Irrotational Field Represented by Scalar Potential: The Gradient Operator and Gradient Integral Theorem
|4.1.1||Surfaces 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
|4.1.2||For 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).
|4.1.3||In Cartesian coordinates, three two-dimensional potential functions
|(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
|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
|(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
|(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
|(i)|| Show that the current, i(t), to this electrode segment
|4.1.5||Inside 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.
|4.1.6||Given 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?
|4.1.7||The 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
|4.1.8||Figure 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
|(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
|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.
|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.
|4.2.2||In Prob. 4.1.5, is given by (a). Use Poisson's
equation to find the charge density.
|4.2.3||Use 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.
|4.2.4||Use 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.
|4.3.1||A 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) +
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.
|4.4.3||A 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
|4.4.4||The charge distribution described in Prob. 4.4.3 is now at infinity
|(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
Solution of Poisoon's Equation for Specified Charge Distributions
|4.5.1||The 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
|(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.]
|4.5.3||All 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
|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
|4.5.5||Charge is distributed with density l = o
x/l coulomb/m along the lines z = a, y = 0, respectively,
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.
|4.5.6||Charge 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
|4.5.8||A 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
|(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
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.
|4.5.11||For 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,
|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
|(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"
in (a).] If this equipotential surface is replaced by a perfectly
conducting electrode, show that the capacitance of the electrode
relative to infinity is
|4.6.2||Charges 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.
|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
|4.6.4||The 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:
|4.6.5||This 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
|(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
|(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
|4.7.2||A 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
|4.7.3||When a bird perches on a dc high-voltage power line and then
flies away, it does so carrying a net charge.
|(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
|4.7.5||The 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
|(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
|4.7.7||The 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?
|4.7.8||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). There is a perfectly conducting material in the plane z = 0.
Determine the potential for z 0. [See part (d) of Prob.
Charge Simulation Approach to Boundary Value Problems
|4.8.1||For 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?