3.3.2 Work, potential difference and the reference point

Work done in moving a charge

A charge q is carried along a curve C from A to B. To move it slowly, without accelerating it, an external agent must apply the force \vec F_{\mathrm{ext}}=-q\vec E that just balances the electric force. The work of this force depends only on the potentials at the end points, not on the shape of C. — **Figure 3.3.4.** A charge $q$ is carried along a curve $C$ from A to B. To move it slowly, without accelerating it, an external agent must apply the force $F_{ext} = - q E$ that just balances the electric force. The work of this force depends only on the potentials at the end points, not on the shape of $C$ .

Let us compute the work needed to move a point charge $q$ against a given electrostatic field $E$ along a curve $C$ from a point A to a point B (Figure 3.3.4). The field exerts the force $q E$ on the charge. To carry the charge slowly, so that it gains no kinetic energy, an external agent (our hand) must apply the opposite force $F = - q E$ . On a small displacement $d l$ it does the work $d W = F \cdot d l = - q E \cdot d l$ . Adding up along $C$ and using Eq. (3.3.5),

W = \int_{A}^{B} F \cdot d l = - q \int_{A}^{B} E \cdot d l = q \int_{A}^{B} \nabla V \cdot d l = q (V_{B} - V_{A}) [J] . (3.3.11)

The last step uses the defining property of the gradient (Section 2.2): along a small displacement $d l = d x a_{x} + d y a_{y} + d z a_{z}$ the change of $V$ is

d V = \frac{\partial V}{\partial x} d x + \frac{\partial V}{\partial y} d y + \frac{\partial V}{\partial z} d z = \nabla V \cdot d l,

so the line integral of $\nabla V$ is just the sum of the small changes of $V$ along the path, $\int_{A}^{B} d V = V_{B} - V_{A}$ , exactly as the total climb of a walk is the final height minus the starting height, however many ups and downs there were in between.

Equation (3.3.11) has several important consequences.

The work does not depend on the path between A and B. Only the values of the potential at the two end points enter. Any curve from A to B (straight, wiggly, or one that wanders far away and comes back) requires the same work (Figure 3.3.5). Home exercise 3.3.6 verifies this for two different paths.
Only potential differences are physically meaningful. The work, the field and the force are all determined by differences or derivatives of $V$ , and the arbitrary constant $V_{0}$ in Eq. (3.3.4) cancels from all of them. The potential difference or voltage between B and A,

$V_{B A} = V_{B} - V_{A} = \frac{W}{q} = - \int_{A}^{B} E \cdot d l [V], (3.3.12)$

is the work per unit charge needed to carry a charge from A to B. This is also what a voltmeter measures. From Eq. (3.3.12), $1 V = 1 J/C$ .
We may choose $V_{0}$ as we wish. The constant $V_{0}$ has no effect on measurable quantities such as the field and the force, because these require a differentiation. Choosing $V_{0}$ amounts to choosing the point at which the potential is zero, the reference point.

Path independence for the field of a positive point charge. The dashed circles are equipotentials (here V_1>V_2>V_3>V_4). Path 1 first follows an equipotential, where no work is done, and then moves straight in, while path 2 wanders in and out. Both start on the same equipotential and end on the same one, so both require the same work q(V_B-V_A). — **Figure 3.3.5.** Path independence for the field of a positive point charge. The dashed circles are equipotentials (here $V_{1} > V_{2} > V_{3} > V_{4}$ ). Path 1 first follows an equipotential, where no work is done, and then moves straight in, while path 2 wanders in and out. Both start on the same equipotential and end on the same one, so both require the same work $q (V_{B} - V_{A})$ .

In plain words

A voltage is a "height difference" for charges: $V_{B} - V_{A}$ volts means that every coulomb carried from A to B needs $V_{B} - V_{A}$ joules. How you carry it does not matter.

Common mistake · Signs and limits in

V = - \int E \cdot d l

The rule is $V_{B} - V_{A} = - \int_{A}^{B} E \cdot d l$ : the lower limit is the starting point A, the upper limit the end point B, and there is a minus sign. Equivalently $V_{A} - V_{B} = + \int_{A}^{B} E \cdot d l$ . Swapping the limits and dropping the minus sign gives an equivalent, correct form, $V_{B} - V_{A} = \int_{B}^{A} E \cdot d l$ , whereas changing only one of them gives the wrong sign. A quick test: moving along $E$ the potential must decrease.

Quick check

A charge $q = 2 μ C$ is carried from a point at $10 V$ to a point at $4 V$ . How much work does the external agent do? And the field?

$W = q (V_{B} - V_{A}) = 2 \times 1 0^{- 6} \times (4 - 10) = - 12 μ J$ . The agent does negative work (it holds the charge back), and the field does $+ 12 μ J$ . A positive charge "rolls downhill" to lower potential by itself.

The reference at infinity

Unless explicitly stated otherwise, we shall fix $V_{0}$ such that $V \to 0$ at infinity, $V (\infty) = 0$ . With $V_{0} = 0$ the integral formulas (3.3.4)–(3.3.6) do exactly this for any charge distribution of finite extent, since every term $d q^{'} / R$ vanishes as $R \to \infty$ . Taking A at infinity and B at the point $r$ in Eq. (3.3.12),

V_{r} - V_{\infty} = V_{r} = \int_{\infty}^{r} \nabla V \cdot d l = - \int_{r}^{\infty} \nabla V \cdot d l = \int_{r}^{\infty} E \cdot d l . (3.3.13)

Physical picture

$V (P)$ can be interpreted as the work that must be done to bring a unit positive point charge from infinity to the point P (slowly, and without disturbing the charges that produce the field). A positive potential means that work must be done against a repelling field, whereas a negative potential means that the field itself pulls the charge in.

Check: the point charge

For a point charge $q$ at the origin, Coulomb's law gives $E = q a_{r} / (4 π ε_{0} r^{2})$ . Integrating along a radial line from $r$ to infinity, where $d l = d r^{'} a_{r}$ ,

V (r) = \int_{r}^{\infty} \frac{q}{4 π ε _{0} r ^{'2}} d r^{'} = \frac{q}{4 π ε _{0}} [- \frac{1}{r ^{'}}]_{r}^{\infty} = \frac{q}{4 π ε _{0} r},

which agrees with Eq. (3.3.6) for $R = r$ : the two definitions of the potential (as an integral over the charges and as a line integral of the field) give the same function when $V (\infty) = 0$ . Figure 3.3.6a compares this $1/ r$ curve with the potentials of a charged spherical shell and of a uniformly charged ball, obtained in Exercise 3.3.3 (by integrating over the charge) and in Exercise 3.3.7 (by integrating the field in exactly this way).

Recipe · Finding

V

by integrating

E

Find $E$ everywhere along a convenient path (Coulomb's or Gauss's law, Sections 3.1 and 3.2).
Choose the reference point where $V = 0$ : infinity for a charge distribution of finite size, a finite point (an earthed conductor, a chosen radius $ρ_{0}$ ) otherwise.
Write $V (P) = \int_{P}^{ref} E \cdot d l$ , integrating from P to the reference along the simplest path (usually radial, since the result does not depend on the path).
If $E$ has different formulas in different regions, split the integral at the boundaries (e.g. $\int_{r}^{b} E_{in} d r^{'} + \int_{b}^{\infty} E_{out} d r^{'}$ ). $V$ is continuous across the boundaries.
Check: $V$ decreases along $E$ , and far from a finite source $V \to Q / (4 π ε_{0} r)$ .

Potential versus distance. (a) Point charge, charged spherical shell and uniformly charged ball, all of total charge Q and radius b, with V(\infty)=0. Outside the sources the three curves coincide. Inside, the shell's potential is constant (zero field) and the ball's is a parabola with V(0)=\tfrac32V(b). — **Figure 3.3.6.** Potential versus distance. (a) Point charge, charged spherical shell and uniformly charged ball, all of total charge $Q$ and radius $b$ , with $V (\infty) = 0$ . Outside the sources the three curves coincide. Inside, the shell's potential is constant (zero field) and the ball's is a parabola with $V (0) = \frac{3}{2} V (b)$ .

(b) An infinitely long line charge, V=-\frac{\rho_L}{2\pi\varepsilon_0}\ln(\rho/\rho_0), for three choices of the reference radius \rho_0 (in arbitrary length units). The curves differ only by a constant, so differences and slopes (the field) are the same. But every curve goes to -\infty as \rho\to\infty, so V(\infty)=0 cannot be imposed. — (b) An infinitely long line charge, $V = - \frac{ρ _{L}}{2 π ε _{0}} ln (ρ / ρ_{0})$ , for three choices of the reference radius $ρ_{0}$ (in arbitrary length units). The curves differ only by a constant, so differences and slopes (the field) are the same. But every curve goes to $- \infty$ as $ρ \to \infty$ , so $V (\infty) = 0$ cannot be imposed.

When infinity cannot be the reference

For charge distributions that extend to infinity, such as an infinitely long line charge or an infinite charged plane, the integral (3.3.13) diverges and $V (\infty) = 0$ cannot be imposed. For the infinite line, $E_{ρ} = ρ_{L} / (2 π ε_{0} ρ)$ gives

V (ρ) - V (ρ_{0}) = - \int_{ρ_{0}}^{ρ} \frac{ρ _{L}}{2 π ε _{0} ρ ^{'}} d ρ^{'} = - \frac{ρ _{L}}{2 π ε _{0}} ln \frac{ρ}{ρ _{0}},

whose magnitude grows without bound as $ρ \to \infty$ (Figure 3.3.6b). One then chooses a finite reference point $ρ_{0}$ , where $V = 0$ , for instance the earthed screen of a coaxial cable (Problem 3.3.2) or the mid-plane of a two-wire line (Exercise 3.3.4). In circuits the reference is the "ground" node. Home exercise 3.3.1 shows how the potential of a finite line becomes unbounded as its length grows. The trouble is not physical: real lines are finite. It only says that an idealised infinite source contains infinite charge, which makes "the work to come in from infinity" infinite.

Common mistake · Infinity as the reference for an infinite line or plane

Writing $V (ρ) = \int_{ρ}^{\infty} \frac{ρ _{L}}{2 π ε _{0} ρ ^{'}} d ρ^{'}$ gives $\infty$ , not a potential. For sources that extend to infinity always choose a finite reference, $V (ρ_{0}) = 0$ .

Common mistake · Forgetting that only differences matter

A statement such as "this wire is at 5 V" has a meaning only together with a reference ("with respect to the chassis"). Adding the same constant to $V$ everywhere changes no field, no force and no work. Two solutions that differ by a constant describe the same physics, so do not treat them as a contradiction.

Quick check

Somebody adds $1000 V$ to the potential at every point of space. What happens to $E$ , and to the work needed to move a charge from A to B?

Nothing: $E = - \nabla V$ and $W = q (V_{B} - V_{A})$ involve only derivatives and differences, from which the constant drops out. Only the reference point has moved.