Electric Potential

AP Physics C: Electricity and Magnetism· AP Physics C: E&M CED — Electrostatics· 14 min read

1. What is Electric Potential?★★☆☆☆⏱ 3 min

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. Denoted $V$ , it has units of volts ( $1 V = 1 J/C = 1 N \cdotp m/C$ ), and is often called "voltage" when referring to potential difference between two points.

Unlike electric field, which is a vector, electric potential is a signed scalar. This simplifies calculations for charge distributions because you 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 the 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.

📘 Definition

Electric Potential

$V$

A signed scalar quantity representing electric potential energy per unit test charge at a point in space, with units of volts.

2. Potential Difference and Potential from Point Charges★★☆☆☆⏱ 4 min

Potential difference $Δ V$ between two points $a$ and $b$ is defined as the change in electric potential energy per unit test charge moving between the points:

ΔV = V_b - V_a = ΔU/q_0 = -W_{a→b}/q_0

where $W_{a \to b}$ is the work done by the electric field on the test charge $q_{0}$ , and $Δ U$ 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 $Q$ :

V(r) = \frac{kQ}{r} = \frac{1}{4πε_0} \frac{Q}{r}

Potential is positive for positive source charges 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, with no vector components required: $V_{t o t a l} = \sum_{i} \frac{k Q _{i}}{r _{i}}$ .

📐 Worked Example

What is the total electric potential at the center of a square of side length $s = 0.2 m$ with four point charges at the corners: two $+ 1 μ C$ and two $- 1 μ C$ 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 $r = \frac{s 2}{2} = 0.1 2 m$ for all charges.
2
Apply scalar superposition for total potential:
3
$V = \frac{k}{r}\left(Q_1 + Q_2 + Q_3 + Q_4\right)$
4
Substitute the charge values to find the sum:
5
$Q_1 + Q_2 + Q_3 + Q_4 = (+1\ \mu\text{C}) + (-1\ \mu\text{C}) + (+1\ \mu\text{C}) + (-1\ \mu\text{C}) = 0$
6
The total potential simplifies to $V = 0$ , regardless of the distance $r$ . Note that the electric field at this point is non-zero, since vector electric fields add instead of canceling.

3. Potential from Continuous Charge Distributions★★★☆☆⏱ 5 min

For a continuous charge distribution, we split the distribution into infinitesimal point charges $d q$ , use the point charge potential for each $d q$ , then integrate to find the total potential:

V = \int \frac{1}{4πε_0} \frac{dq}{r}

where $r$ is the distance from the infinitesimal charge $d q$ 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.

📐 Worked Example

A thin non-conducting rod of length $L$ carries a uniform linear charge density $λ$ . 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 $d q = λ d x$ .
2
The distance from $d q$ to $P$ is $r = (L + d) - x$ .
3
Substitute into the continuous potential formula:
4
$V = \int_0^L \frac{1}{4πε_0} \frac{\lambda dx}{L + d - x}$
5
Use substitution $u = L + d - x$ , $d u = - d x$ to solve the integral:
6
$V = \frac{\lambda}{4πε_0} \int_d^{L+d} \frac{du}{u} = \frac{\lambda}{4πε_0} \ln\left(\frac{L+d}{d}\right)$
7
This is the final potential at point $P$ .

4. Relation Between Electric Potential and Electric Field★★★☆☆⏱ 5 min

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 $E = - \nabla V$ , 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.

📐 Worked Example

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

1
Use the 1D relation between $E$ and $V$ : $E_{x} = - \frac{d V}{d x}$ .
2
Calculate the derivative of $V (x)$ :
3
$\frac{dV}{dx} = 6x - 2$
4
Add the negative sign:
5
$E_x = -(6x - 2) = -6x + 2$
6
Evaluate at $x = 2 m$ :
7
$E_x = -6(2) + 2 = -10\ \text{N/C}$
8
The negative sign indicates the electric field points in the negative $x$ -direction at this point.

✓ Quick check

Test your understanding of the relation between E and V:

The electric potential in a region of space along the x-axis is given by $V (x) = 4 x - x^{2}$ , 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) $x = 1 m$ only
- B) $x = 3 m$ only
- C) $x = 1 m$ and $x = 3 m$
- D) $x = 0 m$ and $x = 4 m$
Reveal answer
2 —
First apply $E_{x} = - d V / d x$ . Taking the derivative gives $d V / d x = 4 - 2 x$ , so $E_{x} = 2 x - 4$ . Setting $∣ E_{x} ∣ = 2$ gives solutions at $x = 1$ and $x = 3$ , so option C is correct.

5. Additional Exam-Style Worked Examples★★★★☆⏱ 6 min

📐 Worked Example

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 ( $r < R$ ) and outside ( $r > R$ ). (b) Calculate $V (r)$ for $r > R$ , with $V = 0$ at infinity. (c) Calculate $V$ at the center of the sphere ( $r = 0$ ).

1
Part (a): For $r > R$ , enclosed charge is $Q$ , so Gauss's law gives:
2
$E(r) = \frac{Q}{4πε_0 r^2} \text{ radially outward}$
3
For $r < R$ , enclosed charge is $Q_{e n c} = Q (r^{3} / R^{3})$ , so:
4
$E(r) = \frac{Q r}{4πε_0 R^3} \text{ radially outward}$
5
Part (b): Potential for $r > R$ is the integral of E from $r$ to infinity:
6
$V(r) = \int_r^\infty E(r') dr' = \int_r^\infty \frac{Q}{4πε_0 r'^2} dr' = \frac{Q}{4πε_0 r}$
7
Part (c): To find $V (0)$ , split the integral into inside and outside the sphere:
8
$V(0) = \int_0^R E_{inside} dr + \int_R^\infty E_{outside} dr$
9
Evaluate the integrals:
10
$= \int_0^R \frac{Q r}{4πε_0 R^3} dr + \frac{Q}{4πε_0 R} = \frac{Q}{8πε_0 R} + \frac{Q}{4πε_0 R} = \frac{3Q}{8πε_0 R}$

📐 Worked Example

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 before air breakdown is $3 \times 1 0^{6} V/m$ , what is the maximum electric potential of the generator relative to infinity?

1
For a conducting spherical dome, all charge is on the surface. The electric field at the surface is $E = \frac{Q}{4 π ϵ _{0} R ^{2}}$ , and the potential at the surface (equal to the potential of the entire dome) is $V = \frac{Q}{4 π ϵ _{0} R}$ .
2
Relate V and E by eliminating $Q / (4 π ϵ_{0})$ : $V = E R$ .
3
Substitute values: $R = 0.15 m$ , $E = 3 \times 1 0^{6} V/m$ :
4
$V = (3 \times 10^6)(0.15) = 4.5 \times 10^5\ \text{V} = 450\ \text{kV}$

6. Common Pitfalls

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 $E = - d V / d x$ —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 $E = - d V / d x = 0$ , 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.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Potential Difference	$Δ V_{ab} = V_{b} - V_{a} = - W_{a \to b} / q_{0}$	Work done by electric field; $Δ V$ is path-independent
Potential from Point Charge	$V (r) = \frac{1}{4 π ϵ _{0}} \frac{Q}{r}$	$V = 0$ at infinity; keep sign of $Q$
Superposition for Multiple Points	$V_{t o t a l} = \sum_{i} \frac{1}{4 π ϵ _{0}} \frac{Q _{i}}{r _{i}}$	Scalar sum, always keep charge signs
Potential from Continuous Charge	$V = \int \frac{1}{4 π ϵ _{0}} \frac{d q}{r}$	Integrate scalar directly, no components needed
E from Potential (1D)	$E_{x} = - \frac{d V}{d x}$	Negative sign gives direction of E along x
E from Potential (3D)	$E = - \nabla V$	E points in direction of decreasing potential
Equipotential Property	$E ⊥$ equipotential surface	No work done moving charge along an equipotential
Potential Inside Conductor	$V = constant$	$E = 0$ inside, so derivative of V is zero; V not necessarily zero
Potential Outside Spherical Charge	$V (r) = \frac{Q}{4 π ϵ _{0} r}$	Applies to solid spheres and spherical shells

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
FRQ: Potential of charged solid sphere
2022 · 1
MCQ: E from potential derivative

Going deeper

syllabusOfficial AP Physics C: E&M CEDOfficial course learning objectives

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, 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 that simplifies many complex calculations.