AP Physics C: E&M · Topic 8.3

Topic 8.3: Electric Fields

Unit 8: Electric Charges, Fields, and Gauss's Law15-25% of the multiple-choice section

An electric field is the electric force per unit charge at a point. It is a vector, it points away from isolated positive charges and toward isolated negative ones, and fields from several sources add as vectors. Inside a conductor in electrostatic equilibrium the field is zero.

AP Physics: Unit 8 (topics 8.3 Electric Fields). AP Physics C: Electricity and Magnetism Unit 8, Topic 8.3. Two learning objectives. 8.3.A, describe the electric field produced by a charged object or configuration of point charges, with 8.3.A.1 (electric fields may originate from charged objects), 8.3.A.2 (the field at a point is the ratio of the electric force on a test charge at the point to the charge of the test charge, relevant equation E equals F sub E over q) and its three sub-statements, and 8.3.A.3 (the field is a vector quantity representable using vector field maps) and its three sub-statements. 8.3.B, describe the electric field generated by charged conductors or insulators, with 8.3.B.1 (in electrostatic equilibrium the excess charge of a conductor is on its surface and the field within is zero), 8.3.B.1.i (at the surface the field is perpendicular to it), 8.3.B.1.ii (outside an isolated sphere with spherically symmetric charge the field equals that of a point charge of the same net charge at the centre) and 8.3.B.2 (in an insulator the excess charge is distributed through the interior as well as the surface and the field within may be nonzero). Topic 8.3 prints NO boundary statement. Its AP Physics 2 sibling, Topic 10.3, does print one, restricting field calculations to four or fewer charged objects, allowing more only in situations of high symmetry, and limiting analysis of fields within insulators to qualitative work; none of those restrictions appears in AP Physics C, because Topics 8.4 and 8.6 remove the need for them. Physics 2's 10.3.B.1 says a solid conductor where 8.3.B.1 says a conductor. Suggested skills are 1.B, 2.A, 2.D, 3.A and 3.B, five rather than the four listed for most Unit 8 topics; 1.B and 3.A each appear exactly once in Unit 8, both here. The CED's sample multiple-choice Question 15 aligns to 8.3.A and essential knowledge 8.3.A.2 at skill 2.C, answer B, and 8.3.A also appears on sample free-response Question 4. Unit 8 is weighted 15 to 25% of the multiple-choice section over about 12 to 24 class periods.

What Topic 8.3 requires

Topic 8.3 has two learning objectives and, unusually for Unit 8, five suggested skills rather than four.

8.3.A, describe the electric field produced by a charged object or configuration of point charges.

8.3.A.1 electric fields may originate from charged objects.
8.3.A.2 the electric field at a given point is the ratio of the electric force exerted on a test charge at the point to the charge of the test charge. The relevant equation is $\vec{E} = \vec{F}_E / q$ .
8.3.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.
8.3.A.2.ii an electric field points away from isolated positive charges and toward isolated negative charges.
8.3.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.
8.3.A.3 the electric field is a vector quantity and can be represented in space using vector field maps.
8.3.A.3.i the net electric field at a given location is the vector sum of individual electric fields created by nearby charged objects.
8.3.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.
8.3.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.

8.3.B, describe the electric field generated by charged conductors or insulators.

8.3.B.1 while in electrostatic equilibrium, the excess charge of a conductor is distributed on the surface of the conductor, and the electric field within the conductor is zero.
8.3.B.1.i at the surface of a charged conductor, the electric field is perpendicular to the surface.
8.3.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.
8.3.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.

Topic 8.3 prints no boundary statement. That is the headline of this page and it gets its own section below.

The five suggested skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Skill 3.A is listed exactly once in Unit 8, here. So is 1.B. Neither appears under any other topic in the unit.

The boundary statement that is not there

AP Physics 2 Topic 10.3 is the algebra-based sibling of this topic, and the two frameworks track each other closely: same title, same two learning objectives, essential-knowledge statements that match nearly word for word. Then Physics 2 prints a boundary statement and AP Physics C prints none at all.

The Physics 2 boundary statement says three things. 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. And students will only be expected to perform qualitative analysis of electric fields within insulators.

None of those three sentences appears in AP Physics C. That is not an oversight, and the two restrictions it drops are both load-bearing.

The four-charge ceiling on fields is gone, because Topic 8.4 asks you to compute the field of a continuous charge distribution by integration, which is a sum over infinitely many charge elements. The ceiling would contradict the next topic. Note that a four-charge ceiling on force does survive, in Topic 8.1's boundary statement, and its third sentence points forward to exactly this.
The qualitative-only restriction on insulators is gone, because Topic 8.6 gives you Gauss's law, which computes the field inside a charged insulator in one line. Statement 8.3.B.2 tells you the field inside an insulator may be nonzero and stops there. Physics 2 stops there permanently. Physics C stops there for three topics.

One more textual difference, and it points the same way. Physics 2's statement 10.3.B.1 says the excess charge of a solid conductor is distributed on the surface. The C course's 8.3.B.1 drops the word "solid" and says simply "a conductor". The C course goes on to handle hollow conductors, cavities and shells in Unit 10, so the narrower word would have been wrong there.

Which page is for you. The Physics 2 page is written for students in the algebra-based course and is the better read if you want the concept with worked field-line reading. This page is for students in AP Physics C: Electricity and Magnetism, and its job is to say which parts of Topic 8.3 are the last stop and which are a staging post for the calculus topics that follow.

Field is force per unit charge, and the ratio is the definition

Statement 8.3.A.2 defines the field as a ratio, and the sheet prints it as one:

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

Three things follow from the fact that it is a ratio rather than a formula for a source.

The field exists whether or not you put anything there. Divide out the test charge and what is left describes the region. Statement 8.3.A.2.i tells you the test charge must be small enough that its presence does not significantly affect the field in its vicinity, which is a statement about the measuring procedure, not about the field. A large test charge would redistribute the source charges and you would measure something else.

Direction comes with the sign of the test charge. Statement 8.3.A.2.iii: the force on a positive test charge is in the same direction as the field. On a negative charge it is opposite. That is the only sign rule in the topic, and it accounts for most of the sign errors.

Units. Newtons per coulomb, straight from the ratio. You will later meet volts per metre for the same quantity, which is the same unit written differently and comes from Unit 9.

What the sheet does not print here. There is no $E = kq/r^2$ line anywhere on the AP Physics C: E&M sheet. What is printed, one line below the ratio, is the integral form for a distribution:

\vec{E} = \frac{1}{4\pi\varepsilon_0} \int \frac{dq}{r^2} \hat{r}

The point-charge field is what that integral gives when all the charge sits at one place, and you get it by combining the ratio above with Coulomb's law from Topic 8.1:

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

Derive it once, in that order, and you will never wonder which of the printed lines to start from on a free-response question that says "begin your derivation by writing a fundamental physics principle or an equation from the reference information".

Superposition, with no ceiling

Statement 8.3.A.3.i is the whole of the multi-source machinery: the net electric field at a given location is the vector sum of individual electric fields created by nearby charged objects.

The procedure never changes.

Compute each source's field magnitude at the point of interest, using $E = k|q|/r^2$ with the distance from that source to that point.
Draw each field vector at the point, away from a positive source and toward a negative one, per 8.3.A.2.ii.
Resolve into components on declared axes.
Add the components. Recombine.

Step 2 is where the sign of the source charge is used, and it is used once. Do not also put a minus sign into the magnitude. The magnitude formula takes $|q|$ .

Two results worth having in advance. For two charges of the same sign, there is a point between them where the fields cancel, and it lies closer to the smaller charge. For two charges of opposite sign, there is no such point between them, because both fields point the same way there; the cancellation point lies outside the pair, beyond the smaller charge. The third worked example on this page finds one.

Symmetry first. Before computing anything, look for a pair of sources that contribute equal and opposite components at your point. Three equal charges at the corners of an equilateral triangle give zero field at the centroid, and the first worked example shows what happens when one of them is changed. Cancellation you can see is faster and more reliable than cancellation you compute.

Where the ceiling would have been. In the algebra-based course, this procedure is capped at four charges. Here it is not capped at all, and the reason is the next topic: replace the sum with an integral and the same superposition principle handles a continuous rod, ring or arc. Statement 8.4.A.1 calls that "integration and the principle of superposition", naming both halves.

Field maps and field line diagrams are not the same thing

Statements 8.3.A.3.ii and 8.3.A.3.iii distinguish two representations, and the CED's wording is careful in a way that is worth copying.

A vector field map (8.3.A.3.ii) uses vectors to depict the magnitude and direction of the field at many locations within a region. Every arrow is a measurement at a point: its length is the magnitude there, its direction is the direction there.

A field line diagram (8.3.A.3.iii) is a simplified model of a field map, and it can be used to determine the relative magnitude and direction of the field at any position in the diagram. Note the two hedges the CED puts in that sentence: simplified, and relative.

What that licenses and what it does not:

Question	Field map	Field line diagram
Direction of $\vec{E}$ at a point	read the arrow	tangent to the line through that point
Which of two points has the larger $\vec{E}$	compare arrow lengths	compare line density, closer means stronger
Numerical value of $\vec{E}$ at a point	if the map is scaled, yes	no
Sign of the nearest source	not directly	lines leave positive, enter negative
Where $\vec{E} = 0$	arrows shrink to nothing	no lines pass through that point

Three rules about field lines that follow from their being a model of a vector field, and each of which is tested:

Field lines never cross. A crossing would mean two directions for the field at one point, and the field has one value per point.
Lines begin on positive charge and end on negative charge, or run off to infinity. The number of lines drawn at a source is proportional to its charge magnitude, which is how a diagram of $+2Q$ and $-Q$ is drawn with twice as many lines on the positive one.
Density encodes magnitude, and only relatively. Twice as dense means stronger, not twice as strong, because the count of lines drawn is a choice by whoever drew it.

Skill 1.B, creating quantitative graphs with appropriate scales and units, is listed for this topic and for no other in Unit 8. The graph it points at is not a field map: it is $E$ against $r$ , which the next section handles.

Conductors and insulators in electrostatic equilibrium

Objective 8.3.B is three statements and they set up most of Unit 10.

8.3.B.1: in electrostatic equilibrium, excess charge on a conductor sits on the surface, and the field inside the conductor is zero. The argument is short. A conductor is a material in which charge carriers move easily (statement 8.1.C.4.ii). If there were a field anywhere inside the metal, the carriers would feel a force and move, so the arrangement would not be static. Equilibrium therefore requires zero field inside, and zero field inside forces the excess charge out to the surface.

8.3.B.1.i: at the surface of a charged conductor, the field is perpendicular to the surface. Same argument, applied along the surface. A component parallel to the surface would push charge along it, and equilibrium forbids that.

8.3.B.1.ii: outside an isolated sphere with spherically symmetric charge, the field is the same as that of a point charge of the same net charge at the centre. This one is a gift, and it is the only quantitative result in objective 8.3.B. It works for a solid conducting sphere, a conducting shell, a uniformly charged insulating ball, and any layered arrangement with spherical symmetry, so long as you are outside all the charge. Note that the statement licenses it without any calculus. Topic 8.6 proves it in two lines with Gauss's law.

8.3.B.2: in an insulator, the excess charge is distributed through the interior as well as at the surface, and the field within may be nonzero. The word doing the work is may. The statement says the field is not required to vanish; it does not tell you what it is. Getting the value is a Gauss's law problem.

Where the graph is. Put those together and the sketch of $E$ against $r$ for a charged sphere is completely determined outside and half-determined inside, which is exactly the shape of a skill 1.B or 1.C exam task:

Region	Charged conducting sphere, radius $R$ , net charge $Q$	Uniformly charged insulating ball, same $R$ and $Q$
$r < R$	$E = 0$ everywhere, by 8.3.B.1	nonzero, by 8.3.B.2; Topic 8.6 gives $E \propto r$
$r = R$	jumps from $0$ to $kQ/R^2$	rises smoothly to $kQ/R^2$
$r > R$	$E = kQ/r^2$ , by 8.3.B.1.ii	$E = kQ/r^2$ , by 8.3.B.1.ii, identical

The last row is the punchline of the second worked example. From the outside, the two spheres are indistinguishable. No field measurement made outside either one can tell you whether the charge is on a metal shell or spread through an insulating ball. Every difference is inside.

How Topic 8.3 is tested

Of the CED's fifteen sample multiple-choice questions, four align to Unit 8, and one of those is Topic 8.3: sample Question 15, aligned to 8.3.A and essential knowledge 8.3.A.2, at skill 2.C.

The item is worth studying because it needs no arithmetic. Two small spheres carrying $-3Q$ and $+4Q$ sit on a line with four labelled tick marks, equally spaced, and the question asks at which tick mark the field has the greatest magnitude. The answer key gives B, which is the tick mark one space from the $-3Q$ sphere on the side facing the $+4Q$ sphere.

Two ideas settle it. First, the inverse square beats the charge ratio: a point one space from $-3Q$ feels far more from that sphere than from the more distant $+4Q$ , even though $4Q$ is the larger charge. Second, and this is what separates B from the tick mark one space away on the other side, between two charges of opposite sign the two field contributions point the same way and add; outside the pair they point opposite ways and partly cancel. Two points at the same distance from the same charge can therefore have very different net fields.

Learning objective 8.3.A also appears on the CED's sample free-response Question 4, the Qualitative/Quantitative Translation question, alongside 11.1.A, 11.7.A, 11.3.B and 11.8.B. That question is a circuit problem, which tells you something useful: field concepts get cited inside questions about other units.

Patterns worth rehearsing:

Vector-add fields from two or three point charges at a point (skills 2.A, 3.B). Components, then recombine.
Find where the net field is zero and argue whether it is between the charges or outside them.
Predict a factor of change in $E$ when a charge or distance changes (skill 2.D). Doubling $r$ divides $E$ by four.
Read a field line diagram for relative magnitude and direction (8.3.A.3.iii), including which source is larger.
Sketch or plot $E$ against $r$ across the inside and outside of a charged sphere (skill 1.B).
Design a procedure to map or measure a field (skill 3.A, listed only here in Unit 8). The CED's own suggested activity for this topic is a competition using the PhET simulation Electric Field Hockey, scoring a goal with the fewest guiding charges.

For a worked point-charge routine with numbers, the electric field and potential guide covers the procedure. This page covers the framework.

Where Topic 8.3 goes wrong

Putting the source's minus sign into the magnitude. $E = k|q|/r^2$ takes the size of the charge. The sign is used once, when you decide which way the arrow points.

Confusing the field's direction with the force's direction on a negative charge. Statement 8.3.A.2.iii is about a positive test charge. Put an electron in a field pointing right and it accelerates left.

Adding magnitudes instead of vectors. Two fields of $1500 \ \mathrm{N/C}$ at $120^\circ$ to each other sum to $1500 \ \mathrm{N/C}$ , not $3000$ .

Saying the field inside a conductor is zero because there is no charge inside. The field is zero because the carriers moved until it was, which is what electrostatic equilibrium means; the charge going to the surface is a consequence, not the cause. The CED words 8.3.B.1 as one statement with the equilibrium condition attached for exactly this reason.

Applying "the field inside is zero" to an insulator. Statement 8.3.B.2 says the opposite: inside a charged insulator the field may be nonzero. The conductor result is about conductors, and the conductor vs insulator comparison is the short version of why.

Reading field line density as an absolute measure. Statement 8.3.A.3.iii licenses relative magnitude only.

Three charges on an equilateral triangle, field at the centroid

Three small charged spheres sit at the corners of an equilateral triangle of side $0.30 \ \mathrm{m}$ : $+5.0 \ \mathrm{nC}$ at the bottom-left corner, $+5.0 \ \mathrm{nC}$ at the bottom-right corner, and $-10 \ \mathrm{nC}$ at the top corner. Find the magnitude and direction of the net electric field at the centroid of the triangle.

Declare axes: $x$ to the right along the bottom side, $y$ upward, origin at the bottom-left corner. Corners are then at $(0,0)$ , $(0.30, 0)$ and $(0.15, 0.2598)$ , since the height of an equilateral triangle of side $s$ is $s\sqrt{3}/2 = 0.30(0.86603) = 0.25981 \ \mathrm{m}$ .
The centroid sits at the average of the three corners, $(0.15, 0.08660)$ . Its distance to every corner is the same, $r = s/\sqrt{3} = 0.30/1.73205 = 0.17321 \ \mathrm{m}$ , so $r^2 = 0.030 \ \mathrm{m^2}$ exactly. That equal distance is the symmetry that makes the whole problem short.
Magnitudes with $E = k|q|/r^2$ . From each $+5.0 \ \mathrm{nC}$ corner: $E = (9.0 \times 10^9)(5.0 \times 10^{-9})/(0.030) = 45/0.030 = 1500 \ \mathrm{N/C}$ . From the $-10 \ \mathrm{nC}$ corner: $E = (9.0 \times 10^9)(10 \times 10^{-9})/(0.030) = 90/0.030 = 3000 \ \mathrm{N/C}$ .
Directions, from 8.3.A.2.ii. The two positive corners push the field away from themselves, so at the centroid those two vectors point up and to the right, and up and to the left, respectively. The negative corner pulls the field toward itself, so that vector points straight up.
Resolve the two from the bottom corners. The line from the bottom-left corner to the centroid rises $0.08660$ over a run of $0.15$ , and $\arctan(0.08660/0.15) = 30^\circ$ , so that vector points $30^\circ$ above the horizontal. Its unit vector is $(0.15, 0.08660)/0.17321 = (0.86603, 0.5)$ , giving components $(1500)(0.86603) = 1299.0$ and $(1500)(0.5) = 750.0 \ \mathrm{N/C}$ .
By mirror symmetry the other bottom corner gives $(-1299.0, +750.0) \ \mathrm{N/C}$ . The horizontal parts cancel exactly. This is the cancellation to spot before computing anything.
The negative corner contributes $(0, +3000) \ \mathrm{N/C}$ , straight up toward itself.
Add: $E_x = 1299.0 - 1299.0 + 0 = 0$ . $E_y = 750.0 + 750.0 + 3000 = 4500 \ \mathrm{N/C}$ .
Sanity check the direction against the picture. Both positive charges push upward from below and the negative charge pulls upward from above, so everything agrees on up, and the horizontal parts have to cancel by the left-right symmetry of the arrangement. A nonzero $E_x$ would have meant an arithmetic slip.
If all three charges were $+5.0 \ \mathrm{nC}$ the field at the centroid would be exactly zero, by threefold symmetry. Changing one charge from $+5.0$ to $-10$ changes that corner's contribution from $1500 \ \mathrm{N/C}$ pointing down to $3000 \ \mathrm{N/C}$ pointing up, a swing of $4500 \ \mathrm{N/C}$ , which is the whole answer.

$E = 4500 \ \mathrm{N/C}$ , directed straight up along the perpendicular bisector of the bottom side, toward the negative charge. The two positive charges cancel each other horizontally and reinforce vertically.

A charged conducting sphere, inside and out, and the graph

A solid metal sphere of radius $0.15 \ \mathrm{m}$ carries a net charge of $+12 \ \mathrm{nC}$ and is isolated and in electrostatic equilibrium. (a) Find the field magnitude at $r = 0.10 \ \mathrm{m}$ , $r = 0.15 \ \mathrm{m}$ and $r = 0.30 \ \mathrm{m}$ from the centre. (b) Describe the graph of $E$ against $r$ . (c) An insulating ball of the same radius carries the same $+12 \ \mathrm{nC}$ spread uniformly through its volume. State what changes at each of the three radii.

(a) At $r = 0.10 \ \mathrm{m}$ you are inside the metal. Statement 8.3.B.1 says the field within a conductor in electrostatic equilibrium is zero, and the excess charge is all on the surface. So $E = 0$ , with no calculation.
At $r = 0.15 \ \mathrm{m}$ you are at the surface. Use 8.3.B.1.ii: outside an isolated sphere with spherically symmetric charge, the field equals that of a point charge of the same net charge at the centre. $E = kQ/R^2 = (9.0 \times 10^9)(12 \times 10^{-9})/(0.15)^2 = 108/0.0225 = 4800 \ \mathrm{N/C}$ , directed radially outward and, per 8.3.B.1.i, perpendicular to the surface.
At $r = 0.30 \ \mathrm{m}$ : $E = 108/(0.30)^2 = 108/0.090 = 1200 \ \mathrm{N/C}$ . The radius doubled and the field fell by a factor of four, which is the inverse square and the standard skill 2.D check.
(b) The graph is flat at zero from $r = 0$ out to $r = R$ , jumps discontinuously to $4800 \ \mathrm{N/C}$ at the surface, and then decays as $1/r^2$ . The discontinuity is real: it reflects the surface charge layer, which has zero thickness in this model. A sketch that ramps up smoothly inside the metal is wrong.
A useful plotting note for skill 1.B. If you are asked to plot data to get a straight line, plot $E$ against $1/r^2$ for points outside the sphere. The slope is $kQ$ , from which $Q$ follows. Plotting $E$ against $r$ gives a curve you cannot fit by eye.
(c) At $r = 0.30 \ \mathrm{m}$ , nothing changes: $1200 \ \mathrm{N/C}$ , identically. Statement 8.3.B.1.ii applies to any spherically symmetric charge distribution, conducting or not, as long as you are outside all of it.
At $r = 0.15 \ \mathrm{m}$ , also $4800 \ \mathrm{N/C}$ , and now the field is continuous rather than jumping, because the charge is spread through the volume instead of concentrated in a surface layer.
At $r = 0.10 \ \mathrm{m}$ the answer changes completely. Statement 8.3.B.2 says the field inside an insulator may be nonzero, and here it is. Topic 8.3 gives you no way to compute it. Gauss's law does, in Topic 8.6, and it gives $E = kQr/R^3$ , which at $r = 0.10 \ \mathrm{m}$ is $(9.0 \times 10^9)(12 \times 10^{-9})(0.10)/(0.15)^3 = 10.8/(3.375 \times 10^{-3}) = 3200 \ \mathrm{N/C}$ .
This is where the two courses split. AP Physics 2's Topic 10.3 boundary statement restricts students to qualitative analysis of electric fields within insulators, permanently. AP Physics C prints no boundary statement here, and three topics later hands you the tool.

(a) Conducting sphere: $E = 0$ at $r = 0.10 \ \mathrm{m}$ ; $E = 4800 \ \mathrm{N/C}$ at the surface; $E = 1200 \ \mathrm{N/C}$ at $r = 0.30 \ \mathrm{m}$ . (b) Zero inside, a jump at $r = R$ , then a $1/r^2$ decay. (c) For the uniformly charged insulating ball the two outside values are identical, and only the interior differs, rising linearly from zero at the centre to $4800 \ \mathrm{N/C}$ at the surface, which is $3200 \ \mathrm{N/C}$ at $r = 0.10 \ \mathrm{m}$ .

Where two unlike charges give zero field

A charge $q_1 = +9.0 \ \mathrm{nC}$ sits at $x = 0$ and a charge $q_2 = -4.0 \ \mathrm{nC}$ sits at $x = 0.60 \ \mathrm{m}$ . Find every point on the $x$ -axis where the net electric field is zero.

Argue the location before computing it. Split the axis into three regions and check the directions with 8.3.A.2.ii.
Between the charges, $0 < x < 0.60$ : the field from $+9.0 \ \mathrm{nC}$ points in $+x$ (away from it) and the field from $-4.0 \ \mathrm{nC}$ also points in $+x$ (toward it). Same direction, so they add and can never cancel. No solution here. This is the general rule for two charges of opposite sign.
Left of the positive charge, $x < 0$ : the field from $+9.0$ points in $-x$ , the field from $-4.0$ points in $+x$ , so cancellation is possible in principle. But every point here is closer to the $+9.0$ , which is also the larger charge, so its contribution always wins. No solution.
Right of the negative charge, $x > 0.60$ : the field from $+9.0$ points in $+x$ , the field from $-4.0$ points in $-x$ . Here the smaller charge is the nearer one, so the two can balance. This is the only candidate region, and noticing that before doing algebra saves you from an extraneous root.
Set the magnitudes equal, with $x$ measured from the origin: $\dfrac{k(9.0 \times 10^{-9})}{x^2} = \dfrac{k(4.0 \times 10^{-9})}{(x - 0.60)^2}$ .
Cancel $k$ and $10^{-9}$ : $\dfrac{9.0}{x^2} = \dfrac{4.0}{(x-0.60)^2}$ . Cross-multiply and take the square root of both sides, keeping the root that puts $x > 0.60$ : $\dfrac{3.0}{x} = \dfrac{2.0}{x - 0.60}$ .
Cross-multiply: $3.0(x - 0.60) = 2.0x$ , so $3.0x - 1.8 = 2.0x$ , giving $x = 1.8 \ \mathrm{m}$ .
Verify both contributions numerically. From $+9.0 \ \mathrm{nC}$ at a distance of $1.8 \ \mathrm{m}$ : $E = (9.0 \times 10^9)(9.0 \times 10^{-9})/(1.8)^2 = 81/3.24 = 25.0 \ \mathrm{N/C}$ in $+x$ . From $-4.0 \ \mathrm{nC}$ at a distance of $1.2 \ \mathrm{m}$ : $E = (9.0 \times 10^9)(4.0 \times 10^{-9})/(1.2)^2 = 36/1.44 = 25.0 \ \mathrm{N/C}$ in $-x$ . They cancel exactly.
Note what the other square root would have given: $3.0(x-0.60) = -2.0x$ , so $x = 0.36 \ \mathrm{m}$ , which is between the charges. The magnitudes there are equal at $625 \ \mathrm{N/C}$ each, but the directions are the same, so they add to $1250 \ \mathrm{N/C}$ rather than cancelling. Squaring the equation created a root the physics rejects, and the region argument in step 4 is what catches it.

One point, at $x = 1.8 \ \mathrm{m}$ , which is $1.2 \ \mathrm{m}$ beyond the negative charge on the far side from the positive one. For two charges of opposite sign the null point always lies outside the pair and beyond the smaller charge; for two of the same sign it lies between them.

Frequently asked questions

What is an electric field in AP Physics C?

It is the electric force per unit charge at a point in space. Essential knowledge 8.3.A.2 defines the electric field at a given point as the ratio of the electric force exerted on a test charge at the point to the charge of the test charge, and the equation sheet prints it as E equals F over q. Because it is a ratio, the field describes the region itself rather than any particular object placed in it, and it exists whether or not a charge is there to feel it. Its units are newtons per coulomb, and it is a vector.

Does AP Physics C Topic 8.3 have a boundary statement?

No, and that is the main difference from the algebra-based course. AP Physics 2's Topic 10.3 does print one, and it restricts students to fields from four or fewer charged objects, allows more charges only in situations of high symmetry, and limits analysis of fields within insulators to qualitative work. AP Physics C prints none of those restrictions. The reason is the two topics that follow: Topic 8.4 asks for the field of a continuous charge distribution by integration, and Topic 8.6 gives Gauss's law, which computes the field inside a charged insulator directly.

Why is the electric field zero inside a conductor?

Because a conductor is a material in which charge carriers move easily, so any field inside it would push those carriers until they stopped feeling one. Essential knowledge 8.3.B.1 states that while in electrostatic equilibrium, the excess charge of a conductor is distributed on the surface of the conductor and the electric field within the conductor is zero. The two halves are one fact: the carriers redistribute to the surface precisely because that is the arrangement that cancels the field everywhere inside. The condition electrostatic equilibrium is essential, since a conductor carrying a current does have a field inside it.

Is the electric field inside a charged insulator zero too?

No. Essential knowledge 8.3.B.2 states 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 carriers in an insulator cannot move easily, so the charge stays where it was put instead of migrating to the surface. Topic 8.3 tells you the field may be nonzero without telling you its value; Gauss's law in Topic 8.6 supplies the value, giving a field that rises in proportion to the distance from the centre inside a uniformly charged ball.

What is the difference between a field map and a field line diagram?

A vector field map draws an arrow at each of many locations, and each arrow's length and direction give the magnitude and direction of the field at that spot. A field line diagram is described in essential knowledge 8.3.A.3.iii as a simplified model of a field map, and it can be used to determine the relative magnitude and direction of the field at any position. The word relative is the limit: line spacing tells you where the field is stronger, but not by how much, because the number of lines drawn is a choice of the person drawing. Lines never cross, and they start on positive charge and end on negative charge.

Is E equals k q over r squared on the AP Physics C equation sheet?

No. The AP Physics C: Electricity and Magnetism sheet prints the definition E equals the electric force divided by the charge, and it prints the integral form for a charge distribution, one over four pi epsilon zero times the integral of dq over r squared in the radial direction. The point-charge field is not printed as a separate line. You get it in one step by combining the definition with Coulomb's law, which is printed, or as the special case of the integral in which all the charge sits at one place. Deriving it takes a line, and free-response questions often ask you to start from a printed equation anyway.

Where is the electric field zero between two point charges?

It depends on the signs. For two charges of the same sign there is a null point on the line between them, closer to the smaller charge, because between them the two fields point in opposite directions. For two charges of opposite sign there is no null point between them, since both fields point the same way there; the null lies outside the pair, on the far side of the smaller charge. Set the two magnitudes equal, take the square root of both sides, and keep only the root that falls in the region your direction argument allows.