Electric Potential — AP Physics C: E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: E&M.

Covers: Definition of electric potential and potential difference, electric potential from point charges and continuous charge distributions, the relation between electric potential and electric field, equipotential surfaces, and integration techniques for calculating potential.

You should already know: Coulomb's law and electric field from point and continuous charge distributions; work-energy theorem for conservative forces; path independence of work for conservative forces.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: E&M style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Electric Potential?

Electric potential is a scalar quantity that describes the electric potential energy per unit test charge at a point in space, created by a source charge distribution. It is denoted $V$ and has units of volts ($1\text{V}=1\text{J/C}=1\text{Ncdotpm/C}$), and is often called "voltage" when referring to potential difference between two points. According to the AP Physics C: E&M Course and Exam Description (CED), electrostatics (including electric potential) makes up 15-20% of the total exam score, with electric potential appearing regularly in both multiple-choice (MCQ) and free-response questions (FRQ).

Unlike electric field, which is a vector, electric potential is a signed scalar, which often simplifies calculations for charge distributions because we only add algebraic values instead of resolving vector components. Electric potential is defined relative to a reference point; for finite charge distributions, we almost always take $V=0$ at infinity, which aligns with our convention for electric potential energy. Because the electric force is conservative, potential difference between two points is path-independent, meaning it only depends on the endpoints of any path between them, not the route taken.

2. Potential Difference and Potential from Point Charges

Potential difference $\Delta V$ between two points $a$ and $b$ is defined as the change in electric potential energy per unit test charge moving between the points: $$\Delta V=V_b-V_a=\frac{\Delta U}{q_0}=-\frac{W_{a\to b}}{q_0}$$ where $W_{a\to b}$ is the work done by the electric field on the test charge $q_0$, and $\Delta U$ is the change in potential energy. If we take the reference potential $V_{\infty }=0$, we can derive the absolute potential at a distance $r$ from a point charge $Q$ by integrating the electric field from infinity to $r$, giving: $$V(r)=\frac{kQ}{r}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}$$ Potential is positive for positive source charges (it takes positive work to bring a positive test charge from infinity near a positive source) and negative for negative source charges. For a system of point charges, potential follows scalar superposition: the total potential is just the algebraic sum of potentials from each individual charge, no vector components required: $V_{total}={\sum }_i\frac{k{Q}_i}{r_i}$.

Worked Example

What is the total electric potential at the center of a square of side length $s=0.2\text{m}$ with four point charges at the corners: two $+1\mu \text{C}$ and two $-1\mu \text{C}$ arranged opposite each other?

1. All four corners are the same distance from the center of the square. Half the diagonal of the square gives $r=\frac{s\sqrt{2}}{2}=0.1\sqrt{2}\text{m}$ for all charges.
2. Apply scalar superposition: $V=\frac{k}{r}\left(Q_1+Q_2+Q_3+Q_4\right)$.
3. Substitute charge values: $Q_1+Q_2+Q_3+Q_4=(+1\mu \text{C})+(-1\mu \text{C})+(+1\mu \text{C})+(-1\mu \text{C})=0$.
4. The total potential simplifies to $V=0$, regardless of the distance $r$. Note that the electric field at this point is non-zero, as vector electric fields add instead of canceling.

Exam tip: When calculating total potential for multiple point charges, always keep the sign of each charge when adding—potential is a signed scalar, not a magnitude.

3. Potential from Continuous Charge Distributions

For a continuous charge distribution, we split the distribution into infinitesimal point charges $dq$, use the point charge potential for each $dq$, then integrate to find the total potential: $$V=\int \frac{1}{4\pi {ϵ}_0}\frac{dq}{r}$$ where $r$ is the distance from the infinitesimal charge $dq$ to the point where we calculate potential. Because this is a scalar integral, it is almost always simpler than integrating to find electric field, which requires resolving vector components first. For example, for a thin charged ring of radius $R$ and total charge $Q$, every $dq$ on the ring is the same distance $\sqrt{x^2+R^2}$ from a point on the central axis $x$ from the center, so the integral immediately simplifies to $V=\frac{Q}{4\pi {ϵ}_0\sqrt{x^2+R^2}}$ with no further work needed.

Worked Example

A thin non-conducting rod of length $L$ carries a uniform linear charge density $\lambda$. Find the electric potential at a point $P$ along the axis of the rod, a distance $d$ from the nearest end of the rod.

1. Set up coordinates: let the rod span $x=0$ to $x=L$, so point $P$ is at $x=L+d$. An infinitesimal slice of the rod at position $x$ has charge $dq=\lambda dx$.
2. The distance from $dq$ to $P$ is $r=(L+d)-x$.
3. Substitute into the continuous potential formula: $$V={\int }_0^L\frac{1}{4\pi {ϵ}_0}\frac{\lambda dx}{L+d-x}$$
4. Use substitution $u=L+d-x$, $du=-dx$ to solve the integral: $$V=\frac{\lambda }{4\pi {ϵ}_0}{\int }_d^{L+d}\frac{du}{u}=\frac{\lambda }{4\pi {ϵ}_0}\ln \left(\frac{L+d}{d}\right)$$ This is the final potential at point $P$.

Exam tip: Always confirm your coordinate system before setting up the integral for continuous potential—double-check that the distance $r$ from $dq$ to your target point is written correctly before integrating.

4. Relation Between Electric Potential and Electric Field

Potential difference is the negative integral of electric field over a path, so we can invert this relationship to find electric field from the derivative of potential. For the one-dimensional case where $V$ only depends on $x$, this gives: $$E_x=-\frac{dV}{dx}$$ In three dimensions, this generalizes to $\vec{E}=-\nabla V$, meaning electric field is the negative gradient of potential. This tells us that electric field always points in the direction of decreasing potential, and its magnitude equals the rate of change of potential with distance. Equipotential surfaces (surfaces of constant potential) are always perpendicular to electric field lines, since no work is done moving a charge along an equipotential. This relation is especially useful: if you have potential as a function of position, you can differentiate to get electric field much more easily than integrating Coulomb's law directly.

Worked Example

The electric potential in a region of space is given by $V(x)=3x^2-2x$ volts, where $x$ is in meters. What is the electric field at $x=2\text{m}$?

1. Use the 1D relation between $E$ and $V$: $E_x=-\frac{dV}{dx}$.
2. Calculate the derivative of $V(x)$: $\frac{dV}{dx}=6x-2$.
3. Add the negative sign: $E_x=-(6x-2)=-6x+2$.
4. Evaluate at $x=2\text{m}$: $E_x=-6(2)+2=-10\text{N/C}$. The negative sign indicates the electric field points in the negative $x$-direction at this point.

Exam tip: Never forget the negative sign in $E=-dV/dx$—the sign tells you the direction of the electric field, which is almost always tested in MCQ problems.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Adding magnitudes of potential for multiple charges, dropping the sign of negative charges when calculating total potential. Why: Students confuse scalar superposition for potential with vector addition for electric field, where they add magnitudes of components. Correct move: Always keep the sign of each charge when writing $V_i=k{Q}_i/r_i$, so negative charges contribute negative potential to the total sum.
• Wrong move: Assuming $V=0$ implies $E=0$, or $E=0$ implies $V=0$. Why: Students assume that if the sum of potential is zero, the derivative (which gives E) must also be zero, which is not true. Correct move: Always calculate $E$ separately from $V$ using $E=-dV/dx$—never infer E is zero or non-zero from just the value of $V$ at a single point.
• Wrong move: Setting the reference potential $V=0$ at the origin (or at the surface of a finite charge distribution) instead of infinity when calculating absolute potential. Why: Students get confused between potential difference (which can use any reference) and absolute potential for finite charge distributions. Correct move: For any calculation of absolute potential for a finite charge distribution, always use $V=0$ at infinity unless explicitly told otherwise.
• Wrong move: Forgetting that $V$ is constant inside a conducting object, but not necessarily zero. Why: Students mix up $E=0$ inside a conductor with the derivative relation $E=-dV/dx=0$, which implies $V$ is constant, not zero. Correct move: If you need potential inside a conductor, it equals the potential at the surface of the conductor, which you calculate from the surrounding source charges.
• Wrong move: When integrating for potential from a continuous charge distribution, treating potential as a vector and adding components before integration. Why: Students are used to calculating E from continuous distributions which requires components, so they carry that habit over to potential. Correct move: Always use scalar superposition for potential, so you can integrate the scalar potential directly without resolving components.

6. Practice Questions (AP Physics C: E&M Style)

Question 1 (Multiple Choice)

The electric potential in a region of space along the x-axis is given by $V(x)=4x-x^2$, where $V$ is in volts and $x$ is in meters. At what position $x$ is the magnitude of the electric field equal to 2 N/C? A) $x=1\text{m}$ only B) $x=3\text{m}$ only C) $x=1\text{m}$ and $x=3\text{m}$ D) $x=0\text{m}$ and $x=4\text{m}$

Worked Solution: First apply the relation $E_x=-dV/dx$. Take the derivative of $V(x)$ to get $dV/dx=4-2x$, so $E_x=-(4-2x)=2x-4$. We need the magnitude of $E_x$ to equal 2, so we solve two cases: $2x-4=2$ gives $x=3\text{m}$, and $2x-4=-2$ gives $x=1\text{m}$. Both positions satisfy the condition. The correct answer is C.

Question 2 (Free Response)

A solid non-conducting sphere of radius $R$ carries a total charge $Q$ uniformly distributed throughout its volume. (a) Use Gauss's law to find the electric field inside the sphere ($r<R$) and outside the sphere ($r>R$). (b) Calculate the electric potential as a function of $r$ for $r>R$, taking $V=0$ at infinity. (c) Calculate the electric potential at the center of the sphere ($r=0$), taking $V=0$ at infinity.

Worked Solution: (a) For $r>R$, the enclosed charge is $Q$, so Gauss's law gives $E\cdot 4\pi r^2=Q/{ϵ}_0$, so $E(r)=\frac{Q}{4\pi {ϵ}_0{r}^2}$ radially outward. For $r<R$, the enclosed charge is $Q_{enc}=Q(r^3/R^3)$, so Gauss's law gives $E\cdot 4\pi r^2=Q{r}^3/({ϵ}_0{R}^3)$, so $E(r)=\frac{Qr}{4\pi {ϵ}_0{R}^3}$ radially outward.

(b) Potential for $r>R$ is $V(r)={\int }_r^{\infty }E(r^{\prime })d{r}^{\prime }={\int }_r^{\infty }\frac{Q}{4\pi {ϵ}_0{r}^{\prime 2}}d{r}^{\prime }=\frac{Q}{4\pi {ϵ}_{0}r}$, matching the expected point charge potential.

(c) To find $V(0)$, integrate from $0$ to $R$ inside the sphere, then from $R$ to infinity outside: $$V(0)={\int }_0^R{E}_{inside}dr+{\int }_R^{\infty }{E}_{outside}dr={\int }_0^R\frac{Qr}{4\pi {ϵ}_0{R}^3}dr+\frac{Q}{4\pi {ϵ}_{0}R}$$ $$=\frac{Q}{8\pi {ϵ}_{0}R}+\frac{Q}{4\pi {ϵ}_{0}R}=\frac{3Q}{8\pi {ϵ}_{0}R}$$

Question 3 (Application / Real-World Style)

In a typical classroom Van de Graaff generator, the hollow metal spherical dome has a radius of 15 cm. If the maximum electric field at the surface of the dome before air breaks down and sparks occur is $3\times 10^6\text{V/m}$, what is the maximum electric potential (maximum voltage) of the generator, measured relative to infinity?

Worked Solution: For a conducting spherical dome, all charge resides on the surface, so the electric field at the surface is $E=\frac{Q}{4\pi {ϵ}_0{R}^2}$, and the potential at the surface (equal to the potential of the entire dome, since it is a conductor) is $V=\frac{Q}{4\pi {ϵ}_{0}R}$. Comparing these two expressions gives $V=ER$. Substitute values: $R=0.15\text{m}$, $E=3\times 10^6\text{V/m}$, so $V=(3\times 10^6)(0.15)=4.5\times 10^5\text{V}=450\text{kV}$. This matches the typical maximum voltage of 15 cm radius classroom Van de Graaff generators, which are commonly used to demonstrate electrostatic effects.

7. Quick Reference Cheatsheet

8. What's Next

Mastering electric potential is a critical prerequisite for the next topics in the AP C E&M electrostatics unit: electric potential energy, capacitance, and analysis of conductors in electrostatic equilibrium. Without a solid understanding of the relation between potential and electric field, you will struggle to derive capacitance for common configurations and calculate energy stored in capacitors, which is a regularly tested topic on the exam. Electric potential also forms the foundation for circuit analysis later in the course, where potential difference (voltage) across circuit components is the core quantity for applying Kirchhoff's laws. This topic also reinforces the scalar superposition principle, which simplifies many calculations that would be much more complex with vector electric fields.

Gauss's Law for Electrostatics Electric Potential Energy Capacitance and Dielectrics DC Circuits