Electric Potential

Electric Potential Energy

Part 1 of 7 — Work and Energy in Electric Fields

Potential Energy of Point Charges

$U = k\frac{q_1 q_2}{r} = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r}$

Work and the Conservative Field

The electric force is conservative, so the work it does is path-independent and equals the drop in potential energy:

$W_{\text{electric}} = -\Delta U = -(U_f - U_i)$

Equivalently, potential energy is built up by integrating the force along the path:

$\Delta U = -\int_{a}^{b} \vec{F}\cdot d\vec{l}$

By the work–energy theorem, with only the electric force acting, $W_{\text{electric}} = \Delta KE$.

Work done by the electric field equals the decrease in PE. Positive charges naturally move from high PE to low PE.

Worked Example — Energy of a Three-Charge Configuration

Problem. Three point charges sit at the corners of an equilateral triangle of side $a = 0.10\,\text{m}$: $q_1 = +2.0\,\mu\text{C}$, $q_2 = +2.0\,\mu\text{C}$, and $q_3 = -3.0\,\mu\text{C}$. Find the total electric potential energy of the configuration. Use $k = 9.0\times10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2$.

Step 1 — Sum over all distinct pairs. Total PE is the sum of the pairwise energies:

$U = \frac{k}{a}\left(q_1 q_2 + q_1 q_3 + q_2 q_3\right)$

since every pair is separated by the same distance $a$.

Step 2 — Compute the products (in units of $\mu\text{C}^2$, i.e. $10^{-12}\,\text{C}^2$):

$q_1 q_2 = (+2)(+2) = +4, \quad q_1 q_3 = (+2)(-3) = -6, \quad q_2 q_3 = (+2)(-3) = -6$

Step 3 — Add them.

$q_1 q_2 + q_1 q_3 + q_2 q_3 = 4 - 6 - 6 = -8\ (\times10^{-12}\,\text{C}^2)$

Step 4 — Multiply by $k/a$.

$U = \frac{(9.0\times10^{9})}{0.10}\,(-8\times10^{-12}) = (9.0\times10^{10})(-8\times10^{-12})$

$U = -0.72\,\text{J}$

Takeaway. The configuration energy is the scalar sum over distinct pairs — the negative result means net energy was released assembling these charges (the attractive pairs dominate).

Potential Energy vs. Force: Why the Integral Matters

The electric force between two charges, $F = kq_1q_2/r^2$, falls off as $1/r^2$, while the potential energy $U = kq_1q_2/r$ falls off as $1/r$. These are not independent — one is the integral of the other.

Starting from the conservative-force definition,

$U(r) = -\int_{\infty}^{r} F\,dr' = -\int_{\infty}^{r}\frac{kq_1q_2}{r'^2}\,dr'$

$U(r) = -kq_1q_2\left[-\frac{1}{r'}\right]_{\infty}^{r} = \frac{kq_1q_2}{r}$

with the convention $U(\infty) = 0$. Reversing the operation, the force is the negative derivative of $U$:

$F = -\frac{dU}{dr} = \frac{kq_1q_2}{r^2}$

Why the reference point matters. Only differences in potential energy are physical. Choosing $U = 0$ at infinity is convenient for isolated charges but is just a choice — what enters the work–energy theorem is always $\Delta U$.

$U$ is the integral of $F$ over distance; $F$ is the (negative) derivative of $U$. Same physics, two viewpoints.

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Electric Potential (Voltage)

Part 2 of 7 — Potential = Energy per Charge

Definition

$V = \frac{U}{q} = \frac{kQ}{r}$

Units: Volts (V) = Joules per Coulomb (J/C).

Potential Difference

$\Delta V = V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}$

$W = q\,\Delta V = q(V_B - V_A)$

Equipotential Surfaces

• Surfaces where $V$ is constant
• Always perpendicular to electric field lines
• No work is done moving a charge along an equipotential

Electric field points from high potential to low potential (for the force on a positive charge).

Worked Example — Potential Difference from a Field by Integration

Problem. In a region the electric field points along $+x$ with magnitude $E(x) = bx$, where $b = 500\,\text{V/m}^2$. Find the potential difference $V_A - V_B$ between $x_A = 0$ and $x_B = 2.0\,\text{m}$, and determine which point is at higher potential.

Step 1 — Use the line-integral definition. Moving along $x$ with $\vec{E}\cdot d\vec{l} = E\,dx$:

$V_B - V_A = -\int_{x_A}^{x_B} E(x)\,dx = -\int_0^{2} b x\,dx$

Step 2 — Integrate.

$V_B - V_A = -b\left[\frac{x^2}{2}\right]_0^{2} = -b\,\frac{(2)^2}{2} = -2b$

Step 3 — Plug in $b$.

$V_B - V_A = -2(500) = -1000\,\text{V}$

So $V_A - V_B = +1000\,\text{V}$.

Step 4 — Interpret. Since $V_A - V_B > 0$, point A (at $x=0$) is at the higher potential. This makes sense: the field points from high to low potential ($+x$ direction), so potential decreases as $x$ increases.

Takeaway. Potential difference is minus the line integral of $\vec{E}$. Even a position-dependent field is handled by a single definite integral.

Potential vs. Potential Energy — Keep Them Straight

These two ideas are easy to conflate but answer different questions:

The potential $V$ exists at a point in space whether or not a charge is placed there — it is "energy available per unit charge." Multiply by an actual charge $q$ to get the energy $U = qV$ that charge would have.

A sign caution. The field points toward lower potential, but a negative charge has higher potential energy where the potential is lower (since $U = qV$ with $q < 0$). So an electron is pushed toward regions of higher potential — opposite to a proton.

$V$ is a property of the space (set by the sources); $U = qV$ is what a specific charge "feels."

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Potential and Electric Field

Part 3 of 7 — $\vec{E}$ from $V$ and $V$ from $\vec{E}$

Finding E from V

$E_x = -\frac{\partial V}{\partial x}, \quad E_y = -\frac{\partial V}{\partial y}, \quad E_z = -\frac{\partial V}{\partial z}$

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

For spherical symmetry: $E_r = -\dfrac{dV}{dr}$.

Finding V from E

$V(r) - V(\infty) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{l}$

$\vec{E}$ points in the direction of steepest decrease of $V$.

Worked Example — The Gradient in Two Dimensions

Problem. In a region the potential is $V(x,y) = (4.0\,\text{V/m}^2)\,x^2 y$. Find the electric-field vector $\vec{E}$ at the point $(x, y) = (1.0\,\text{m}, 2.0\,\text{m})$.

Step 1 — Take partial derivatives. Treat the other variable as a constant each time.

$\frac{\partial V}{\partial x} = 4\cdot 2x\,y = 8xy, \qquad \frac{\partial V}{\partial y} = 4x^2$

Step 2 — Apply $\vec{E} = -\nabla V$.

$E_x = -\frac{\partial V}{\partial x} = -8xy, \qquad E_y = -\frac{\partial V}{\partial y} = -4x^2$

Step 3 — Evaluate at $(1, 2)$.

$E_x = -8(1)(2) = -16\,\text{V/m}, \qquad E_y = -4(1)^2 = -4\,\text{V/m}$

Step 4 — Combine into a vector and find the magnitude.

$\vec{E} = (-16\,\hat{x} - 4\,\hat{y})\,\text{V/m}$

$|\vec{E}| = \sqrt{(-16)^2 + (-4)^2} = \sqrt{256 + 16} = \sqrt{272} \approx 16.5\,\text{V/m}$

Takeaway. Each component of $\vec{E}$ is minus the partial derivative of $V$ in that direction; differentiate, then negate, then evaluate.

Which Direction Should You Compute — V from E, or E from V?

Both directions appear on the AP exam; pick the easier path for the given information.

Use $V = -\int \vec{E}\cdot d\vec{l}$ (integration) when you know the field. This is natural right after a Gauss's-law problem: you already have $E(r)$, so integrate inward from infinity to get $V(r)$.

Use $\vec{E} = -\nabla V$ (differentiation) when you know the potential. This is natural after a superposition problem: scalar potentials are easy to add, and differentiating the sum gives the (otherwise messy) vector field.

Equipotentials and field lines. Because $\vec{E} = -\nabla V$:

• Field lines are always perpendicular to equipotential surfaces.
• $\vec{E}$ points "downhill" — from high $V$ to low $V$.
• Where equipotentials are densely spaced, $|\vec{E}|$ is large (steep gradient).

Field known → integrate to get $V$. Potential known → differentiate to get $\vec{E}$. The gradient is the bridge between the two.

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Potential of Charge Distributions

Part 4 of 7 — Superposition of Potentials

Superposition Principle

For multiple point charges:

$V = \sum_i \frac{kq_i}{r_i}$

Unlike the electric field, potential is a scalar — no vector components to track.

Continuous Distributions

$V = \int \frac{k\,dq}{r}$

Common results:

Set up $dq = \lambda\,dl$ (line), $\sigma\,dA$ (surface), or $\rho\,dV$ (volume), then integrate $k\,dq/r$.

Worked Example — Potential on the Axis of a Charged Ring

Problem. A thin ring of radius $R$ carries total charge $Q$ uniformly distributed. Derive the potential at a point on the axis a distance $x$ from the center, then evaluate $E_x$ from it.

Step 1 — Set up the charge element. A small arc carries charge $dq$. Every point on the ring is the same distance from the axial field point:

$r = \sqrt{R^2 + x^2}$

Step 2 — Write the potential integral. Because $r$ is constant for all $dq$, it comes out of the integral:

$V = \int \frac{k\,dq}{r} = \frac{k}{\sqrt{R^2 + x^2}}\int dq$

Step 3 — Integrate. The total charge is $\int dq = Q$:

$V(x) = \frac{kQ}{\sqrt{R^2 + x^2}}$

Step 4 — Get the axial field by differentiating. By symmetry the field on the axis is purely along $x$, so $E_x = -dV/dx$. Write $V = kQ(R^2 + x^2)^{-1/2}$:

$E_x = -\frac{dV}{dx} = -kQ\cdot\left(-\tfrac12\right)(R^2 + x^2)^{-3/2}(2x)$

$E_x = \frac{kQ\,x}{(R^2 + x^2)^{3/2}}$

Step 5 — Sanity checks. At the center $x = 0$: $V = kQ/R$ and $E_x = 0$ (symmetry). For $x \gg R$: $V \approx kQ/x$ and $E_x \approx kQ/x^2$ — the ring looks like a point charge.

Takeaway. When all source charge is equidistant from the field point, the scalar potential integral is trivial — and differentiating $V$ gives the field for free.

Setting Up the Charge Element $dq$

The hardest part of a continuous-distribution integral is writing $dq$ correctly. Match the geometry:

Then express the distance $r$ from that element to the field point in the same variable you integrate over, and evaluate

$V = \int \frac{k\,dq}{r}$

When is the potential integral easy? If every $dq$ is the same distance from the field point (like the ring's axis), $r$ is constant and pulls out of the integral, leaving just $\int dq = Q$. If $r$ varies, you get a genuine single-variable integral — for a finite rod it gives a logarithm.

Symmetry shortcut. For an axial field, once $V(x)$ is known you can get the on-axis field from $E_x = -dV/dx$ instead of doing a separate vector integral.

Pick the right $dq$, write $r$ in the integration variable, integrate the scalar $k\,dq/r$, then differentiate if you also need the field.

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Conductors and Potential

Part 5 of 7 — Electrostatics of Conductors

Key Facts About Conductors

Just outside a conductor, the surface field is $E = \sigma/\epsilon_0$.

Charge Distribution on Conductors

• Charge concentrates at sharp points (small radius of curvature)
• The field is strongest near sharp points
• This is the principle behind lightning rods

A conductor in equilibrium is a single equipotential — surface and interior are all at the same voltage.

Worked Example — Field and Potential of a Charged Conducting Sphere

Problem. A solid conducting sphere of radius $R$ carries total charge $Q$. Use Gauss's law to find $E(r)$ for $r > R$, then integrate to find $V(r)$ outside and the potential of the sphere itself.

Step 1 — Apply Gauss's law outside ($r > R$). Choose a concentric spherical Gaussian surface of radius $r$. By symmetry $\vec{E}$ is radial and uniform on it:

$\oint \vec{E}\cdot d\vec{A} = E\,(4\pi r^2) = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{Q}{\epsilon_0}$

Step 2 — Solve for the field.

$E(r) = \frac{Q}{4\pi\epsilon_0 r^2} = \frac{kQ}{r^2} \qquad (r > R)$

This is identical to a point charge — all the charge "looks" concentrated at the center.

Step 3 — Integrate the field to get the potential (taking $V(\infty) = 0$):

$V(r) = -\int_{\infty}^{r} E\,dr' = -\int_{\infty}^{r}\frac{kQ}{r'^2}\,dr' = -kQ\left[-\frac{1}{r'}\right]_{\infty}^{r} = \frac{kQ}{r} \qquad (r \ge R)$

Step 4 — Potential of the sphere. Inside a conductor $E = 0$, so $V$ does not change from the surface inward. Evaluate the outside result at $r = R$:

$V_{\text{sphere}} = \frac{kQ}{R} \qquad (r \le R)$

Takeaway. Gauss's law gives the field from symmetry; integrating $-\int E\,dr$ gives the potential; and because $E = 0$ inside a conductor, the entire sphere sits at the constant value $kQ/R$.

Graphing $E(r)$ and $V(r)$ for a Charged Conducting Sphere

Knowing the shapes of these graphs is a frequent AP free-response expectation.

Field $E(r)$:

• For $r < R$ (inside the conductor): $E = 0$ (flat line on the axis).
• At $r = R$: $E$ jumps discontinuously to $kQ/R^2$.
• For $r > R$: $E = kQ/r^2$, decaying as $1/r^2$.

Potential $V(r)$:

• For $r \le R$: $V = kQ/R$, a constant (the conductor is an equipotential).
• For $r > R$: $V = kQ/r$, decaying as $1/r$ and joining the constant value continuously at $r = R$.

The key contrast: the field is discontinuous at the surface (because of the surface charge), but the potential is always continuous. This follows from $V = -\int E\,dr$: integrating even a jump in $E$ produces a smooth $V$.

Field can jump across a charged surface; potential never does. Sketch the flat-then-$1/r$ shape for $V$ and the zero-then-$1/r^2$ shape for $E$.

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Electric Potential Workshop

Part 6 of 7 — Strategies and Practice

Problem-Solving Framework

1. Identify the charge configuration and the symmetry
2. Choose between $V$ from $\vec{E}$ (integration) or $V$ from the charges (superposition)
3. Calculate the potential at the points of interest
4. Use $\Delta V$ to find work, KE changes, or $\vec{E}$ via the gradient

Key Relationships

$W_{\text{field}} = q(V_i - V_f), \qquad \Delta KE = q(V_i - V_f), \qquad \vec{E} = -\nabla V$

Sign rule of thumb: a positive charge speeds up moving to lower potential; an electron speeds up moving to higher potential.

A useful energy unit: $1\,\text{eV}$ is the energy gained by a charge of magnitude $e$ moving through $1\,\text{V}$.

Worked Example — Accelerating an Electron Through a Potential Difference

Problem. An electron (mass $m = 9.11\times10^{-31}\,\text{kg}$, charge magnitude $e = 1.60\times10^{-19}\,\text{C}$) starts from rest and is accelerated toward a region of higher potential, crossing $\Delta V = 100\,\text{V}$. Find its final kinetic energy (in eV and in joules) and its final speed.

Step 1 — Decide whether it speeds up. The electron is negative, so the electric force pushes it toward higher potential. Moving that way, it gains kinetic energy.

Step 2 — Compute the energy gained. The magnitude of the kinetic-energy change equals the charge magnitude times the potential difference crossed:

$\Delta KE = e\,|\Delta V| = (1.60\times10^{-19})(100) = 1.60\times10^{-17}\,\text{J}$

In electron-volts this is simply $\Delta KE = 100\,\text{eV}$, because one electron through $100\,\text{V}$ gains $100\,\text{eV}$ by definition.

Step 3 — Relate energy to speed. Starting from rest, all of $\Delta KE$ becomes $\tfrac12 m v^2$:

$\frac{1}{2}m v^2 = 1.60\times10^{-17}\,\text{J}$

Step 4 — Solve for $v$.

$v = \sqrt{\frac{2(1.60\times10^{-17})}{9.11\times10^{-31}}} = \sqrt{3.51\times10^{13}} \approx 5.9\times10^{6}\,\text{m/s}$

Takeaway. The eV is built for this: a charge of magnitude $e$ through $\Delta V$ volts gains $\Delta V$ eV. Convert to joules before using $\tfrac12 m v^2$ to get a speed.

A Decision Tree for Potential Problems

Faced with an unfamiliar configuration, route yourself quickly:

1. Are you given the field $\vec{E}$?

• Yes, and it is symmetric → integrate $V = -\int \vec{E}\cdot d\vec{l}$.
• Yes, uniform field → $\Delta V = -E\,d$ along the field.

2. Are you given the charges (points or a distribution)?

• Point charges → superpose: $V = \sum kq_i/r_i$ (just add numbers).
• Continuous distribution → set up $dq$ and integrate $\int k\,dq/r$.

3. Do you need the field but already have $V$?

• Differentiate: $\vec{E} = -\nabla V$.

4. Do you need work, speed, or KE?

• Use $W_{\text{field}} = q(V_i - V_f)$ and the work–energy theorem $\Delta KE = W$.
• Convert eV to joules ($1\,\text{eV} = 1.6\times10^{-19}\,\text{J}$) before any $\tfrac12 mv^2$.

Scalar tools (potential, energy) are almost always less error-prone than vector tools (field, force) — reach for them first.

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Electric Potential Review

Part 7 of 7 — Summary

Key Formulas

Potential is a scalar — superpose by adding numbers. Field is a vector — get it from $-\nabla V$ when you already know $V$.

Worked Example — Two-Sphere Capstone

Problem. A small conducting sphere of radius $R_1 = 0.020\,\text{m}$ carries charge $Q = +4.0\,\text{nC}$. (a) Find the potential at its surface. (b) A test charge $q = +1.0\,\text{nC}$ is brought from infinity to a point $r = 0.10\,\text{m}$ from the sphere's center. How much work does an external agent do? Use $k = 9.0\times10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2$.

Step 1 — Surface potential of the sphere.

$V(R_1) = \frac{kQ}{R_1} = \frac{(9.0\times10^{9})(4.0\times10^{-9})}{0.020} = \frac{36}{0.020} = 1800\,\text{V}$

Step 2 — Potential at $r = 0.10\,\text{m}$. Outside the sphere it behaves like a point charge:

$V(r) = \frac{kQ}{r} = \frac{(9.0\times10^{9})(4.0\times10^{-9})}{0.10} = \frac{36}{0.10} = 360\,\text{V}$

Step 3 — Work to bring in the test charge. The external work equals the change in potential energy, $W_{\text{ext}} = q\,[V(r) - V(\infty)]$ with $V(\infty) = 0$:

$W_{\text{ext}} = q\,V(r) = (1.0\times10^{-9})(360) = 3.6\times10^{-7}\,\text{J}$

Step 4 — Interpret the sign. $W_{\text{ext}} > 0$: positive work is required to push a positive test charge toward the positive sphere, against the repulsion.

Takeaway. Compute the potential the source produces at the field point, then multiply by the test charge to get energy or work — no vector integral needed once $V$ is known.

Connecting the Whole Unit

The electric potential unit ties together every tool from the course. Here is the web of relationships in one place:

$\vec{F} = q\vec{E} \quad\longleftrightarrow\quad U = qV$

$\vec{E} = -\nabla V \quad\longleftrightarrow\quad V = -\int \vec{E}\cdot d\vec{l}$

$\vec{E} \;\text{(vector, superpose with components)} \quad\text{vs.}\quad V \;\text{(scalar, superpose by adding)}$

Reading the column on the left vs. right: the left column is everything about force and field (vectors); the right column is everything about energy and potential (scalars). Multiplying a field quantity by the charge $q$ moves you to the corresponding energy quantity.

Gauss's law sits upstream of all of it: from symmetry it hands you $\vec{E}$, which you then integrate to $V$ or use directly in $\vec{F} = q\vec{E}$.

Master the two-column map — force/field on the left, energy/potential on the right, joined by $\times q$ and by the gradient — and any problem becomes a matter of choosing the easier column.

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