Topic 8.4: Electric Fields of Charge Distributions

What Topic 8.4 requires

Topic 8.4 is one of the sixteen topics in AP Physics C: Electricity and Magnetism whose title has no counterpart at all in the algebra-based course. There is no sibling page to compare it against and nothing to disentangle: the whole topic is new.

It is also the shortest topic in Unit 8 on paper. One learning objective, two essential-knowledge statements, one equation, and a boundary statement that is longer than the rest of the topic combined.

8.4.A, describe the electric field resulting from a given charge distribution.

• 8.4.A.1 states that expressions for the electric field of specified charge distributions can be found using integration and the principle of superposition. The relevant equation printed with it is

• 8.4.A.2 states that symmetry considerations of certain charge distributions can simplify analysis of the electric field resulting from those charge distributions.

The boundary statement, in full. "AP Physics C: Electricity & Magnetism only expects students to use calculus to find the electric field resulting from the following charge distributions and locations: an infinitely long, uniformly charged wire or cylinder at a distance from its central axis, a thin ring of charge at a location along the axis of the ring, a semicircular arc or part of a semicircular arc at its center, and a finite wire or line charge at a point collinear with the line charge or at a location along its perpendicular bisector."

That is a closed list, and it is worth reading against what people assume is on it. The uniformly charged disc is not there. The spherical shell is not there. The solid sphere is not there. A ring at a point off its axis is not there. An arc at a point other than its centre is not there. Spheres and cylinders with volume charge belong to Topic 8.6, where Gauss's law does them in two lines instead of a hard integral.

The four 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.

Notice what is missing from that list: 2.B, the numerical-computation skill, is not listed for Topic 8.4. It is listed for Topics 8.1 and 8.6. This topic is about deriving a symbolic expression and sketching what it does, which is exactly how the CED's own sample free-response question uses it.

The integral, term by term

Every symbol in the printed equation is doing something, and misreading one of them is the usual cause of a stalled derivation.

• $dq$ is an infinitesimal piece of the charge. It is a scalar carrying a sign, since charge is a scalar (statement 8.1.A.1.i). Your first job in every problem is to write $dq$ in terms of a coordinate you can integrate over.
• $r$ is the distance from that piece to the field point, not from the origin, not from the centre of the object. It changes as the piece moves, which is what makes this an integral rather than a product.
• $\hat{r}$ is the unit vector pointing from the charge element toward the field point. It also changes as the piece moves, and it is the reason the integral is a vector integral.
• $1/(4\pi\varepsilon_0)$ is $k$, and the scoring note on the CED's own sample question says an answer may be expressed in terms of either.

The vector part is the hard part, and symmetry is what removes it. You cannot integrate a rotating unit vector directly. What you do instead, every time, is pick a direction, take the component of $d\vec{E}$ along it, and integrate a scalar. Statement 8.4.A.2 is the licence for the other half: the components that cancel do not need integrating at all.

So the printed equation is really used in this form:

with $\theta$ the angle between the direction you chose and the line from the element to the field point, and with a separate one-sentence argument that every perpendicular component has a partner that cancels it.

The CED's scoring guidelines make that structure explicit. For the free-response part built on this objective, one point is awarded for a multi-step derivation starting with an equation for the electric field of a charge distribution, and a separate point is awarded for indicating that only one component of the field need be considered, with the example given as including a cosine term in the integral. The symmetry argument is worth a mark on its own. Write it down.

The five cases, and what each integral looks like

The boundary statement names four kinds of distribution and, for two of them, the specific locations. Counting the finite line's two allowed locations separately gives five distinct standard problems.

Two of those five are easy for the same structural reason: the distance $r$ from every charge element to the field point is the same. On the axis of a ring and at the centre of an arc, $r$ is constant, so $1/r^2$ leaves the integral and all that remains is $\int dq = Q$. That is why those two are the standard exam cases: the calculus is honest but short.

The collinear finite line is the opposite kind of easy: $r$ varies, but every element pushes in the same direction, so there are no components to manage and the integral is one-dimensional and elementary.

The perpendicular bisector case is the only one on the list where both things happen at once, and it is the one worth practising.

On the infinite wire. The boundary statement licenses calculus for it, and the direct integral is doable: it is the perpendicular-bisector result with the length taken to infinity. But this is the one case where the tool from Topic 8.6 is strictly better, and the third worked example below takes the limit so you can see the two answers meet.

What is not on the list is as informative as what is. No disc, no sphere, no shell, no field off the axis of a ring. If a problem hands you one of those, either it is a Gauss's law problem, or the field expression is given to you and you are being asked to do something else with it, such as integrate it to get a potential in Unit 9.

Turning charge into a coordinate: lambda, sigma and rho

The step that stalls most derivations is writing $dq$. It is mechanical once you name the density.

The C: E&M sheet's symbol list defines $\rho$ as "resistivity or charge density", so the letter is shared with a circuits quantity and context decides which is meant. The sheet does not define $\lambda$ or $\sigma$ on the E&M page; $\lambda$ appears in the reprinted Mechanics table as linear mass density. Define whichever you use, in words, the first time it appears in your answer.

On an arc, use the angle. A piece of a circular arc of radius $R$ subtending $d\theta$ has arc length $R \, d\theta$, which the Geometry and Trigonometry table on the sheet gives as $s = r\theta$. So $dq = \lambda R \, d\theta$, and the limits are angles rather than lengths. This substitution is what makes the arc case a two-line problem.

Non-uniform densities are allowed and appear. If a problem gives $\lambda(x) = \lambda_0 x / L$ or $\rho(r) = \rho_0 r/R$, the density goes inside the integral rather than out in front. Nothing about the method changes; the integral is just no longer trivial. Statement 8.6.A.5 in the next topic states the general rule for getting a total charge from a density function, and it is the same idea.

A units check that catches most slips. Whatever you write for $dq$ must come out in coulombs. If $\lambda$ is in C/m then $\lambda \, d\theta$ is wrong and $\lambda R \, d\theta$ is right, because $d\theta$ is dimensionless.

What the equation sheet gives you, and what it does not

The AP Physics C: E&M equation sheet prints the field integral itself, third line in the Electricity and Magnetism column, right after Coulomb's law and the definition of the field. So the starting point of every Topic 8.4 derivation is on the sheet, which matters because free-response questions in this course routinely instruct you to begin from a fundamental principle or an equation from the reference information.

The booklet's appendix also carries three tables that `equations.ts` transcriptions of the physics equations leave out, and two of them earn their keep here.

Geometry and Trigonometry prints, among others: circle $A = \pi r^2$ and $C = 2\pi r$; arc length $s = r\theta$; sphere $V = \frac{4}{3}\pi r^3$ and $S = 4\pi r^2$; cylinder $V = \pi r^2 \ell$ and $S = 2\pi r\ell + 2\pi r^2$; and the right-triangle relations. Arc length is the one you will reach for in this topic.

Calculus prints eleven rules: the chain rule, and the derivatives of $x^n$, $e^{ax}$, $\ln ax$, $\sin(ax)$ and $\cos(ax)$; then five integrals, $\int x^n dx$, $\int e^{ax} dx$, $\int \frac{dx}{x+a}$, $\int \cos(ax)\,dx$ and $\int \sin(ax)\,dx$.

Read that list again against the table of five cases above. The antiderivative you need for the perpendicular-bisector case is not printed. There is no $\int \frac{dx}{(x^2+a^2)^{3/2}}$ on the sheet, and there is no trigonometric-substitution rule that would get you there. Neither is the closely related $\int \frac{x \, dx}{(x^2+a^2)^{3/2}}$.

Two consequences worth planning around:

1. For the ring and the arc, the integral collapses to $\int dq = Q$ or to $\int \cos\theta \, d\theta$, both of which the printed rules cover. Those cases need nothing you do not have.
2. For the perpendicular bisector of a finite line, either you know the result $\int_{-L}^{L} \frac{dx}{(x^2+d^2)^{3/2}} = \frac{2L}{d^2\sqrt{L^2+d^2}}$, or you set the integral up correctly and go as far as you can. The CED's own scoring pattern awards a point for the setup and a point for the symmetry argument before any antiderivative appears, so a correct unevaluated integral is not a zero.

A vector fact from the same appendix. The Vectors table prints $\vec{A} \cdot \vec{B} = AB\cos\theta$ and $|\vec{A} \times \vec{B}| = AB\sin\theta$, neither of which appears in the physics-equation columns. The dot product is the formal justification for the $\cos\theta$ that appears in every one of these derivations.

The CED's own sample question is a ring on its axis

The first question on the AP Physics C: Electricity and Magnetism free-response section is the Mathematical Routines question, worth 10 points with a suggested time of 20 to 25 minutes. The CED's sample of that question aligns to four learning objectives: 8.4.A, 9.2.A, 9.2.B and 9.3.A.

The setup: a thin nonconducting ring of radius $R$ lies in the $xy$-plane, centred at the origin, carrying a positive charge $Q$ uniformly distributed around its circumference. A small sphere of mass $m$ and charge $-Q$ sits on the $z$-axis. Part A uses energy and potential. Part B, worth 3 of the 10 points, asks for the magnitude of the electric field due to the ring at a point on the axis at $z = 2R$, in terms of $Q$, $R$ and physical constants, beginning from a fundamental principle or a reference-information equation.

The three points are awarded like this:

1. For a multi-step derivation starting with an equation for the electric field of a charge distribution.
2. For an indication that only the $z$-component of the field need be considered, the example given being the inclusion of a $\cos\theta$ term in the integral.
3. For a correct final answer, which the scoring note says may be in terms of either $\varepsilon_0$ or $k$.

Read the middle point again. A third of the credit is for the symmetry sentence, the one that says the radial components from opposite sides of the ring cancel. It is a line of writing, not a calculation, and it is the line students leave out.

The scoring guidelines also give an alternate solution that earns the same three points by a different route: write the potential $V = k\int dq/r = kQ/\sqrt{R^2+z^2}$, differentiate, and use $E_z = -dV/dz$. That is worth knowing, because $V$ is a scalar and needs no symmetry argument at all. Both $V = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r}$ and $E_x = -\frac{dV}{dx}$ are printed on the sheet. The catch is that the route only exists once you have Unit 9; inside Unit 8 the vector integral is the only tool you have.

The first worked example below derives the general on-axis expression and then finds where the field is strongest, which is a natural skill 1.C follow-up: sketch $E$ against $z$ and the graph has to start at zero, rise to a peak and fall away as $1/z^2$.

Limits and checks, which are where the marks are protected

A symbolic answer can be checked in ways a number cannot, and skill 2.C, comparing quantities across scenarios, is listed for this topic. Three checks catch nearly every algebra error.

The far-field check. Stand far enough away from any bounded charge distribution and it has to look like a point charge. Every expression on the allowed list must reduce to $kQ/r^2$ as the distance grows. For the ring, $E = \dfrac{kQz}{(R^2+z^2)^{3/2}} \to \dfrac{kQz}{z^3} = \dfrac{kQ}{z^2}$ when $z \gg R$. If your expression does not do that, it is wrong.

The symmetry-point check. Where an argument says the field must vanish, your expression must give zero. At the centre of a full ring, and at the centre of a ring measured along its axis at $z = 0$, the field is zero: every element has a partner directly opposite. Substituting $z = 0$ into the ring result gives zero, as it must.

The units check. Charge over $\varepsilon_0$ over an area gives newtons per coulomb. A stray factor of $R$ shows up immediately.

A fourth check for the infinite cases. An infinite line has no far field to reduce to, because it is not bounded, and it must instead fall as $1/r$ rather than $1/r^2$. That difference is diagnostic: if a distribution is effectively infinite in one direction the field falls more slowly, and if it is effectively infinite in two directions the field does not fall at all. The third worked example shows a finite rod sitting between those two behaviours.

Sketching, which is skill 1.C. Qualitative graph sketches are listed first among this topic's skills. The features that get marked are the ones you can read off the checks above:

• the value at the symmetric point, usually zero,
• whether the curve rises or falls initially,
• the location of any maximum, if there is one,
• the power law far away, $1/z^2$ for a bounded distribution,
• and whether the curve is continuous everywhere.

Draw the asymptote before the curve.

Where Topic 8.4 goes wrong

Pulling $r$ out of the integral when it varies. Legal on the axis of a ring and at the centre of an arc, where every element is the same distance away. Illegal for a line charge, where $r$ is a function of the integration variable.

Integrating the vector. $\hat{r}$ turns as you move along the distribution. Take a component first, then integrate a scalar. Every correct derivation on this list has a $\cos\theta$ or a $\sin\theta$ in it, or has no components to worry about at all.

Forgetting to say why the perpendicular components cancel. The CED's scoring guidelines award a separate point for it.

Writing $dq = \lambda \, d\theta$ on an arc. It is $\lambda R \, d\theta$. The units check catches this instantly.

Using the total charge where the enclosed or contributing charge is meant. On a half-ring, $\int dq$ over the half is $Q$ if $Q$ is the charge of the half, and $Q/2$ if $Q$ is the charge of the whole ring. Say which.

Reaching for Gauss's law on the allowed list. A ring, an arc and a finite rod have no useful symmetry: there is no surface on which the field has constant magnitude and a constant angle to the surface. Gauss's law is still true for them and still useless. Conversely, reaching for the integral on a uniformly charged sphere is legal but slow, and the CED does not ask for it.

Losing the sign of the charge. The integral gives a magnitude and a direction along the axis you chose. A negative distribution reverses the direction. Nothing in the algebra does that for you.