### Irrotational Field Represented by Scalar Potential: The Gradient Operator and Gradient Integral Theorem

4.1.1Surfaces of constant that are spherical are given by
For example, the surface at radius a has the potential Vo.
 (a) In Cartesian coordinates, what is grad( )? (b) By the definition of the gradient operator, the unit normal n to an equipotential surface is Evaluate n in Cartesian coordinates for the spherical equipotentials given by (a) and show that it is equal to ir, the unit vector in the radial direction in spherical coordinates.
4.1.2For Example 4.1.1, carry out the integral of E ds from the origin to (x, y) = (a, a) along the line y = x and show that it is indeed equal to (0, 0) - (a, a).

4.1.3In Cartesian coordinates, three two-dimensional potential functions are
 (a) Determine E for each potential. (b) For each function, make a sketch of and E using the conventions of Fig. 4.1.3. (c) For each function, make a sketch using conventions of Fig. 4.1.4.
4.1.4*A cylinder of rectangular cross-section is shown in Fig.~P4.1.4. The electric potential inside this cylinder is
where o(t) is a given function of time.
 (a) Show that the electric field intensity is (b) By direct evaluation, show that E is irrotational. (c) Show that the charge density is (d) Show that the tangential E is zero on the boundaries. (e) Sketch the distributions of , , and E using conventions of Figs. 2.7.3 and 4.1.3. (f) Compute the line integral of E ds between the center and corner of the rectangular cross-section (points shown in Fig.~P4.1.4) and show that it is equal to (a/2, b/2, t). Why would you expect the integration to give the same result for any path joining the point (a) to any point on the wall? (g) Show that the net charge inside a length d of the cylinder in the z direction is first by integrating the charge density over the volume and then by using Gauss' integral law and integrating o E da over the surface enclosing the volume. (h) Find the surface charge density on the electrode at y = 0 and use your result to show that the net charge on the electrode segment between x = a/4 and x = 3a/4 having depth d into the paper is (i) Show that the current, i(t), to this electrode segment is
4.1.5Inside the cylinder of rectangular cross-section shown in Fig.~P4.1.4, the potential is given as
where o(t) is a given function of time.
 (a) Find E. (b) By evaluating the curl, show that E is indeed irrotational. (c) Find . (d) Show that E is tangential to all of the boundaries. (e) Using the conventions of Figs. 2.7.3 and 4.1.3, sketch , , and E. (f) Use E as found in part (a) to compute the integral of E ds from (a) to (b) in Fig.~P4.1.4. Check your answer by evaluating the potential difference between these points. (g) Evaluate the net charge in the volume by first using Gauss' integral law and integrating o E da over the surface enclosing the volume and then by integrating over the volume.
4.1.6Given the potential
where A, m, and are given constants.
 (a) Find E. (b) By direct evaluation, show that E is indeed irrotational. (c) Determine the charge density . (d) Can you adjust m so that = 0 throughout the volume?
4.1.7The system, shown in cross-section in Fig.~P4.1.7, extends to in the z direction. It consists of a cylinder having a square cross-section with sides which are resistive sheets (essentially many resistors in series). Thus, the voltage sources V at the corners of the cylinder produce linear distributions of potential along the sides. For example, the potential between the corners at (a, 0) and (0, a) drops linearly from V to -V.

 (a) Show that the potential inside the cylinder can match that on the walls of the cylinder if it takes the form A(x2 - y2. What is A? (b) Determine E and show that there is no volume charge density within the cylinder. (c) Sketch the equipotential surfaces and lines of electric field intensity.

4.1.8Figure P4.1.8 shows a cross-sectional view of a model for a "capacitance" probe designed to measure the depth h of penetration of a tool into a metallic groove. Both the "tool" and the groove can be considered constant potential surfaces having the potential difference v(t) as shown. An insulating segment at the tip of the tool is used as a probe to measure h. This is done by measuring the charge on the surface of the segment. In the following, we start with a field distribution that can be made to fit the problem, determine the charge and complete some instructive manipulations along the way.
 (a) Given that the electric field intensity between the groove and tool takes the form show that E is irrotational and evaluate the coefficient C by computing the integral of E ds between point (a) and the origin. (b) Find the potential function consistent with (a) and evaluate C by inspection. Check with part (a). (c) Using the conventions of Figs. 2.7.3 and 4.1.3, sketch lines of constant potential and electric field E for the region between the groove and the tool surfaces. (d) Determine the total charge on the insulated segment, given v(t). (Hint: Use the integral form of Gauss' law with a convenient surface S enclosing the electrode.)
4.1.9*In cylindrical coordinates, the incremental displacement vector, given in Cartesian coordinates by (9), is
Using arguments analogous to (7)-(12), show that the gradient operator in cylindrical coordinates is as given in Table I at the end of the text.

4.1.10*Using arguments analogous to those of (7)-(12), show that the gradient operator in spherical coordinates is as given in Table I at the end of the text.

### Poisson's Equation

4.2.1*In Prob. 4.1.4, the potential is given by (a). Use Poisson's equation to show that the associated charge density is as given by (c) of that problem.

4.2.2In Prob. 4.1.5, is given by (a). Use Poisson's equation to find the charge density.

4.2.3Use the expressions for the divergence and gradient in cylindrical coordinates from Table I at the end of the text to show that the Laplacian operator is as summarized in that table.

4.2.4Use the expressions from Table I at the end of the text for the divergence and gradient in spherical coordinates to show that the Laplacian operator is as summarized in that table.

### Superposition Principle

4.3.1A current source I(t) is connected in parallel with a capacitor C and a resistor R. Write the ordinary differential equation that can be solved for the voltage v(t) across the three parallel elements. Follow steps analogous to those used in this section to show that if Ia(t) va(t) and Ib(t) vb(t), then Ia(t) + Ib(t) va(t) + vb(t).

### Fields Associated with Charge Singularities

4.4.1*A two-dimensional field results from parallel uniform distributions of line charge, +l at x = d/2, y = 0 and -l at x = -d/2, y = 0, as shown in Fig.~P4.4.1. Thus, the potential distribution is independent of z.
 (a) Start with the electric field of a line charge, (1.3.13), and determine . (b) Define the two-dimensional dipole moment as p = dl and show that in the limit where d 0 (while this moment remains constant), the electric potential is
4.4.2*For the configuration of Prob. 4.4.1, consider the limit in which the line charge spacing d goes to infinity. Show that, in polar coordinates, the potential distribution is of the form
Express this in Cartesian coordinates and show that the associated E is uniform.

4.4.3A two-dimensional charge distribution is formed by pairs of positive and negative line charges running parallel to the z axis. Shown in cross-section in Fig.~P4.4.3, each line is at a distance d/2 from the origin. Show that in the limit where d r, this potential takes the form A cos 2 /rn. What are the constants A and n?
4.4.4The charge distribution described in Prob. 4.4.3 is now at infinity (d r).
 (a) Show that the potential in the neighborhood of the origin takes the form A(x2 - y2). (b) How would you position the line charges so that in the limit where they moved to infinity, the potential would take the form of (4.1.18)?

### Solution of Poisoon's Equation for Specified Charge Distributions

4.5.1The only charge is restricted to a square patch centered at the origin and lying in the x - y plane, as shown in Fig.~P4.5.1.
 (a) Assume that the patch is very thin in the z direction compared to other dimensions of interest. Over its surface there is a given surface charge density s (x, y). Express the potential along the z axis for z > 0 in terms of a two-dimensional integral. (b) For the particular surface charge distribution s = o |xy|/a2 where o and a are constants, determine along the positive z axis. (c) What is at the origin? (d) Show that has a z dependence for z a that is the same as for a point charge at the origin. In this limit, what is the equivalent point charge for the patch? (e) What is E along the positive z axis?
4.5.2*The highly insulating spherical shell of Fig.~P4.5.2 has radius R and is "coated" with a surface charge density s = o cos , where o is a given constant.
(a) Show that the distribution of potential along the z axis in the range z > R is
[Hint: Remember that for the triangle shown in the figure, the law of cosines gives c = (b2 + a2 - 2ab cos )1/2.]
 (b) Show that the potential distribution for the range z < R along the z axis inside the shell is (c) Show that along the z axis, E is (d) By comparing the z dependence of the potential to that of a dipole polarized in the z direction, show that the equivalent dipole moment is qd = (4 /3)o R3.
4.5.3All of the charge is on the surface of a cylindrical shell having radius R and length 2l, as shown in Fig.~P4.5.3. Over the top half of this cylinder at r = R the surface charge density is o (coulomb/m2), where o is a positive constant, while over the lower half it is -o.
 (a) Find the potential distribution along the z axis. (b) Determine E along the z axis. (c) In the limit where z l, show that becomes that of a dipole at the origin. What is the equivalent dipole moment?
4.5.4*A uniform line charge of density l and length d is distributed parallel to the y axis and centered at the point (x, y, z) = (a, 0, 0), as shown in Fig.~P4.5.4. Use the superposition integral to show that the potential (x, y, z) is
4.5.5Charge is distributed with density l = o x/l coulomb/m along the lines z = a, y = 0, respectively, between the points x = 0 and x = l, as shown in Fig.~P4.5.5. Take o as a given charge per unit length and note that l varies from zero to o over the lengths of the line charge distributions. Determine the distribution of along the z axis in the range 0 < z < a.

4.5.6Charge is distributed along the z axis such that the charge per unit length l (z) is given by
Determine and E at a position z > a on the z axis.

4.5.7*A strip of charge lying in the x - z plane between x = -b and x = b extends to in the z direction. On this strip the surface charge density is
where d > b. Show that at the location (x, y) = (d, 0), the potential is
4.5.8A pair of charge strips lying in the x - z plane and running from z = + to z = - are each of width 2d with their left and right edges, respectively, located on the z axis. The one between the z axis and (x, y) = (2d, 0) has a uniform surface charge density o, while the one between (x, y) = (-2d, 0) and the z axis has s = -o. (Note that the symmetry makes the plane x = 0 one of zero potential.) What must be the value of o if the potential at the center of the right strip, where (x, y) = (d, 0), is to be V?
4.5.9*Distributions of line charge can be approximated by piecing together uniformly charged segments. Especially if a computer is to be used to carry out the integration by summing over the fields due to the linear elements of line charge, this provides a convenient basis for calculating the electric potential for a given line distribution of charge. In the following, you determine the potential at an arbitrary observer coordinate r due to a line charge that is uniformly distributed between the points r + b and r + c, as shown in Fig.~P4.5.9a. The segment over which this charge (of line charge density l) is distributed is denoted by the vector a, as shown in the figure.

Viewed in the plane in which the position vectors a , b, and c lie, a coordinate denoting the position along the line charge is as shown in Fig.~P4.5.9b. The origin of this coordinate is at the position on the line segment collinear with a that is nearest to the observer position r.
 (a) Argue that in terms of , the base and tip of the a vector are as designated in Fig.~P4.5.9b along the axis. (b) Show that the superposition integral for the potential due to the segment of line charge at r' is where (c) Finally, show that the potential is (d) A straight segment of line charge has the uniform density o between the points (x, y, z) = (0, 0, d) and (x, y, z) = (d, d, d). Using (c), show that the potential (x, y, z) is
4.5.10*Given the charge distribution, (r ), the potential follows from (3). This expression has the disadvantage that to find E, derivatives of must be taken. Thus, it is not enough to know at one location if E is to be determined. Start with (3) and show that a superposition integral for the electric field intensity is
where ir' r is a unit vector directed from the source coordinate r' to the observer coordinate r. (Hint: Remember that when the gradient of is taken to obtain E, the derivatives are with respect to the observer coordinates with the source coordinates held fixed.) A similar derivation is given in Sec. 8.2, where an expression for the magnetic field intensity H is obtained from a superposition integral for the vector potential A.

4.5.11For a better understanding of the concepts underlying the derivation of the superposition integral for Poisson's equation, consider a hypothetical situation where a somewhat different equation is to be solved. The charge density is assumed in part to be a predetermined density s(x, y, z), and in part to be induced at a given point (x, y, z) in proportion to the potential itself at that same point. That is,

 (a) Show that the expression to be satisfied by is then not Poisson's equation but rather where s(x, y, z) now plays the role of . (b) The first step in the derivation of the superposition integral is to find the response to a point source at the origin, defined such that Because the situation is then spherically symmetric, the desired response to this point source must be a function of r only. Thus, for this response, (b) becomes Show that for r 0, a solution is and use (c) to show that A = Q/4o. (c) What is the superposition integral for ?

4.5.12*Because there is a jump in potential across a dipole layer, given by (31), there is an infinite electric field within the layer.
 (a) With n defined as the unit normal to the interface, argue that this internal electric field is (b) In deriving the continuity condition on E, (1.6.12), using (4.1.1), it was assumed that E was finite everywhere, even within the interface. With a dipole layer, this assumption cannot be made. For example, suppose that a nonuniform dipole layer s(x) is in the plane y = 0. Show that there is a jump in tangential electric field, Ex, given by

### Electroquasistatics in the Presence of Perfect Conductors

4.6.1*A charge distribution is represented by a line charge between z = c and z = b along the z axis, as shown in Fig.~P4.6.1a. Between these points, the line charge density is given by
and so it has the distribution shown in Fig.~P4.6.1b. It varies linearly from the value o where z = c to o (a - b)/(a - c) where z = b. The only other charges in the system are at infinity, where the potential is defined as being zero.

An equipotential surface for this charge distribution passes through the point z = a on the z axis. [This is the same "a" as appears in (a).] If this equipotential surface is replaced by a perfectly conducting electrode, show that the capacitance of the electrode relative to infinity is

4.6.2Charges at "infinity" are used to impose a uniform field E = Eo iz on a region of free space. In addition to the charges that produce this field, there are positive and negative charges, of magnitude q, at z = +d/2 and z = -d/2, respectively, as shown in Fig.~P4.6.2. Spherical coordinates (r, , ) are defined in the figure.
 (a) The potential, radial coordinate and charge are normalized such that Show that the normalized electric potential can be written as (b) There is an equipotential surface = 0 that encloses these two charges. Thus, if a "perfectly conducting" object having a surface taking the shape of this = 0 surface is placed in the initially uniform electric field, the result of part (a) is a solution to the boundary value problem representing the potential, and hence electric field, around the object. The following establishes the shape of the object. Use (b) to find an implicit expression for the radius r at which the surface intersects the z axis. Use a graphical solution to show that there will always be such an intersection with r > d/2. For q = 2, find this radius to two-place accuracy. (c) Make a plot of the surface = 0 in a = constant plane. One way to do this is to use a programmable calculator to evaluate given r and . It is then straightforward to pick a and iterate on r to find the location of the surface of zero potential. Make q = 2. (d) We expect E to be largest at the poles of the object. Thus, it is in these regions that we expect electrical breakdown to first occur. In terms of Eo and with q = 2, what is the electric field at the north pole of the object? (e) In terms of Eo and d, what is the total charge on the northern half of the object. [Hint: A numerical calculation is not required.]
4.6.3*For the disk of charge shown in Fig. 4.5.3, there is an equipotential surface that passes through the point z = d on the z axis and encloses the disk. Show that if this surface is replaced by a perfectly conducting electrode, the capacitance of this electrode relative to infinity is
4.6.4The purpose of this problem is to get an estimate of the capacitance of, and the fields surrounding, the two conducting spheres of radius R shown in Fig.~P4.6.4, with the centers separated by a distance h. We construct an approximate field solution for the field produced by charges Q on the two spheres, as follows:
 (a) First we place the charges at the centers of the spheres. If R h, the two equipotentials surrounding the charges at r1 R and r2 R are almost spherical. If we assume that they are spherical, what is the potential difference between the two spherical conductors? Where does the maximum field occur and how big is it? (b) We can obtain a better solution by noting that a spherical equipotential coincident with the top sphere is produced by a set of three charges. These are the charge -Q at z = -h/2 and the two charges inside the top sphere properly positioned according to (33) of appropriate magnitude and total charge +Q. Next, we replace the charge -Q by two charges, just like we did for the charge +Q. The net field is now due to four charges. Find the potential difference and capacitance for the new field configuration and compare with the previous result. Do you notice that you have obtained higher-order terms in R/h? You are in the process of obtaining a rapidly convergent series in powers of R/h.
4.6.5This is a continuation of Prob. 4.5.4. The line distribution of charge given there is the only charge in the region 0 x. However, the y - z plane is now a perfectly conducting surface, so that the electric field is normal to the plane x = 0.
 (a) Determine the potential in the half-space 0 x. (b) For the potential found in part (a), what is the equation for the equipotential surface passing through the point (x, y, z) = (a/2, 0, 0)? (c) For the remainder of this problem, assume that d = 4a. Make a sketch of this equipotential surface as it intersects the plane z = 0. In doing this, it is convenient to normalize x and y to a by defining = x/a and = y/a. A good way to make the plot is then to compute the potential using a programmable calculator. By iteration, you can quickly zero in on points of the desired potential. It is sufficient to show that in addition to the point of part (a), your curve passes through three well-defined points that suggest its being a closed surface. (d) Suppose that this closed surface having potential V is actually a metallic (perfect) conductor. Sketch the lines of electric field intensity in the region between the electrode and the ground plane. (e) The capacitance of the electrode relative to the ground plane is defined as C = q/V, where q is the total charge on the surface of the electrode having potential V. For the electrode of part (c), what is C?

### Method of Images

4.7.1*A point charge Q is located on the z axis a distance d above a perfect conductor in the plane z = 0.
 (a) Show that above the plane is (b) Show that the equation for the equipotential surface = V passing through the point z = a < d is (c) Use intuitive arguments to show that this surface encloses the point charge. In terms of a, d, and o, show that the capacitance relative to the ground plane of an electrode having the shape of this surface is
4.7.2A positive uniform line charge is along the z axis at the center of a perfectly conducting cylinder of square cross-section in the x - y plane.
 (a) Give the location and sign of the image line charges. (b) Sketch the equipotentials and E lines in the x - y plane.
4.7.3When a bird perches on a dc high-voltage power line and then flies away, it does so carrying a net charge.
 (a) Why? (b) For the purpose of measuring this net charge Q carried by the bird, we have the apparatus pictured in Fig.~P4.7.3. Flush with the ground, a strip electrode having width w and length l is mounted so that it is insulated from ground. The resistance, R, connecting the electrode to ground is small enough so that the potential of the electrode (like that of the surrounding ground) can be approximated as zero. The bird flies in the x direction at a height h above the ground with a velocity U. Thus, its position is taken as y = h and x = Ut. (c) Given that the bird has flown at an altitude sufficient to make it appear as a point charge, what is the potential distribution? (d) Determine the surface charge density on the ground plane at y = 0. (e) At a given instant, what is the net charge, q, on the electrode? (Assume that the width w is small compared to h so that in an integration over the electrode surface, the integration in the z direction is simply a multiplication by w.) (f) Sketch the time dependence of the electrode charge. (g) The current through the resistor is dq/dt. Find an expression for the voltage, v, that would be measured across the resistance, R, and sketch its time dependence.
4.7.4*Uniform line charge densities +l and -l run parallel to the z axis at x = a, y = 0 and x = b, y = 0, respectively. There are no other charges in the half-space 0 < x. The y - z plane where x = 0 is composed of finely segmented electrodes. By connecting a voltage source to each segment, the potential in the x = 0 plane can be made whatever we want. Show that the potential distribution you would impose on these electrodes to insure that there is no normal component of E in the x = 0 plane, Ex (0, y, z), is

4.7.5The two-dimensional system shown in cross-section in Fig. P4.7.5 consists of a uniform line charge at x = d, y = d that extends to infinity in the z directions. The charge per unit length in the z direction is the constant . Metal electrodes extend to infinity in the x = 0 and y = 0 planes. These electrodes are grounded so that the potential in these planes is zero.
 (a) Determine the electric potential in the region x > 0, y > 0. (b) An equipotential surface passes through the line x = a, y = a (a < d). This surface is replaced by a metal electrode having the same shape. In terms of the given constants a, d, and o, what is the capacitance per unit length in the z direction of this electrode relative to the ground planes?
4.7.6*The disk of charge shown in Fig. 4.5.3 is located at z = s rather than z = 0. The plane z = 0 consists of a perfectly conducting ground plane.
 (a) Show that for 0 < z, the electric potential along the z axis is given by (b) Show that the capacitance relative to the ground plane of an electrode having the shape of the equipotential surface passing through the point z = d < s on the z axis and enclosing the disk of charge is
4.7.7The disk of charge shown in Fig. P4.7.7 has radius R and height h above a perfectly conducting plane. It has a surface charge density s = o r/R. A perfectly conducting electrode has the shape of an equipotential surface that passes through the point z = a < h on the z axis and encloses the disk. What is the capacitance of this electrode relative to the plane z = 0?
4.7.8A straight segment of line charge has the uniform density o between the points (x, y, z) = (0, 0, d) and (x, y, z) = (d, d, d). There is a perfectly conducting material in the plane z = 0. Determine the potential for z 0. [See part (d) of Prob. 4.5.9.]

### Charge Simulation Approach to Boundary Value Problems

4.8.1For the six-segment approximation to the fields of the parallel plate capacitor in Example 4.8.1, determine the respective strip charge densities in terms of the voltage V and dimensions of the system. What is the approximate capacitance?