Part 3 · Chapter 24

Electric Potential

The electric field is a vector at every point in space — powerful, but cumbersome. Electric potential captures the same information as a single number at each point, a scalar landscape of "electric height." Charges roll downhill in it, energy is read off by simple subtraction, and the field itself can be recovered as the slope. It is the most useful bookkeeping device in all of electromagnetism.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • Electric potential energy \(\Delta U = -W\) — the work done by the electric force as a charge moves, with \(U = 0\) chosen at infinity.
  • Electric potential as energy per unit charge, \(V = \dfrac{U}{q}\), measured in volts (1 V = 1 J/C), and potential difference \(\Delta V = -\dfrac{W}{q}\).
  • Equipotential surfaces, on which no work is done to move a charge, always perpendicular to the field lines.
  • Going both ways between field and potential: \(V = -\displaystyle\int \vec{E}\cdot d\vec{s}\) and \(\vec{E} = -\nabla V\).
  • The potential of a point charge \(V = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{q}{r}\), of a group (a simple scalar sum), a dipole, and a continuous distribution.
  • The energy of a system of point charges, and why a charged conductor is an equipotential volume.
Section 24-1

What Is Physics?

One of the recurring goals of physics is to find an easier way to do something. A central achievement of the last two chapters was the electric field — but a field is a vector at every point in space, and adding vectors (with their components and directions) is tedious. In this chapter we trade the vector field for a scalar partner: the electric potential. It stores the same physics, yet it is just one number at each point, so combining contributions becomes ordinary addition.

The idea grows directly out of energy. Because the electrostatic force is conservative — just like gravity — we can define a potential energy for a charge in a field, and from it a potential energy per unit charge. That quotient is the electric potential, and it is the quantity your wall socket, your battery, and every voltmeter actually report. Mastering it lets us answer questions about work and energy with arithmetic instead of integrals, and to recover the field whenever we need it as a simple slope of the potential.

Section 24-2

Electric Potential Energy

When a charged particle moves through an electric field, the electric force does work \(W\) on it. Because that force is conservative, the work depends only on the start and end points, not on the path — exactly the condition that lets us define a potential energy \(U\). The change in potential energy is the negative of the work done by the field:

Change in electric potential energy
\[ \Delta U = U_{f} - U_{i} = -W \]
If we adopt the usual convention that \(U = 0\) when the particle is infinitely far away, then the potential energy at a given point is \(U = -W_{\infty}\), where \(W_{\infty}\) is the work the field does as the charge is brought in from infinity. The SI unit of \(U\) is the joule.
The gravity analogy. Lifting a ball raises its gravitational potential energy; releasing it lets the field do positive work as it falls. A positive charge behaves the same way in an electric field — it "falls" toward lower potential energy. A negative charge is the contrarian: it gains energy by moving the way a positive charge would lose it.
Section 24-3

Electric Potential

The potential energy \(U\) of a test charge in a field is proportional to the charge itself — double the charge and you double the energy. Dividing it out gives a quantity that belongs to the field alone, independent of whatever charge we use to probe it. That quantity is the electric potential \(V\).

Electric potential and potential difference
\[ V = \frac{U}{q}, \qquad \Delta V = V_{f} - V_{i} = \frac{\Delta U}{q} = -\frac{W}{q} \]
The SI unit is the volt: \(1\,\mathrm{V} = 1\,\mathrm{J/C}\). Potential is a scalar — it has sign but no direction. Only differences in potential have physical meaning; the zero level is ours to choose, usually at infinity or at ground.

Rearranging gives the work an external agent must do to move a charge between two points with no change in kinetic energy, \(W_{\text{app}} = q\,\Delta V\). For energies of single particles a convenient unit is the electronvolt: the energy an electron gains crossing a potential difference of one volt.

The electronvolt
\[ 1\,\mathrm{eV} = e(1\,\mathrm{V}) = (1.602\times10^{-19}\,\mathrm{C})(1\,\mathrm{V}) = 1.602\times10^{-19}\,\mathrm{J} \]
A natural energy unit for electrons, ions, and atomic-scale physics, where the joule is awkwardly large.
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What "voltage" really means
Potential is electric potential energy per unit charge — the field's own scalar landscape.

A 12 V battery does not store "12 volts of charge." It maintains a 12 J/C difference in potential between its terminals, so each coulomb that flows through the circuit delivers 12 joules of energy. Positive charge spontaneously moves from high potential to low, just as water flows downhill.

Section 24-4

Equipotential Surfaces

The points that share a single value of \(V\) form an equipotential surface. Moving a test charge from one point to another on the same surface requires no net work, because \(\Delta V = 0\) — and this holds for any path between those points, whether or not it stays on the surface. Equipotentials are the contour lines of the electric landscape, the exact analogue of the elevation contours on a topographic map.

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A geometric rule worth memorising
The electric field is always perpendicular to the equipotential surfaces.

If the field had a component along a surface, it would do work as a charge slid across it — contradicting "equipotential." So \(\vec{E}\) must point straight across the contours, from high \(V\) to low. For a point charge the equipotentials are concentric spheres; for a uniform field they are parallel planes; field lines pierce both at right angles.

Section 24-5

Calculating the Potential from the Field

If we know the field \(\vec{E}\) everywhere, we can find the potential difference between two points by integrating the field along any convenient path that joins them. The work per unit charge done by the field, with a minus sign, is the potential difference:

Potential difference from the field
\[ V_{f} - V_{i} = -\int_{i}^{f} \vec{E}\cdot d\vec{s} \]
The line integral is path-independent, so choose the easiest route — often one running straight along (or straight across) the field. Taking \(V_{i} = 0\) at a reference point gives the potential at any point as \(V = -\int_{\text{ref}} \vec{E}\cdot d\vec{s}\).
Reading the sign. Walk against the field (toward where field lines originate) and \(\vec{E}\cdot d\vec{s}\) is negative, so \(V\) rises — you are climbing the electric hill. Walk with the field and \(V\) falls. The field always points "downhill," from high potential to low.
Section 24-6

Potential Due to a Point Charge

To find the potential a distance \(r\) from an isolated point charge \(q\), integrate its field \(E = kq/r^{2}\) radially from \(r\) out to infinity (where \(V = 0\)). The path is along the field, so the dot product is trivial, and the integral gives a clean inverse-distance result.

Potential of a point charge
\[ V = -\int_{\infty}^{r} \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r'^{2}}\,dr' = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r} \]
The potential falls off as \(1/r\) — more slowly than the field's \(1/r^{2}\). Its sign matches the sign of \(q\): a positive charge raises the surrounding potential, a negative charge lowers it. This same expression gives \(V\) outside any spherically symmetric distribution, by the shell theorem.
Section 24-7

Potential Due to a Group of Point Charges

Here is where potential earns its keep. Because \(V\) is a scalar, the potential from several charges is found by ordinary addition of numbers — no components, no angles, no vector diagrams. Using superposition with each charge's own sign:

Potential of n point charges
\[ V = \sum_{i=1}^{n} V_{i} = \frac{1}{4\pi\varepsilon_{0}}\sum_{i=1}^{n} \frac{q_{i}}{r_{i}} \]
An algebraic sum: include the sign of every charge and the distance \(r_{i}\) from that charge to the field point. Summing scalars is far easier than summing the vector fields of the same charges — the great computational advantage of potential over field.
Why this is so much easier. To get the field of four charges you would compute four vectors and add them component by component. To get the potential you compute four numbers and add them. When you finally need the field, take a derivative of \(V\) at the end (Section 24-10) — often the shortest road of all.
Section 24-8

Potential Due to an Electric Dipole

An electric dipole is two equal and opposite charges \(\pm q\) separated by a distance \(d\), with dipole moment \(p = qd\). Adding the two point-charge potentials and taking the far-field limit (\(r \gg d\)) gives a compact expression that depends on the angle \(\theta\) measured from the dipole axis:

Far-field potential of a dipole
\[ V = \frac{1}{4\pi\varepsilon_{0}}\frac{p\cos\theta}{r^{2}} \]
Two features stand out. The potential falls as \(1/r^{2}\), faster than a single charge, because the two opposite charges nearly cancel at a distance. And it is positive on the \(+q\) side (\(\theta = 0\)), negative on the \(-q\) side (\(\theta = 180^{\circ}\)), and exactly zero on the perpendicular bisector (\(\theta = 90^{\circ}\)).
Section 24-9

Potential Due to a Continuous Charge Distribution

For charge spread continuously over a line, surface, or volume, the sum becomes an integral. Slice the distribution into infinitesimal elements \(dq\), treat each as a point charge contributing \(dV = k\,dq/r\), and integrate over the whole body:

Potential of a continuous distribution
\[ V = \frac{1}{4\pi\varepsilon_{0}}\int \frac{dq}{r} \]
Because \(V\) is a scalar, this is a single scalar integral — no components to track, unlike the field integral of Chapter 22. Express \(dq\) through the appropriate density: \(\lambda\,dx\) for a line, \(\sigma\,dA\) for a surface, \(\rho\,dV\) for a volume.
A clean special case. For a uniformly charged ring of radius \(R\) and charge \(q\), every element on the ring is the same distance \(\sqrt{z^{2}+R^{2}}\) from a point on its axis, so \(r\) comes straight out of the integral: \(V = \dfrac{kq}{\sqrt{z^{2}+R^{2}}}\). The field calculation needed an extra \(\cos\) factor; the potential did not.
Section 24-10

Calculating the Field from the Potential

Section 24-5 went from field to potential by integrating. We can reverse the trip: if we know \(V\) everywhere, the field is the negative rate of change of potential with position. The component of \(\vec{E}\) in any direction is how steeply \(V\) drops in that direction.

Field as the gradient of the potential
\[ E_{s} = -\frac{\partial V}{\partial s}, \qquad \vec{E} = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right) = -\nabla V \]
The field points in the direction of steepest decrease of \(V\), with magnitude equal to that steepest slope. Parallel to an equipotential surface the slope is zero, so the field has no component there — confirming again that \(\vec{E}\perp\) equipotentials.
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The two-way street
Integrate the field to get the potential; differentiate the potential to get the field.

\(V = -\int \vec{E}\cdot d\vec{s}\) going one way, \(\vec{E} = -\nabla V\) coming back. Closely spaced equipotentials mean a steep slope and a strong field; widely spaced ones mean a weak field. The contour map and the field map are the same information in two languages.

Section 24-11

Electric Potential Energy of a System of Point Charges

The potential energy of a collection of charges is the work needed to assemble them — bringing each one in from infinity, one at a time, against the field of those already in place. For a single pair separated by \(r\):

Energy of a pair of charges
\[ U = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{r} \]
Positive for like charges (you must do work to push them together — they store energy ready to fly apart) and negative for opposite charges (they assemble themselves, releasing energy). For more than two charges, sum this over every distinct pair, counting each pair once.
Count each pair exactly once. Three charges have three pairs (1–2, 1–3, 2–3); four charges have six. A reliable recipe: bring in charge 1 (no work, empty space), then 2 (against 1's potential), then 3 (against the potential of 1 and 2), and so on. Each new arrival's work is \(q\) times the potential already present at its destination.
Section 24-12

Potential of a Charged Isolated Conductor

In Chapter 23 we found that the field inside a conductor in equilibrium is zero and that excess charge lives on the outer surface. Potential adds a clean corollary. Since \(\vec{E} = 0\) throughout the metal, \(V_{f} - V_{i} = -\int \vec{E}\cdot d\vec{s} = 0\) between any two interior points — so the entire conductor sits at a single potential.

A conductor is an equipotential volume
\[ V_{\text{conductor}} = \text{constant}, \qquad V_{\text{surface of a sphere}} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{R} \]
Every point — interior and surface alike — shares the same \(V\). For an isolated conducting sphere of radius \(R\), the outside potential is that of a point charge at the centre, and it holds that constant value all the way in. The surface is necessarily an equipotential, which is why the field meets it perpendicularly.
Charge crowds the sharp spots. On a conductor of uneven shape the charge density — and the field just outside — is largest where the surface curves most sharply. The potential is the same everywhere on the metal, but the tightly packed equipotentials near a point or edge mean intense fields there. This is the principle behind the lightning rod and the corona discharge.
Worked Examples

Putting It to Work

1 Potential near a point charge

Problem. Find the electric potential at a distance \(r = 0.30\,\mathrm{m}\) from a point charge \(q = +4.0\times10^{-9}\,\mathrm{C}\), with \(V = 0\) at infinity.

Solution. Use the point-charge potential directly.

V = kq / r
\[ V = \frac{(8.99\times10^{9})(4.0\times10^{-9})}{0.30} \approx 120\,\mathrm{V} \]

Positive, as it must be for a positive charge. Double the distance and the potential halves — the \(1/r\) signature, gentler than the field's \(1/r^{2}\).

2 Potential at the centre of a square

Problem. Four charges sit at the corners of a square whose centre is a distance \(r = 0.919\,\mathrm{m}\) from each corner: \(q_{1}=+12\,\mathrm{nC}\), \(q_{2}=-24\,\mathrm{nC}\), \(q_{3}=+31\,\mathrm{nC}\), \(q_{4}=+17\,\mathrm{nC}\). Find \(V\) at the centre.

Solution. Potential is a scalar, so just add. All four are equidistant, so \(r\) factors out.

V = k(Σq) / r
\[ V = \frac{(8.99\times10^{9})(36\times10^{-9})}{0.919} \approx 350\,\mathrm{V} \]

The orientation of the charges never entered — a vivid demonstration of how much simpler the scalar potential is than the vector field.

3 Speeding up an electron

Problem. An electron, starting from rest, is accelerated through a potential difference of \(V = 1000\,\mathrm{V}\). Find its kinetic energy and speed.

Solution. The field does work \(W = -q\,\Delta V = e\,V\) on the electron (it moves toward higher potential, against the field).

K = eV, then K = ½mv²
\[ K = eV = 1000\,\mathrm{eV} = 1.6\times10^{-16}\,\mathrm{J} \;\Longrightarrow\; v = \sqrt{\frac{2K}{m}} \approx 1.9\times10^{7}\,\mathrm{m/s} \]

The electronvolt makes the energy bookkeeping effortless: 1000 V across an electron is simply 1000 eV. This is the working principle of an electron gun in an old cathode-ray tube.

4 Field from a potential function

Problem. In a region the potential is \(V(x) = 1500 - 200x^{2}\) (volts, with \(x\) in metres). Find the field at \(x = 2.0\,\mathrm{m}\).

Solution. The field is the negative slope of \(V\).

Eₓ = −dV/dx
\[ E_{x} = -\frac{dV}{dx} = -(-400x) = 400x \;\Longrightarrow\; E_{x}(2.0) = 800\,\mathrm{N/C} \]

Positive, so the field points in the \(+x\) direction — toward decreasing potential, as always. Knowing \(V\) as a function alone was enough to recover the field.

5 Assembling three charges

Problem. Three charges \(q = +2.0\,\mu\mathrm{C}\) each are placed at the corners of an equilateral triangle of side \(a = 0.10\,\mathrm{m}\). Find the potential energy of the configuration.

Solution. There are three equal pairs, each separated by \(a\).

U = 3 · kq² / a
\[ U = 3\cdot\frac{(8.99\times10^{9})(2.0\times10^{-6})^{2}}{0.10} \approx 1.1\,\mathrm{J} \]

Positive, because all charges are alike — energy you supplied is stored, ready to drive the charges apart the instant they are released.

Review

Chapter Summary

✓ Potential energy

\(\Delta U = -W\); with \(U=0\) at infinity, \(U = -W_{\infty}\), measured in joules.

✓ Electric potential

\(V = U/q\), in volts (J/C); only differences \(\Delta V = -W/q\) are physical.

✓ Equipotentials

Surfaces of constant \(V\); no work to move along them; always \(\perp\vec{E}\).

✓ Field ↔ potential

\(V = -\int \vec{E}\cdot d\vec{s}\) one way; \(\vec{E} = -\nabla V\) the other.

✓ Point charge

\(V = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{q}{r}\); sign of \(V\) matches sign of \(q\).

✓ Many charges

Scalar sum \(V = k\sum q_{i}/r_{i}\); continuous: \(V = k\int dq/r\).

✓ System energy

\(U = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{q_{1}q_{2}}{r}\) per pair; sum over all pairs.

✓ Conductor

Whole conductor is one equipotential; \(V\) constant inside and on the surface.

Practice

Problems

Take \(k = 8.99\times10^{9}\,\mathrm{N\cdot m^{2}/C^{2}}\), \(\varepsilon_{0} = 8.85\times10^{-12}\,\mathrm{C^{2}/N\cdot m^{2}}\), \(e = 1.602\times10^{-19}\,\mathrm{C}\), and \(V = 0\) at infinity unless stated otherwise. Remember that potential is a scalar — keep track of signs, not directions.

  1. How much work does the electric field do on a charge \(q = 2.0\,\mu\mathrm{C}\) as it moves through a potential drop of 150 V?
  2. A proton is moved from a point at \(V = +60\,\mathrm{V}\) to a point at \(V = -40\,\mathrm{V}\). Find the change in its potential energy, in joules and in electronvolts.
  3. Find the potential 0.50 m from a point charge \(q = -3.0\,\mathrm{nC}\).
  4. At what distance from a \(+5.0\,\mathrm{nC}\) charge is the potential equal to 100 V?
  5. Two charges \(+4.0\,\mathrm{nC}\) and \(-4.0\,\mathrm{nC}\) are 0.20 m apart. Find the potential at the midpoint between them, and at a point 0.10 m beyond the negative charge along the line joining them.
  6. Three charges \(+q\), \(+q\), and \(-q\) (with \(q = 2.0\,\mathrm{nC}\)) sit at three corners of a square of side 0.10 m. Find the potential at the empty fourth corner.
  7. An electron is released from rest at a point where \(V = 0\) and drifts to a point where \(V = +200\,\mathrm{V}\). Find its kinetic energy and speed there.
  8. In a region the potential is \(V(x) = 3.0x - 2.0x^{3}\) (volts, \(x\) in metres). Find \(E_{x}\) at \(x = 1.0\,\mathrm{m}\).
  9. A uniform field of 600 N/C points in the \(+x\) direction. Find the potential difference \(V_{B} - V_{A}\) between \(A\) at the origin and \(B\) at \(x = 0.40\,\mathrm{m}\).
  10. A charged ring of radius 4.0 cm carries \(q = 8.0\,\mathrm{nC}\). Find the potential at a point on its axis 3.0 cm from the centre.
  11. Compute the potential energy of two protons separated by \(1.0\times10^{-15}\,\mathrm{m}\) (roughly a nuclear diameter), in joules and in MeV.
  12. Four equal charges \(+q\) are placed at the corners of a square of side \(a\). Find the total potential energy of the configuration in terms of \(k\), \(q\), and \(a\).
  13. An isolated conducting sphere of radius 0.15 m is raised to a potential of 200 V. Find the charge on it and its surface charge density.
  14. Show, starting from \(\vec{E} = -\nabla V\), that the field is everywhere perpendicular to an equipotential surface, and explain why this guarantees the surface of a conductor is an equipotential.
Tip: three habits keep potential problems clean. First, decide your zero level (usually infinity) and stick with it. Second, when adding contributions, remember potential is a scalar — carry each charge's sign but forget about direction entirely. Third, move fluently between the two pictures: integrate the field to get \(V\) when the field is simple, and differentiate \(V\) to get the field when the potential is simple — pick whichever derivative or integral is the easier one.