9.2 Electric Potential

Electric potential is the electric potential energy per unit charge at a point in space, measured in volts (J/C). It is a scalar, so you add contributions from different charges without worrying about direction, and it connects to the electric field through derivatives and integrals.

Why This Matters for the AP Physics C: E&M Exam

Electric Potential sits in a unit worth a meaningful part of the exam, and the ideas here show up across the whole course because energy methods often beat force methods for solving problems.

This topic gives you the tools for the Translation Between Representations free-response question, where you might sketch an equipotential diagram from an electric field map, build an energy diagram for a charge moving through a field, and explain why those representations agree. Being fluent with V, E, and equipotentials lets you move between graphical, verbal, and mathematical models quickly. Calculating potential from charge distributions using integration also supports multi-step problems where you derive a symbolic expression and then plug in numbers.

Key Takeaways

• Electric potential V is energy per charge (scalar, measured in volts), so contributions from multiple charges add algebraically using superposition.
• For a point charge, $V=\frac{q}{4\pi\varepsilon_0 r}$; for distributions, integrate $V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{r}$.
• Potential difference is $\Delta V=\frac{\Delta U_E}{q}$ and is path independent because the electric field is conservative.
• Field comes from potential by the negative gradient: $E_x=-\frac{dV}{dx}$, and potential comes from field by $\Delta V=-\int_a^b \vec{E}\cdot d\vec{r}$.
• Equipotential lines (isolines) are always perpendicular to electric field vectors, and E points toward decreasing potential.
• A battery's potential difference comes from chemical processes that separate positive and negative charges.

Electric Potential of Charged Objects

Electric Potential Energy per Unit Charge

Electric potential tells you the electric potential energy per unit charge at any point in space. It gives you a scalar field that maps the energy landscape around charges, which makes it easier to analyze how charges will move and interact.

• Measured in volts (V), equal to joules per coulomb (J/C)
• Represents the work needed to move a unit positive charge from infinity to that point
• Unlike electric field (a vector), electric potential is a scalar
• The reference point (zero potential) is usually set at infinity

Integration and Superposition for Potential

You can calculate the potential from charge distributions using integration and the principle of superposition, which means individual contributions add algebraically (no vector components to track).

For a single point charge, the electric potential is:

$V=\frac{q}{4\pi\varepsilon_0 r}$

Where:

• $q$ is the charge
• $r$ is the distance from the charge
• $\varepsilon_0$ is the permittivity of free space

For multiple point charges, use scalar superposition to find the total potential:

$V=\frac{1}{4\pi\varepsilon_0}\sum_{i}\frac{q_i}{r_i}$

For continuous charge distributions, integrate over the entire distribution:

$V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{r}$

The integration process works like this:

1. Break the distribution into infinitesimal elements $dq$
2. Find the potential due to each element
3. Integrate over the entire distribution

Potential Difference Between Points

The electric potential difference $\Delta V$ between two points is the change in electric potential energy per unit charge when a test charge moves between those points.

$\Delta V=\frac{\Delta U_E}{q}$

Key properties:

• Independent of the path taken between the points (conservative field property)
• Measured in volts (V), with 1 V equal to 1 J/C
• Positive $\Delta V$ means electric potential energy increases when you move a positive test charge from the initial to the final point
• In circuits, potential difference drives the flow of charge

Chemical Processes and Potential Difference

Potential differences often come from chemical processes that separate positive and negative charges, converting chemical energy into electrical potential energy.

• Batteries use chemical reactions to create a potential difference between terminals
• During discharge, electrons flow from the negative terminal to the positive terminal through the external circuit
• The chemical reaction continuously separates charges, maintaining the potential difference
• Similar charge-separation processes appear in biological systems like nerve cell membranes

Relationship of Potential and Field

Spatial Rate of Change in Potential

You can get the electric field from the potential using spatial derivatives. At any point, the field component in a given direction equals the negative rate of change of potential in that direction.

For the field components:

$E_x = -\frac{dV}{dx}$ $E_y = -\frac{dV}{dy}$ $E_z = -\frac{dV}{dz}$

In vector notation:

$\vec{E} = -\nabla V$

This tells you:

• The electric field points in the direction of steepest decrease in potential
• The field magnitude is larger where the potential changes more rapidly with distance
• You can calculate fields from potential functions by taking derivatives

Integrating Field and Displacement

Going the other way, you can find the potential difference between two points by integrating the field along a path:

$\Delta V = V_b - V_a = -\int_a^b \vec{E} \cdot d\vec{r}$

Properties of this line integral:

• The negative sign reflects that the field points toward decreasing potential
• The dot product $\vec{E} \cdot d\vec{r}$ picks out the component of the field along the path
• The result is independent of the specific path between points a and b
• This path independence is a signature of conservative fields

Field Maps and Equipotential Lines

Electric field vector maps and equipotential lines are two ways to picture the same field.

Equipotential lines (isolines) connect points of equal potential and have these properties:

• They are always perpendicular to electric field vectors
• There is no component of the electric field along an equipotential line, so the field is entirely perpendicular to it
• No work is done moving a charge along an equipotential line
• Spacing tells you field strength:
  • Closely spaced lines mean a strong field
  • Widely spaced lines mean a weak field
• You can construct an isoline map from a field vector map, and a field map from an isoline map, because the field is always perpendicular to equipotentials and points toward decreasing potential

How field vectors relate to equipotential lines:

• They point toward decreasing potential (perpendicular to equipotential lines)
• Their magnitude is set by how steeply the potential changes
• They predict charge motion: a positive charge accelerates in the field direction (toward lower potential), while a negative charge accelerates opposite the field (toward higher potential)

These visual tools let you reason about complex field setups qualitatively, without doing every calculation.

How to Use This on the AP Physics C: E&M Exam

Free Response

For the Translation Between Representations question, practice turning an electric field vector map into an equipotential (isoline) map and back. Remember two rules every time: isolines are perpendicular to field vectors, and field vectors point toward lower potential. If asked to build an energy diagram for a charge moving through a field, use $\Delta U_E = q\Delta V$ to connect position to energy, then justify why your graphs agree with the field map.

Problem Solving

• To find V from a continuous charge distribution, set up $V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{r}$. Write $dq$ in terms of a charge density ($\lambda$, $\sigma$, or $\rho$), express $r$ in terms of your integration variable, then integrate. Because V is a scalar, you never split into components.
• To get E from V, take the negative derivative in each direction. To get V from E, integrate $-\int \vec{E}\cdot d\vec{r}$ along a convenient path.
• Set V = 0 at infinity unless a problem tells you otherwise, then read potential differences relative to that.

Common Trap

Watch the signs on charges and on $\Delta V$. A negative charge moving to lower potential gains potential energy and loses kinetic energy, which feels backwards if you only memorized the positive-charge case. Work through $\Delta U_E = q\Delta V$ with the actual sign of q every time.

Practice Problem 1: Point Charge Potential

Solution: First, calculate the electric potential at point P due to the 5.0 nC charge using the point-charge formula:

$V = \frac{q}{4\pi\varepsilon_0 r}$

Given:

• $q = 5.0 \times 10^{-9}$ C
• $r = 0.30$ m
• $\varepsilon_0 = 8.85 \times 10^{-12}$ C²/N·m²

$V = \frac{5.0 \times 10^{-9}}{4\pi \times 8.85 \times 10^{-12} \times 0.30}$ $V = \frac{5.0 \times 10^{-9}}{3.33 \times 10^{-11}}$ $V = 150$ V

To find the external work to bring the second charge slowly from infinity to point P, note that it equals the increase in electric potential energy of the system:

$W_{ext} = \Delta U = q_2 V_P$

Since $V = 0$ at infinity: $W_{ext} = \Delta U = q_2 \times V_P = (2.0 \times 10^{-9})(150) = 3.0 \times 10^{-7}$ J = 300 nJ

The work done by the electric field is the negative of this value: $W_{field} = -3.0 \times 10^{-7}$ J.

Practice Problem 2: Electric Field from Potential

Solution: To find the field from the potential, use: $\vec{E} = -\nabla V$

Calculate the negative gradient of the potential function: $E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(5x^2 - 3y^2 + 2z) = -10x$

$E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(5x^2 - 3y^2 + 2z) = 6y$

$E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(5x^2 - 3y^2 + 2z) = -2$

At the point (1, 2, 3): $E_x = -10 \times 1 = -10$ V/m $E_y = 6 \times 2 = 12$ V/m $E_z = -2$ V/m

Therefore, the electric field at point (1, 2, 3) is: $\vec{E} = (-10\hat{i} + 12\hat{j} - 2\hat{k})$ V/m

Practice Problem 3: Potential Difference and Work

Solution: To find the change in electric potential energy, use: $\Delta U_E = q\Delta V = q(V_f - V_i)$

Given:

• $q = -1.6 \times 10^{-19}$ C (electron charge)
• $V_i = 25$ V
• $V_f = 10$ V

$\Delta U_E = (-1.6 \times 10^{-19})(10 - 25)$

$\Delta U_E = (-1.6 \times 10^{-19})(-15)$ $\Delta U_E = +2.4 \times 10^{-18}$ J

Because the change in electric potential energy is positive, the electron gains electric potential energy. If only electric forces act, conservation of energy means its kinetic energy decreases by $2.4 \times 10^{-18}$ J. So the electron gains potential energy and loses kinetic energy during this motion.

This makes sense once you track the directions: the field points from higher potential to lower potential. A negative charge like an electron feels a force opposite the field, that is, toward higher potential. So when the electron moves from 25 V to 10 V, it moves against the electric force on it, which is why it slows down and gains potential energy, just as a ball thrown upward against gravity slows and gains gravitational potential energy.

Common Misconceptions

• Potential and potential energy are not the same. Potential V is energy per charge (volts), while potential energy U is in joules. They are connected by $U_E = qV$.
• Zero potential does not mean zero field, and zero field does not mean zero potential. For example, midway between two equal positive charges the field is zero but the potential is not.
• A negative charge does not always lose energy moving to lower potential. Because $\Delta U_E = q\Delta V$ depends on the sign of q, a negative charge moving to lower potential actually gains potential energy.
• Potential is a scalar, so you add the values directly, including their signs. Do not try to break potential into x and y components like you would with a field.
• Equipotential lines are not the same as field lines. Field lines show field direction; equipotentials connect equal-potential points and are always perpendicular to the field.
• High potential alone does not move charges. It is the potential difference between two points that drives charge motion and does work.

Related AP Physics C: E&M Guides

• 9.1 Electric Potential Energy
• 9.3 Conservation of Electric Energy