# 4.5 Electric Scalar Potential

The operation of an electric circuit usually is described in terms of the currents flowing through its branches and the voltages at its nodes. The voltage difference V between two points in a circuit represents the amount of work, or potential energy, required to move a unit charge from one to the other.

Even though when analyzing a circuit we may not consider the electric fields present in the circuit, it is in fact the existence of these fields that gives rise to voltage differences across circuit elements such as resistors or capacitors. The relationship between the electric field E and the electric potential V is the subject of this section.

# 4.5.1 Electric Potential as a Function of Electric Field

We begin by considering the simple case of a positive charge q in a uniform electric field E=−y ˆ E in the −y direction (Fig. 4-12). The presence of the field E exerts a force F e=q E on the charge in the −y direction. To move the charge along the positive y direction (against the force F e), we need to provide an external force F ext to counteract F e, which requires the expenditure of energy. To move q without acceleration (at constant speed), the net force acting on the charge must be zero, which means that F ext+F e=0, or

F ext=−F e=−q E.(4.34)

# Figure 4-12 Work done in moving a charge q a distance d y against the electric field E is d W=q E d y.

Figure 4 12 Full Alternative Text

The work done (or energy expended) in moving any object a vector differential distance d l while exerting a force F ext is

d W=F ext⋅d l=−q E⋅d l(J).(4.35)

Work (or energy) is measured in joules (J). If the charge is moved a distance d y along y ˆ, then

d W=−q(−y ˆ E)⋅y ˆ d y=q E d y.(4.36)

The differential electric potential energy d W per unit charge is called the differential electric potential (or differential voltage) d V. That is,

d V=d W q=−E⋅d l(J/C or V).(4.37)

The unit of V is the volt (V) with 1 V=1 J/C, and since V is measured in volts, the electric field is expressed in volts per meter (V/m).

The potential difference corresponding to moving a point charge from point P 1 to point P 2 (Fig. 4-13) is obtained by integrating Eq. (4.37) along any path between them. That is,

∫P 1 P 2 d V=−∫P 1 P 2 E⋅d l,(4.38)

or

# Figure 4-13 In electrostatics, the potential difference between P 2 and P 1 is the same irrespective of the path used for calculating the line integral of the electric field between them.

Figure 4 13 Full Alternative Text

where V 1 and V 2 are the electric potentials at points P 1 and P 2, respectively. The result of the line integral on the right-hand side of Eq. (4.39) is independent of the specific integration path that connects points P 1 and P 2. This follows immediately from the law of conservation of energy. To illustrate with an example, consider a particle in Earth’s gravitational field. If the particle is raised from a height h 1 above Earth’s surface to height h 2, the particle gains potential energy in an amount proportional to (h 2−h 1). If instead we were to first raise the particle from height h 1 to a height h 3 greater than h 2, thereby giving it potential energy proportional to (h 3−h 1), and then let it drop back to height h 2 by expending an energy amount proportional to (h 3−h 2), its net gain in potential energy would again be proportional to (h 2−h 1).

The same principle applies to the electric potential energy W and to the potential difference (V 2−V 1). The voltage difference between two nodes in an electric circuit has the same value regardless of which path in the circuit we follow between the nodes. Moreover, Kirchhoff’s voltage law states that the net voltage drop around a closed loop is zero. If we go from P 1 to P 2 by path 1 in Fig. 4-13 and then return from P 2 to P 1 by path 2, the right-hand side of Eq. (4.39) becomes a closed contour, and the left-hand side vanishes. In fact, the line integral of the electrostatic field E around any closed contour C is zero:

As we will see later in Chapter 6, if E is a time-varying function, it is no longer conservative, and its line integral along a closed path is not necessarily zero.

The conservative property of the electrostatic field can be deduced from Maxwell’s second equation, Eq. (4.1b). If ∂/∂t=0, then

∇×E=0.(4.41)

If we take the surface integral of ∇×E over an open surface S and then apply Stokes’s theorem expressed by Eq. (3.107) to convert the surface integral into a line integral, we obtain

∫S(∇×E)⋅d s=∮C E⋅d l=0,(4.42)

where C is a closed contour surrounding S. Thus, Eq. (4.41) is the differential-form equivalent of Eq. (4.40).

We now define what we mean by the electric potential V at a point in space. Before we do so, however, let us revisit our electric-circuit analogue. Just as a node in a circuit cannot be assigned an absolute voltage, a point in space cannot have an absolute electric potential. The voltage of a node in a circuit is measured relative to that of a conveniently chosen reference point to which we have assigned a voltage of zero, which we call ground. The same principle applies to the electric potential V. Usually (but not always), the reference point is chosen to be at infinity. That is, in Eq. (4.39) we assume that V 1=0 when P 1 is at infinity. Therefore, the electric potential V at any point P is

# Example 4-8: Computing V from E along Two Paths

A vector field is said to be conservative if its line integral between two points is the same—irrespective of the path taken between them. In a given region of space, the field E is given by

E=x ˆ x 2+y ˆ y 2+z ˆ z 2.(4.44)

(a) Confirm that E is conservative by demonstrating that ∇×E=0. (b) Compute the potential difference V 21 between points 1 and 2 in Fig. 4-14 following the direct path between them. (c) Compute V 21 by following the path A B C D between points 1 and 2.

# Figure 4-14 Computing V 21 along two paths (Example 4-8).

Figure 4 14 Full Alternative Text

Solution: (a) The given electric field has components E x=x 2, E y=y 2, and E z=z 2. Applying the curl operator to E gives

∇×E=x ˆ y ˆ z ˆ∂∂x∂∂y∂∂z x 2 y 2 z 2=x ˆ∂z 2∂y−∂y 2∂z−y ˆ∂z 2∂x−∂x 2∂z+z ˆ∂y 2∂x−∂x 2∂y=x ˆ(0−0)−y ˆ(0−0)+z ˆ(0−0)=0.

(b) Voltage V 21 is given by

V 21=−∫P 1 P 2 E⋅d l.

The straight-line path resides in the x–y plane, so it is described by the linear form y=a x+b. At point 1, x 1=1 and y 1=−2. Hence,

−2=a+b.

Similarly, at point 2, x 2=3 and y 2=2, so

2=3 a+b.

The two equations lead to a=2, b=−4, and

y=2 x−4.(4.45)

Since path P 1–P 2 is entirely in the x–y plane, we can set z=0 in the expression for E. Also, we can use the relation given by Eq. (4.45) to reduce E to a single variable:

E=x ˆ x 2+y ˆ y 2+z ˆ z 2 z=0 and y=2 x−4=x ˆ x 2+y ˆ(2 x−4)2.(4.46)

In general,

d l=x ˆ d x+y ˆ d y+z ˆ d z.

In the x–y plane, d z=0, and along the straight-line path given by y=2 x−4,

d y=2 d x.(4.47)

Hence,

d l=x ˆ d x+y ˆ 2 d x.(4.48)

The potential difference is then

V 21=−∫P 1 P 2 E⋅d l=−∫x=1 x=3[x ˆ x 2+y ˆ(2 x−4)2]⋅[x ˆ d x+y ˆ 2 d x]=−∫x=1 3[x 2+2(2 x−4)2]d x=−∫x=1 3(9 x 2−32 x+32)d x=−14(V).(4.49)

(c) Path A B C D in Fig. 4-14 consists of three segments.

A to B:

E=x ˆ x 2+y ˆ y 2+z ˆ z 2 x=1, z=0=x ˆ 1+y ˆ y 2, (4.50a)

and

d l=y ˆ d y. (4.50b)

B to C:

E=x ˆ x 2+y ˆ y 2+z ˆ z 2 y=0, z=0=x ˆ x 2, (4.51a)

and

d l=x ˆ d x. (4.51b)

C to D:

E=x ˆ x 2+y ˆ y 2+z ˆ z 2 x=3, z=0=x ˆ 9+y ˆ y 2, (4.52a)

and

d l=y ˆ d y. (4.52b)

Hence,

V 21=−∫P 1 P 2 E⋅d l=−[∫A@x=1, y=−2 B@x=1, y=0(x^ l+y^ y 2)⋅y^ d y+∫B@x=1, y=0 C@x=3, y=0 x^ x 2⋅x^ d x+∫C@x=3, y=0 D@x=3, y=2(x^ 9+y^ y 2)⋅y^ d y]=−[y 3 3|−2 0+x 3 3|1 3+y 3 3|0 2]=−[+8 3+27 3−1 3+8 3]=−14 (V),(4.53)

which is identical with the result given by Eq. (4.49) for the line integral along the straight-line path between points 1 and 2.

# 4.5.2 Electric Potential Due to Point Charges

The electric field due to a point charge q located at the origin is given by Eq. (4.32) as

E=R ˆ q 4 π ε R 2(V/m).(4.54)

The field is radially directed and decays quadratically with the distance R from the observer to the charge.

As was stated earlier, the choice of integration path between the end points in Eq. (4.43) is arbitrary. Hence, we can conveniently choose the path to be along the radial direction R ˆ, in which case d l=R ˆ d R and

V=−∫∞R R ˆ q 4 π ε R 2⋅R ˆ d R=q 4 π ε R(V).(4.55)

If the charge q is at a location other than the origin, say at position vector R 1, then V at observation position vector R becomes

V=q 4 π ε|R−R 1|(V),(4.56)

where |R−R 1| is the distance between the observation point and the location of the charge q. The principle of superposition applied previously to the electric field E also applies to the electric potential V. Hence, for N discrete point charges q 1,q 2,…,q N residing at position vectors R 1, R 2, …, R N, the electric potential is

# 4.5.3 Electric Potential Due to Continuous Distributions

To obtain expressions for the electric potential V due to continuous charge distributions over a volume υ′, across a surface S′, or along a line l′, we (1) replace q i in Eq. (4.57) with ρ v d υ′, ρ s d s′, and ρ ℓ d l′, respectively; (2) convert the summation into an integration; and (3) define R′=|R−R i| as the distance between the integration point and the observation point. These steps lead to the following expressions:

# 4.5.4 Electric Field as a Function of Electric Potential

In Section 4-5.1, we expressed V in terms of a line integral over E. Now we explore the inverse relationship by reexamining Eq. (4.37):

d V=−E⋅d l.(4.59)

For a scalar function V, Eq. (3.73) gives

d V=∇V⋅d l,(4.60)

where ∇V is the gradient of V. Comparison of Eq. (4.59) with Eq. (4.60) leads to

The expressions for V, given by Eqs. (4.57) to (4.58c), involve scalar sums and scalar integrals, and as such are usually much easier to evaluate than the vector sums and integrals in the expressions for E derived in Section 4-3 on the basis of Coulomb’s law. Thus, even though the electric potential approach for finding E is a two-step process, it is conceptually and computationally simpler to apply than the direct method based on Coulomb’s law.

# Example 4-9: Electric Field of an Electric Dipole

An electric dipole consists of two point charges of equal magnitude but opposite polarity separated by a distance d (Fig. 4-15(a)). Determine V and E at any point P given that P is at a distance R≫d from the dipole center and the dipole resides in free space.

# Figure 4-15 Electric dipole with dipole moment p=q d (Example 4-9).

(a) Electric dipole

Figure 4 15-a Full Alternative Text

(b) Electric-field pattern

Solution: To simplify the derivation, we align the dipole along the z axis and center it at the origin (Fig. 4-15(a)). For the two charges shown in Fig. 4-15(a), application of Eq. (4.57) gives

V=1 4 π ε 0 q R 1+−q R 2=q 4 π ε 0 R 2−R 1 R 1 R 2.

Since d≪R, the lines labeled R 1 and R 2 in Fig. 4-15(a) are approximately parallel to each other, in which case the following approximations apply:

R 2−R 1≈d cos θ,R 1 R 2≈R 2.

Hence,

V=q d cos θ 4 π ε 0 R 2.(4.62)

To generalize this result to an arbitrarily oriented dipole, note that the numerator of Eq. (4.62) can be expressed as the dot product of q d (where d is the distance vector from −q to +q) and the unit vector R ˆ pointing from the center of the dipole toward the observation point P. That is,

q d cos θ=q d⋅R ˆ=p⋅R ˆ,(4.63)

where p=q d is called the dipole moment. Using Eq. (4.63) in Eq. (4.62) then gives

In spherical coordinates, Eq. (4.61) is given by

E=−∇V=−R ˆ∂V∂R+θ ˆ 1 R∂V∂θ+ϕ ˆ 1 R sin θ∂V∂ϕ,(4.65)

where we have used the expression for ∇V in spherical coordinates given in Appendix C. Upon taking the derivatives of the expression for V given by Eq. (4.62) with respect to R and θ and then substituting the results in Eq. (4.65), we obtain

We stress that the expressions for V and E given by Eqs. (4.64) and (4.66) apply only when R≫d. To compute V and E at points in the vicinity of the two dipole charges, it is necessary to perform all calculations without resorting to the far-distance approximations that led to Eq. (4.62). Such an exact calculation for E leads to the field pattern shown in Fig. 4-15(b).

# 4.5.5 Poisson’s Equation

With D=ε E, the differential form of Gauss’s law given by Eq. (4.26) may be cast as

∇⋅E=ρ v ε.(4.67)

Inserting Eq. (4.61) in Eq. (4.67) gives

∇⋅(∇V)=−ρ v ε.(4.68)

Given Eq. (3.110) for the Laplacian of a scalar function V,

∇2 V=∇⋅(∇V)=∂2 V∂x 2+∂2 V∂y 2+∂2 V∂z 2,(4.69)

Eq. (4.68) can be cast in the abbreviated form

This is known as Poisson’s equation. For a volume υ′ containing a volume charge density distribution ρ v, the solution for V derived previously and expressed by Eq. (4.58a) as

V=1 4 π ε∫υ′ρ v R′d υ′(4.71)

satisfies Eq. (4.70). If the medium under consideration contains no charges, Eq. (4.70) reduces to

and it is then referred to as Laplace’s equation. Poisson’s and Laplace’s equations are useful for determining the electrostatic potential V in regions with boundaries on which V is known, such as the region between the plates of a capacitor with a specified voltage difference across it.

Concept Question 4-9: What is a conservative field?

Concept Question 4-10: Why is the electric potential at a point in space always defined relative to the potential at some reference point?

Concept Question 4-11: Explain why Eq. (4.40) is a mathematical statement of Kirchhoff’s voltage law.

Concept Question 4-12: Why is it usually easier to compute V for a given charge distribution and then find E using E=−∇V than to compute E directly by applying Coulomb’s law?

Concept Question 4-13: What is an electric dipole?