Question

Consider a cylindrical capacitor like that shown in Fig. 24.6. Let d = rb − ra be the spacing between the inner and outer conductors. (a) Let the radii of the two conductors be only slightly different, so that d << ra. Show that the result derived in Example 24.4 (Section 24.1) for the capacitance of a cylindrical capacitor then reduces to Eq. (24.2), the equation for the capacitance of a parallel-plate capacitor, with A being the surface area of each cylinder. Use the result that In (1 + z) ≅ z for |z| << 1. (b) Even though the earth is essentially spherical, its surface appears flat to us because its radius is so large. Use this idea to explain why the result of part (a) makes sense from a purely geometrical standpoint.

Figure. 24.6

A long cylindrical capacitor. The linear charge density λ is assumed to be positive in this figure. The magnitude of charge in a length L of either cylinder is λL.

EXAMPLE 24.4 A cylindrical capacitor

A long cylindrical conductor has a radius ra and a linear charge density +λ. It is surrounded by a coaxial cylindrical conducting shell with inner radius rb and linear charge density −λ (Fig. 24.6). Calculate the capacitance per unit length for this capacitor, assuming that there is vacuum in the space between cylinders.

SOLUTION

IDENTIFY: As in Example 24.3, we use the fundamental definition of capacitance.

SET UP: We first find expressions for the potential difference Vab between the cylinders and the charge Q in a length L of the cylinders; we then find the capacitance of a length L using Eq. (24.1). Our target variable is this capacitance divided by L.

EXECUTE: To find the potential difference between the cylinders, we use a result that we worked out in Example 23.10 (Section 23.3). There we found that at a point outside a charged cylinder a distance r from the axis, the potential due to the cylinder is

 

where r0 is the (arbitrary) radius at which V = 0. We can use this same result for the potential between the cylinders in the present problem because, according to Gauss’s law, the charge on the outer cylinder doesn’t contribute to the field between cylinders (see Example 24.3). In our case, we take the radius r0 to be rb, the radius of the inner surface of the outer cylinder, so that the outer conducting cylinder is at V = 0. Then the potential at the outer surface of the inner cylinder (where r = ra) is just equal to the potential Vab of the inner (positive) cylinder a with respect to the outer (negative) cylinder b, or

 

This potential difference is positive (assuming that λ is positive, as in Fig. 24.6) because the inner cylinder is at higher potential than the outer.

The total charge Q in a length L is Q = λL, so from Eq. (24.1) the capacitance C of a length L is

 

The capacitance per unit length is

 

Substituting ϵ0 = 8.85 × 10−12 F/m = 8.85 pF/m, we get

 

EVALUATE: We see that the capacitance of the coaxial cylinders is determined entirely by the dimensions, just as for the parallel-plate case. Ordinary coaxial cables are made like this but with an insulating material instead of vacuum between the inner and outer conductors. A typical cable for TV antennas and VCR connections has a capacitance per unit length of 69 pF/m.

EXAMPLE 24.3 A Spherical capacitor

Two concentric spherical conducting shells are separated by vacuum. The inner shell has total charge +Q and outer radius ra, and the outer shell has charge −Q and inner radius rb (Fig. 24.5). (The inner shell is attached to the outer shell by thin insulating rods that have negligible effect on the capacitance.) Find the capacitance of this spherical capacitor.

SOLUTION

IDENTIFY: This isn’t a parallel-plate capacitor, so we can’t use the relationships developed for that particular geometry. Instead, we’ll go back to the fundamental definition of capacitance: the magnitude of the charge on either conductor divided by the potential difference between the conductors.

SET UP: We use Gauss’s law to find the electric field between the spherical conductors. From this value we determine the potential difference Vab between the two conductors; we then use Eq. (24.1) to find the capacitance C = Q/Vab.

EXECUTE: Using the same procedure as in Example 22.5 (Section 22.4), we take as our Gaussian surface a sphere with radius r between the two spheres and concentric with them. Gauss’s law, Eq. (22.8), states that the electric flux through this surface is equal to the total charge enclosed within the surface, divided by ϵ0:

 

By symmetry, is constant in magnitude and parallel to at every point on this surface, so the integral in Gauss’s law is equal to (E)(4πr2). The total charge enclosed is Qencl = Q, so we have

 

The electric field between the spheres is just that due to the charge on the inner sphere; the outer sphere has no effect. We found in Example 22.5 that the charge on a conducting sphere produces zero field inside the sphere, which also tells us that the outer conductor makes no contribution to the field between the conductors.

The above expression for E is the same as that for a point charge Q, so the expression for the potential can also be taken to be the same as for a point charge, V = Q/4πϵ0r. Hence the potential of the inner (positive) conductor at r = ra with respect to that of the outer (negative) conductor at r = rb is

 

Finally, the capacitance is

 

As an Example, if ra = 9.5 cm and rb = 10.5 cm,

 

EVALUATE: We can relate this result to the capacitance of a parallel-plate capacitor. The quantity 4πrarb is intermediate between the areas and of the two spheres; in fact, it’s the geometric mean of these two areas, which we can denote by Agm. The distance between spheres is d = rb − ra, so we can rewrite the above result as C = ϵ0Agm/d. This is exactly the same form as for parallel plates: C =ϵ0A/d. The point is that if the distance between spheres is very small in comparison to their radii, they behave like parallel plates with the same area and spacing.

Figure. 24.5

A spherical capacitor.

EXAMPLE 22.5 Field of a charged conducting sphere

We place positive charge q on a solid conducting sphere with radius R (Fig. 22.18). Find at any point inside or outside the sphere.

SOLUTION

IDENTIFY: As we discussed earlier in this section, all the charge must be on the surface of the sphere. The system has spherical symmetry.

SET UP: To take advantage of the symmetry, we take as our Gaussian surface an imaginary sphere of radius r centered on the conductor. To calculate the field outside the conductor, we take r to be greater than the conductor’s radius R; to calculate the field inside, we take r to be less than R. In either case, the point where we want to calculate lies on the Gaussian surface.

EXECUTE: The role of symmetry deserves careful discussion before we do any calculations. When we say that the system is spherically symmetric, we mean that if we rotate it through any angle about any axis through the center, the system after rotation is indistinguishable from the original unrotated system. The charge is free to move on the conductor, and there is nothing about the conductor that would make it tend to concentrate more in some regions than others. So we conclude that the charge is distributed uniformly over the surface.

Symmetry also shows that the direction of the electric field must be radial, as shown in Fig. 22.18. If we again rotate the system, the field pattern of the rotated system must be identical to that of the original system. If the field had a component at some point that was perpendicular to the radial direction, that component would have to be different after at least some rotations. Thus there can’t be such a component, and the field must be radial. For the same reason the magnitude E of the field can depend only on the distance r from the center and must have the same value at all points on a spherical surface concentric with the conductor.

Our choice of a sphere as a Gaussian surface takes advantage of these symmetry properties. We first consider the field outside the conductor, so we choose r > R. The entire conductor is within the Gaussian surface, so the enclosed charge is q. The area of the Gaussian surface is 4πr2; is uniform over the surface and perpendicular to it at each point. The flux integral ∮E⊥dA in Gauss’s law is therefore just E(4πr2), and Eq. (22.8) gives

 

This expression for the field at any point outside the sphere (r > R) is the same as for a point charge; the field due to the charged sphere is the same as though the entire charge were concentrated at its center. Just outside the surface of the sphere, where r = R,

 

CAUTION Flux can be positive or negative Remember that we have chosen the charge q to be positive. If the charge is negative, the electric field is radially inward instead of radially outward, and the electric flux through the Gaussian surface is negative. The electric field magnitudes outside and at the surface of the sphere are given by the same expressions as above, except that q denotes the magnitude (absolute value) of the charge.

To find inside the conductor, we use a spherical Gaussian surface with radius r < R. The spherical symmetry again tells us that E(4πr2) = Qencl/ϵ0. But because all of the charge is on the surface of the conductor, our Gaussian surface (which lies entirely within the conductor) encloses no charge. So Qencl = 0 and, therefore, the electric field inside the conductor is zero.

EVALUATE: We already knew that inside the conductor, as it must be inside any solid conductor when the charges are at rest. Figure 22.18 shows E as a function of the distance r from the center of the sphere. Note that in the limit as R → 0, the sphere becomes a point charge; there is then only an “outside,” and the field is everywhere given by E = q/4πϵ0r2. Thus we have deduced Coulomb’s law from Gauss’s law. (In Section 22.3 we deduced Gauss’s law from Coulomb’s law, so this completes the demonstration of their logical equivalence.)

We can also use this method for a conducting spherical shell (a spherical conductor with a concentric spherical hole in the center) if there is no charge inside the hole. We use a spherical Gaussian surface with radius r less than the radius of the hole. If there were a field inside the hole, it would have to be radial and spherically symmetric as before, so E = Qencl/4πϵ0r2. But now there is no enclosed charge, so Qencl = 0 and E = 0 inside the hole.

Can you use this same technique to find the electric field in the interspace between a charged sphere and a concentric hollow conducting sphere that surrounds it?

Figure. 22.18

Calculating the electric field of a conducting sphere with positive charge q. Outside the sphere, the field is the same as if all of the charge were concentrated at the center of the sphere.

EXAMPLE 23.10 An infinite line charge or charged conducting cylinder

Find the potential at a distance r from a very long line of charge with linear charge density (charge per unit length) λ.

SOLUTION

IDENTIFY: One approach to this problem is to divide the line of charge into infinitesimal elements, as we did in Example 21.11 (Section 21.5) to find the electric field produced by such a line. We could then integrate as in Eq. (23.16) to find the net potential V. In this case, however, our task is greatly simplified because we already know the electric field.

SET UP: In both Example 21.11 and Example 22.6 (Section 22.4), we found that the electric field at a distance r from a long straight- line charge (Fig. 23.20a) has only a radial component, given by

 

We use this-expression to find the potential by integrating as in Eq. (23.17).

EXECUTE: Since the field has only a radial component, the scalar product is equal to Erdr. Hence the potential of any point a with respect to any other point b, at radial distances ra, and rb from the line of charge, is

 

If we take point b at infinity and set Vb = 0, we find that Va is infinite:

 

This shows that if we try to define V to be zero at infinity, then V must be infinite at any finite distance from the line charge. This is not a useful way to define V for this problem! The difficulty is that the charge distribution itself extends to infinity.

To get around this difficulty, remember that we can define V to be zero at any point we like. We set Vb = 0 at point b at an arbitrary radial distance r0. Then the potential V = Va at point a at a radial distance r is given by V − 0 = (λ/2πϵ0) ln (r0/r), or

 

EVALUATE: According to our result, if λ is positive, then V decreases as r increases. This is as it should be: V decreases as we move in the direction of .

From Example 22.6, the expression for Er with which we started also applies outside a long charged conducting cylinder with charge per unit length λ (Fig. 23.20b). Hence our result also gives the potential for such a cylinder, but only for values of r (the distance from the cylinder axis) equal to or greater than the radius R of the cylinder. If we choose r0 to be the cylinder radius R, so that V = 0 when r = R, then at any point for which r > R,

 

Inside the cylinder, , and V has the same value (zero) as on the cylinder’s surface.

Figure. 23.20

Electric field outside (a) a long positively charged wire and (b) a long, positively charged cylinder.

EXAMPLE 21.11 Field of a line of charge

Positive electric charge Q is distributed uniformly along a line with length 2a, lying along the y-axis between y = −a and y =+a. (This might represent one of the charged rods in Fig. 21.1.) Find the electric field at point P on the x-axis at a distance x from the origin.

SOLUTION

IDENTIFY: As in Example 21.10, our target variable is the electric field due to a continuous distribution of charge.

SET UP: Figure 21.25 shows the situation. We need to find the electric field at P as a function of the coordinate x. The x-axis is the perpendicular bisector of the charged line, so as in Example 21.10 we will be able to make use of a symmetry argument.

EXECUTE: We divide the line charge into infinitesimal segments, each of which acts as a point charge; let the length of a typical segment at height y be dy. If the charge is distributed uniformly, the linear charge density λ at any point on the line is equal to Q/2a (the total charge divided by the total length). Hence the charge dQ in a segment of length dy is

 

The distance r from this segment to P is (x2 + y2)1/2, so the magnitude of field dE at P due to this segment is

 

We represent this field in terms of its x- and y-components:

 

We note that sin α =y/(x2+ y2)1/2 and cos α = x/(x2 + y2)1/2; combining these with the expression for dE, we find

 

To find the total field components Ex and Ey, we integrate these expressions, noting that to include all of Q, we must integrate from y = −a to y = +a. We invite you to work out the details of the integration; an integral table is helpful. The final results are

 

or, in vector form,

 

EVALUATE: Using a symmetry argument as in Example 21.10, we could have guessed that Ey would be zero; if we place a positive test charge at P, the upper half of the line of charge pushes downward on it, and the lower half pushes up with equal magnitude.

To explore our result, let’s first see what happens in the limit that x is much larger than a. Then we can neglect a in the denominator of Eq. (21.9), and our result becomes

 

This means that if point P is very far from the line charge in comparison to the length of the line, the field at P is the same as that of a point charge. We found a similar result for the charged ring in Example 21.10.

To further explore our exact result for , Eq. (21.9), let’s express it in terms of the linear charge density λ = Q/2a. Substituting Q = 2aλ into Eq. (21.9) and simplifying, we get

 

Now we can answer the question: What is at a distance x from a very long line of charge? To find the answer we take the limit of Eq. (21.10) as a becomes very large. In this limit, the term x2/a2 in the denominator becomes much smaller than unity and can be thrown away. We are left with

 

The field magnitude depends only on the distance of point P from the line of charge. So at any point P at a perpendicular distance r from the line in any direction, has magnitude

 

Thus the electric field due to an infinitely long line of charge is proportional to 1/r rather than to 1/r2 as for a point charge. The direction of is radially outward from the line if λ is positive and radially inward if λ is negative.

There’s really no such thing in nature as an infinite line of charge. But when the field point is close enough to the line, there’s very little difference between the result for an infinite line and the real-life finite case. For Example, if the distance r of the field point from the center of the line is 1 % of the length of the line, the value of E differs from the infinite-length value by less than 0.02%.

Figure. 21.1

Experiments in electrostatics. (a) Negatively charged objects repel each other. (b) Positively charged objects repel each other. (c) Positively charged objects and negatively charged objects attract each other.

Figure 21.25

Our sketch for this problem.

EXAMPLE 21.10 Field of a ring of charge

A ring-shaped conductor with radius a carries a total charge Q uniformly distributed around it (Fig. 21.24). Find the electric field at a point P that lies on the axis of the ring at a distance x from its center.

SOLUTION

IDENTIFY: This is a problem in the superposition of electric fields. The new wrinkle is that the charge is distributed continuously around the ring rather than in a number of point charges.

SET UP: The field point is an arbitrary point on the x-axis in Fig. 21.24. Our target variable is the electric field at such a point as a function of the coordinate x.

EXECUTE: As shown in Fig. 21.24, we imagine the ring divided into infinitesimal segments of length ds. Each segment has charge dQ and acts as a point-charge source of electric field. Let be the electric field from one such segment; the net electric field at P is then the sum of all contributions from all the segments that make up the ring. (This same technique works for any situation in which charge is distributed along a line or a curve.)

The calculation of is greatly simplified because the field point P is on the symmetry axis of the ring. Consider two segments at the top and bottom of the ring: The contributions to the field at P from these segments have the same x-component but opposite y-components. Hence the total y-component of field due to this pair of segments is zero. When we add up the contributions from all such pairs of segments, the total field will have only a component along the ring’s symmetry axis (the x-axis), with no component perpendicular to that axis (that is, no y-component or z-component). So the field at P is described completely by its x-component Ex.

To calculate Ex, note that the square of the distance r from a ring segment to the point P is r2 = x2 + a2. Hence the magnitude of this segment’s contribution to the electric field at P is

 

Using cos α = x/r = x/(x2 + a)1/2, the x-component dEx of this field is

 

To find the total x-component Ex of the field at P, we integrate this expression over all segments of the ring:

 

Since x does not vary as we move from point to point around the ring, all the factors on the right side except dQ are constant and can be taken outside the integral. The integral of dQ is just the total charge Q, and we finally get

 

EVALUATE: Our result for shows that at the center of the ring (x = 0) the field is zero. We should expect this; charges on opposite sides of the ring would push in opposite directions on a test charge at the center, and the forces would add to zero. When the field point P is much farther from the ring than its size (that is, x ≫ a) the denominator in Eq. (21.8) becomes approximately equal to x3, and the expression becomes approximately

 

In other words, when we are so far from the ring that its size a is negligible in comparison to the distance x, its field is the same as that of a point charge. To an observer far from the ring, the ring would appear like a point, and the electric field reflects this.

In this Example we used a symmetry argument to conclude that had only an x-component at a point on the ring’s axis of symmetry. We’ll use symmetry arguments many times in this and subsequent chapters. Keep in mind, however, that such arguments can be used only in special cases. At a point in the xy-plane that is not on the x-axis in Fig. 21.24, the symmetry argument doesn’t apply, and the field has in general both x- and y-components.

Figure. 21.24 Calculating the electric field on the axis of a ring of charge. In this figure, the charge is assumed to be positive.

EXAMPLE 22.6 Field of a line charge

Electric charge is distributed uniformly along an infinitely long, thin wire. The charge per unit length is λ (assumed positive). Find the electric field. (This is an approximate representation of the field of a uniformly charged finite wire, provided that the distance from the field point to the wire is much less than the length of the wire.)

SOLUTION

IDENTIFY: The system has cylindrical symmetry. The field must point away from the positive charges. To determine the direction of more precisely, as well as how its magnitude can depend on position, we use symmetry as in Example 22.5.

SET UP: Cylindrical symmetry means that we can rotate the system through any angle about its axis, and we can shift it by any amount along the axis; in each case the resulting system is indistinguishable from the original. Hence at each point can’t change when either of these operations is carried out. The field can’t have any component parallel to the wire; if it did, we would have to explain why the field lines that begin on the wire pointed in one direction parallel to the wire and not the other. Also, the field can’t have any component tangent to a circle in a plane perpendicular to the wire with its center on the wire. If it did, we would have to explain why the component pointed in one direction around the wire rather than the other. All that’s left is a component radially outward from the wire at each point. So the field lines outside a uniformly charged, infinite wire are radial and lie in planes perpendicular to the wire. The field magnitude can depend only on the radial distance from the wire.

These symmetry properties suggest that we use as a Gaussian surface a cylinder with arbitrary radius r and arbitrary length l, with its ends perpendicular to the wire (Fig. 22.19).

EXECUTE: We break the surface integral for the flux ΦE into an integral over each flat end and one over the curved side walls. There is no flux through the ends because lies in the plane of the surface and E⊥ = 0. To find the flux through the side walls, note that is perpendicular to the surface at each point, so E = E⊥; by symmetry, E has the same value everywhere on the walls. The area of the side walls is 2πrl. (To make a paper cylinder with radius r and height l, you need a paper rectangle with width 2πr, height l, and area 2πrl.) Hence the total flux ΦE through the entire cylinder is the sum of the flux through the side walls, which is (E) (2πrl), and the zero flux through the two ends. Finally, we need the total enclosed charge, which is the charge per unit length multiplied by the length of wire inside the Gaussian surface, or Qencl = λl. From Gauss’s law, Eq. (22.8),

 

This is the same result that we found in Example 21.11 (Section 21.5) by much more laborious means.

We have assumed that λ is positive. If it is negative, is directed radially inward toward the line of charge, and in the above expression for the field magnitude E we must interpret λ as the magnitude (absolute value) of the charge per unit length.

EVALUATE: Note that although the entire charge on the wire contributes to the field, only the part of the total charge that is within the Gaussian surface is considered when we apply Gauss’s law. This may seem strange; it looks as though we have somehow obtained the right answer by ignoring part of the charge and the field of a short wire of length l would be the same as that of a very long wire. But we do include the entire charge on the wire when we make use of the symmetry of the problem. If the wire is short, the symmetry with respect to shifts along the axis is not present, and the field is not uniform in magnitude over our Gaussian surface. Gauss’s law is then no longer useful and cannot be used to find the field; the problem is best handled by the integration technique used in Example 21.11.

We can use a Gaussian surface like that in Fig. 22.19 to show that the field at points outside a long, uniformly charged cylinder is the same as though all the charge were concentrated on a line along its axis. We can also calculate the electric field in the space between a charged cylinder and a coaxial hollow conducting cylinder surrounding it. We leave these calculations to you (see Problems 22.37 and 22.40).

Problem 22.37

The Coaxial Cable. A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b and outer radius c. The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length λ. Calculate the electric field (a) at any point between the cylinders a distance r from the axis and (b) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance r from the axis of the cable, from r = 0 to r = 2c. (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

Problem 22.40

A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume ρ. (a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density ρ. (b) What is the electric field at a point outside the volume in terms of the charge per unit length λ in the cylinder? (c) Compare the answers to parts (a) and (b) for r = R. (d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.

Figure. 22.19

A coaxial cylindrical Gaussian surface is used to find the electric field outside an infinitely long, charged wire.

Answer 1

The capacitance C of a cylindrical capacitor in terms of inner radius and outer radius is given as follows:

Here, L is the length of the cylinder, and is the permittivity of the free space.

(a)

The distance d between the inner and outer conductors is given as follows:

Rearrange the above equation for

Substitute in the above equation and solve for C.

Here,

For the values of

Take as z and compare this with above equation and find the value of

Substitute forin the above equation and solve for C.

The area A of a cylinder of radius and length L is given as follows:

Substitute A for in the above equation and solve for C.

The above expression represents capacitance of a parallel plate capacitor.

Thus, the capacitance of a cylindrical capacitor is reduces to the equation for the capacitance for a parallel plate capacitor.

(b)

The result of part (a) is a capacitance due to a parallel plate capacitor when the separation between inner and outer conductors is very small.

Since the radius of the earth is very large, the outer surface of the earth is appears like a flat surface.

From a purely geometric standpoint, this makes sense if the sample area is small enough, it appears almost flat. Like the earth sampled from the human perspective appears flat.