x)^2$ .
Square root both sides, keeping both distances positive: $(0.50 - x)/x = \sqrt{q_2/q_1} = \sqrt{9.0/4.0} = 1.5$ . So $0.50 - x = 1.5x$ , giving $2.5x = 0.50$ and $x = 0.20 \ \mathrm{m}$ .
Substitute back to check. At $x = 0.20$ : $E_1 = 36 / (0.20)^2 = 36/0.040 = 900 \ \mathrm{N/C}$ right, and $E_2 = 81 / (0.30)^2 = 81/0.090 = 900 \ \mathrm{N/C}$ left. They cancel exactly.

(a) $720 \ \mathrm{N/C}$ in the $-x$ direction, toward $q_1$ . (b) The field is zero at $x = 0.20 \ \mathrm{m}$ , which is closer to the smaller charge, as it must be for the two contributions to match.

A charged metal sphere, inside and out (10.3.B.1)

A solid metal sphere of radius $0.10 \ \mathrm{m}$ carries a net charge of $+8.0 \times 10^{-9} \ \mathrm{C}$ and is in electrostatic equilibrium, isolated from everything else. Find the magnitude of the electric field (a) at a point $0.050 \ \mathrm{m}$ from the center, (b) just outside the surface, and (c) at $0.30 \ \mathrm{m}$ from the center.

(a) The point at $0.050 \ \mathrm{m}$ is inside the metal, since the radius is $0.10 \ \mathrm{m}$ . Statement 10.3.B.1 settles it with no arithmetic: at electrostatic equilibrium the excess charge of a solid conductor sits on the surface and the electric field within the conductor is zero. So $E = 0$ .
For (b) and (c), statement 10.3.B.1.ii licenses treating the sphere as a point charge of $+8.0 \times 10^{-9} \ \mathrm{C}$ at its center, because the distribution is spherically symmetric. Measure every $r$ from that center.
(b) Just outside the surface, $r = 0.10 \ \mathrm{m}$ and $r^2 = 0.010$ : $E = (9.0 \times 10^9)(8.0 \times 10^{-9}) / 0.010 = 72 / 0.010 = 7200 \ \mathrm{N/C}$ . The charge is positive, so the field points radially outward, and by 10.3.B.1.i it is perpendicular to the surface.
(c) At $r = 0.30 \ \mathrm{m}$ , $r^2 = 0.090$ : $E = 72 / 0.090 = 800 \ \mathrm{N/C}$ , again radially outward.
Sense check with the inverse square: tripling the distance should cut the field to one ninth, and $7200/9 = 800$ . The ratio confirms both numbers at once.
Contrast with an insulator. Had the sphere been plastic with the same excess charge, 10.3.B.2 puts charge throughout the interior as well as the surface and allows a nonzero interior field, so part (a) would not be zero. The boundary statement means that is asked qualitatively, not as a calculation.

(a) Zero, because the point is inside a conductor in electrostatic equilibrium. (b) $7200 \ \mathrm{N/C}$ radially outward and perpendicular to the surface. (c) $800 \ \mathrm{N/C}$ radially outward, one ninth of the surface value because the distance tripled.

Frequently asked questions

What is an electric field in AP Physics 2?

It is the electric force per unit charge at a point in space. Essential knowledge 10.3.A.2 defines it as the ratio of the electric force exerted on a test charge at that point to the charge of that test charge, written E = F_E/q. Dividing out the test charge leaves a quantity that describes the location itself, so the field is still there when nothing sits in it. It is a vector, measured in newtons per coulomb.

What is the difference between electric field and electric force?

The force is measured in newtons and needs two charges to exist. The field is measured in newtons per coulomb, comes from the source charges alone, and has a value at every point whether or not a second charge is there. They are linked by F_E = qE, so a field of 2000 N/C exerts 4.0 micronewtons on a 2.0 nanocoulomb charge. Both are vectors, but they point the same way only when the charge sitting in the field is positive.

Which direction do electric field lines point?

Away from positive charge and toward negative charge. Essential knowledge 10.3.A.2.ii says an electric field points away from isolated positive charges and toward isolated negative charges, so lines start on positive charges and end on negative ones. The field at any point is tangent to the line through it, and where lines are packed closer the field is stronger. Lines never cross, because the net field at a point has one direction.

Why is the electric field zero inside a conductor?

Because a conductor has charges free to move, and any interior field would keep pushing them. Once the pushing stops the conductor is in electrostatic equilibrium, and essential knowledge 10.3.B.1 states that the excess charge is then distributed on the surface and the electric field within the conductor is zero. Those are the same fact: charge migrates to the surface because that is the arrangement leaving no field inside.

Is the electric field inside an insulator also zero?

No. Essential knowledge 10.3.B.2 says that while in electrostatic equilibrium, the excess charge of an insulator is distributed throughout the interior of the insulator as well as at the surface, and the electric field within the insulator may have a nonzero value. Charge cannot travel freely through an insulator, so it stays in the bulk instead of collecting on the surface. The Topic 10.3 boundary statement adds that only qualitative analysis of electric fields within insulators is expected.

How many charges can an AP Physics 2 question ask you to combine?

Four or fewer for a calculation, with one exception. The Topic 10.3 boundary statement says AP Physics 2 only expects students to make calculations of the electric field resulting from four or fewer charged objects or systems, then adds that analysis resulting from more charges is allowed in situations of high symmetry. So a uniformly charged sphere is fair game, because the symmetry does the summing.

Do you put the negative sign into E = kq/r squared?

No. The AP Physics 2 sheet prints that equation with absolute value bars on both the field and the charge, so it returns a magnitude only. Use the size of the charge for the size of the field, then set direction separately: away if the charge is positive, toward it if negative. A magnitude cannot be negative, so an answer of minus 2000 N/C means a sign was used twice.