Topic 9.2: Electric Potential

What Topic 9.2 requires

Topic 9.2 carries two learning objectives, the only topic in Unit 9 that does, and it holds the unit's single boundary statement.

9.2.A, describe the electric potential due to a configuration of charged objects.

• 9.2.A.1 states that electric potential describes the electric potential energy per unit charge at a point in space.
• 9.2.A.2 states that expressions for the electric potential of charge distributions can be found using integration and the principle of superposition, with the relevant equation $V = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\int \dfrac{dq}{r}$.
• 9.2.A.2.i states that the electric potential for single point charge is $V = \dfrac{q}{4\pi\varepsilon_0 r}$.
• 9.2.A.2.ii states that 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, with the relevant equation $V = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum_i \dfrac{q_i}{r_i}$.
• 9.2.A.3 states that 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, with the relevant equation $\Delta V = \dfrac{\Delta U_E}{q}$.
• 9.2.A.4 states that electric potential difference may also result from chemical processes that cause positive and negative charges to separate, such as in a battery.

The boundary statement under 9.2.A, in full: "AP Physics C: Electricity & Magnetism only expects students to use calculus to find the electric potential 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."

9.2.B, describe the relationship between electric potential and electric field.

• 9.2.B.1 states that the value of an electric field component in any direction at a given location is equal to the negative of the spatial rate of change in electric potential at that location, with the relevant equation $E_x = -\dfrac{dV}{dx}$.
• 9.2.B.2 states that the change in electric potential between two points can be determined by integrating the dot product of the electric field and the displacement along the path connecting the points, with the relevant equation $\Delta V = V_b - V_a = -\displaystyle\int_a^b \vec{E} \cdot d\vec{r}$.
• 9.2.B.3 states that 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 say that equipotential lines represent lines of equal electric potential and are also referred to as isolines of electric potential (9.2.B.3.i); that isolines are perpendicular to electric field vectors, and an isoline map may be constructed from a field map and a field map from an isoline map (9.2.B.3.ii); that an electric field vector points in the direction of decreasing potential (9.2.B.3.iii); and that there is no component of an electric field along an isoline (9.2.B.3.iv).

Suggested skills: 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.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Against AP Physics 2: an average over an interval versus a value at a point

Topic 10.5 in AP Physics 2 shares this topic's title and about half its content verbatim. Statements 9.2.A.1 and 9.2.A.3 are word for word Physics 2's 10.5.A.1 and 10.5.A.3, and the scalar-superposition sum 9.2.A.2.ii is word for word its 10.5.A.2. So is statement 9.2.B.3, along with three of its four sub-statements; the fourth, 9.2.B.3.i, is Physics 2's 10.5.B.2.i with two words removed, "lines of equal electric potential" here against "lines of equal electric potential in space" there.

The temptation is to say the algebra-based course lacks the field-potential relationship. It does not. What it has is a different version of it, and the contrast is the cleanest single illustration of what calculus does to this course.

Read the Physics 2 equation as a finite difference and the Physics C equation as its limit, because that is exactly what they are. The absolute-value bars in the algebra-based version are there because a finite difference over an interval cannot tell you a direction; you have to look at the picture. The derivative can, and statement 9.2.B.3.iii tells you what it says: the field points in the direction of decreasing potential.

Three more differences, each checkable:

• Physics C adds the potential integral. Statement 9.2.A.2 and its equation $V = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r}$ have no Physics 2 counterpart. Neither does the single-point-charge form 9.2.A.2.i, which the algebra-based course subsumes into its sum.
• Physics C adds the line integral. Statement 9.2.B.2, $\Delta V = -\int_a^b \vec{E} \cdot d\vec{r}$, is new. Physics 2 has nothing that turns a field into a potential difference over a path.
• The boundary statements point in opposite directions. Physics 2 says: "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)." Physics C names four specific extended distributions you are expected to integrate. One statement fences extended charge out; the other fences a particular list of it in.

One statement travels the other way. Physics 2's 10.5.A.4, that charges redistribute so the surfaces of conductors in electrical contact reach the same potential, is not in Physics C's Topic 9.2 at all. Physics C promotes it to its own topic, 10.2, and changes "electrons will be redistributed" to "charges will be redistributed".

If you want the algebra-based treatment, read the AP Physics 2 topic. This page is for the calculus-based course.

The integral that is on the sheet, and the two forms that are not

Check what the exam booklet gives you, because the CED prints more potential equations in Topic 9.2 than the AP Physics C: E&M equation sheet does.

Printed on the sheet:

Not printed on the sheet, though all three are in the CED for this topic: the single point charge $V = \dfrac{q}{4\pi\varepsilon_0 r}$ from 9.2.A.2.i, the scalar sum $V = \dfrac{1}{4\pi\varepsilon_0}\sum_i \dfrac{q_i}{r_i}$ from 9.2.A.2.ii, and the definition $\Delta V = \dfrac{\Delta U_E}{q}$ from 9.2.A.3.

That last group is worth pausing on, because the AP Physics 2 sheet does print the scalar sum. Two students sitting different exams get different reference material for the same equation. Physics C hands you the integral and expects the sum to be an obvious special case of it, which it is: replace $\int dq$ with $\sum q_i$ and the continuum becomes a set of points. The exam booklet's version of $\Delta V = -\int_a^b \vec{E} \cdot d\vec{r}$ also drops the middle term $V_b - V_a$ that the CED prints, so keep track yourself of which endpoint is which.

The practical upshot: on a free-response question, starting from the printed integral and specialising it is a legitimate and safe opening move, and the CED's own instruction on a Mathematical Routines question is to begin a derivation by writing a fundamental physics principle or an equation from the reference information. Writing $V = kq/r$ from memory is fine for a multiple-choice question and is worth one extra line of justification on a derivation.

Two more printed lines you need in this topic: $\Delta U_E = q\Delta V$, which turns a potential difference into an energy, and the constant $k = \dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9\ \mathrm{N \cdot m^2/C^2}$.

Setting up the integral, and the four distributions the CED names

Every potential integral in this course has the same four steps, and the reason it is easier than the corresponding field integral is worth saying out loud: potential is a scalar, so there are no components and nothing cancels by direction.

1. Slice the distribution into elements $dq$ and write $dq$ in terms of a coordinate: $dq = \lambda\,dx$ for a line, $\lambda\,R\,d\theta$ for an arc, $\sigma\,dA$ for a sheet.
2. Write $r$, the distance from the element to the field point, in that same coordinate.
3. Integrate $\dfrac{1}{4\pi\varepsilon_0}\displaystyle\int \dfrac{dq}{r}$ over the whole distribution, in the coordinate you chose.
4. Check the limits. Far away, the answer should collapse to $kQ_{\text{total}}/r$.

Compare the field integral on the same sheet, $\vec{E} = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r^2}\hat{r}$: it has an $r^2$ and a unit vector, so you must resolve into components and argue a symmetry cancellation before you integrate. The potential integral has neither. That is why the usual route on an exam is potential first, then differentiate.

The boundary statement names exactly four cases. Here is what each one is really asking for.

An infinitely long, uniformly charged wire or cylinder, at a distance from its central axis. You cannot integrate $dq/r$ for this one: the integral diverges, because there is charge out at infinity in both directions. Use the line-integral route instead. Get $E(r) = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ from Gauss's law and then integrate the field between two radii. The answer is a logarithm and a difference only. There is no absolute potential here, which is a feature of the geometry and not a mistake.

A thin ring of charge, at a location along the axis of the ring. Every element sits at the same distance $\sqrt{R^2 + z^2}$ from the axial point, so $r$ comes straight out of the integral: $V = \dfrac{kQ}{\sqrt{R^2 + z^2}}$. No actual integration is required, which is the point of the case.

A semicircular arc, or part of a semicircular arc, at its center. Every element is at distance $R$, so again $r$ comes out and $V = \dfrac{kQ}{R}$, whatever the angular extent. Notice that the field at the same point is not zero and does need an integral with components, which makes this the sharpest illustration of the scalar advantage.

A finite wire or line charge, at a point collinear with the line charge or at a location along its perpendicular bisector. These are the two positions where the distance is a clean function of one coordinate, and both give logarithms.

Compare this list with the Topic 8.4 boundary statement for the electric field in Unit 8 and you will find the same four distributions in the same order, with only "electric field" changed to "electric potential". The framework is telling you these are the same problems worked twice.

Getting the field back: the derivative and the gradient

Statement 9.2.B.1 is written carefully. It does not say "the field equals minus the slope of the potential graph". It says the value of a field component in any direction equals the negative of the spatial rate of change of potential at that location, in that direction.

That extra precision is the whole content. In one dimension:

and in three, one such relation per axis, each a partial derivative holding the other coordinates fixed. AP Physics C: E&M expresses it component by component because the sheet gives you the $x$ version only, and you apply it to whichever direction the geometry hands you: $E_r = -dV/dr$ along a radius, $E_z = -dV/dz$ along an axis.

Three habits follow.

Differentiate with respect to the right variable. If your potential came out as a function of $z$ along a ring's axis, then $-dV/dz$ gives you $E_z$ and tells you nothing about the radial component off the axis.

A zero derivative does not mean a zero potential, and a zero potential does not mean a zero derivative. At the centre of a charged ring, $V$ is at its maximum along the axis and the field is exactly zero. On the perpendicular bisector of a dipole, $V$ is zero everywhere on that plane and the field is not zero at all. Both are questions the exam asks, and both are decided by the derivative rather than by the value.

The units are consistent, and this is a real check. A volt per metre is a newton per coulomb. Confirm it: $\mathrm{V/m} = (\mathrm{J/C})/\mathrm{m} = (\mathrm{N \cdot m/C})/\mathrm{m} = \mathrm{N/C}$.

Going the other way, statement 9.2.B.2 gives the line integral:

Two features of it are commonly missed. The minus sign means potential falls as you move along the field, which is the same content as 9.2.B.3.iii in equation form. And the path does not matter, because the electrostatic field is conservative, so you may take whatever route makes the dot product easy: run along a field line where the dot product is just $E\,dr$, or across an isoline where it is zero.

For the special case of a uniform field over a straight path parallel to it, the integral collapses to $\Delta V = -Ed$, and only then. That is the relation the algebra-based course works with as a magnitude, $\lvert \vec{E} \rvert = \lvert \Delta V/\Delta r \rvert$, and importing it into a non-uniform field is the most common error in this unit.

Isolines and field maps, and reading one off the other

Statement 9.2.B.3 and its four sub-statements are the graphical half of this topic, and they are the half the second free-response question is built to test.

• Equipotential lines are lines of equal electric potential, and the CED's preferred word for them is isolines (9.2.B.3.i). Expect that word on the exam.
• Isolines are perpendicular to electric field vectors (9.2.B.3.ii), and the statement is explicit that the construction runs both ways: you can build an isoline map from a field map and a field map from an isoline map.
• Field vectors point toward decreasing potential (9.2.B.3.iii).
• There is no component of the electric field along an isoline (9.2.B.3.iv).

The last two are the same fact stated as a direction and as a component, and together they are why the first one is true. If the field had any component along an isoline, then $-\int \vec{E} \cdot d\vec{r}$ along that isoline would be nonzero, so the potential would change along a line of constant potential. It cannot, so the component is zero, so the field is perpendicular. Nothing here has to be memorised separately.

Two consequences that show up constantly:

Moving a charge along an isoline takes no work. $\Delta V = 0$, so $\Delta U_E = q\Delta V = 0$. A charge released on an isoline still accelerates, though, because the field is perpendicular to it and not zero.

Closely spaced isolines mean a strong field. Draw them at equal potential intervals and the spacing is inversely proportional to the field magnitude, straight from $E = -dV/dx$.

The CED suggests a lab for exactly this: connect two electrodes to a 9 V battery, immerse them in a shallow pan of water, probe the potential with a voltmeter, build an isoline map, and estimate the field strength at various locations from it. The estimate is a finite difference, $E \approx \Delta V/\Delta r$ between neighbouring isolines, which is the algebra-based equation reappearing as a measurement technique rather than as a theory statement.

A conductor is the extreme case of an isoline map. Every point on a conductor in electrostatic equilibrium is at the same potential, so its surface is an equipotential surface, and the external field meets it at right angles. That is statement 10.1.A.3.ii and 10.1.A.5 in Topic 10.1, and it is 9.2.B.3.ii doing the work.

Traps

Adding potentials as vectors. Potential is a scalar. Statement 9.2.A.2.ii calls the superposition "scalar" in the text of the statement itself. Add signed numbers; do not resolve anything into components.

Assuming $V = 0$ implies $E = 0$, or the reverse. Neither follows. The midpoint between two equal positive charges has zero field and a large positive potential. Anywhere on the perpendicular bisector of a dipole has zero potential and a nonzero field. The exam asks this in both directions.

Using $\Delta V = Ed$ in a non-uniform field. That form is the line integral evaluated for a constant field along a straight path parallel to it. Near a point charge, along a ring's axis, or outside a charged sphere, the field varies and you must integrate.

Losing the minus sign in $-\int \vec{E} \cdot d\vec{r}$. A quick check: walking along a field line from a positive charge, you move toward lower potential, so $\Delta V$ must be negative.

Setting $V = 0$ at infinity for an infinite distribution. For an infinite line or an infinite plane the integral diverges and only differences are defined. Answer in differences and say so.

Reading $r$ as a coordinate rather than a distance. In $V = kq/r$, the $r$ is the distance from the charge to the field point, always positive. A charge at $x = -0.050$ m and a field point at $x = +0.150$ m give $r = 0.200$ m, not $0.100$ m.

How Topic 9.2 is assessed

The AP Physics C: Electricity and Magnetism exam runs 3 hours: 42 multiple-choice questions worth 50% in 85 minutes, then 4 free-response questions worth 50% in 95 minutes. A four-function, scientific or graphing calculator is allowed on both sections.

The Unit 9 opener points specifically at the second free-response question, Translation Between Representations, worth 12 points with a suggested time of 25 to 30 minutes. The CED describes its Unit 9 flavour directly: a student might be asked to sketch an equipotential diagram from an electric field map, or to create an energy diagram for a point charge moving inside a region with an electric field, and then to make connections between the two representations, justifying how they are consistent with each other. That is statements 9.2.B.3.i through iv, examined as a drawing task.

The unit opener also names skills 1.A, 1.C, 2.A and 3.C as the ones Unit 9 builds, and the framework's Unit 9 Progress Check is about 18 multiple-choice questions and 4 free-response questions.

Skill 2.A, deriving a symbolic expression by following a logical mathematical pathway, decides derivation marks: name the principle, write the general equation from the reference information, specialise it to the geometry with a stated reason, carry symbols to the end, and only then substitute numbers.

For a wider tour of the field and potential pair, the electric field and potential guide covers the routine; this page covers what the CED requires of the topic.