^* An asterisk on a problem number designates a "show that" problem. These problems are especially designed for self study.

The Lorentz Law in Free Space^*

1.1.1^*

Assuming in Example 1.1.1 that v_i = 0 and that E_x < 0, show that by the time the electron has reached the position x = h, its velocity is -2eE_xh/m. In an electric field of only E_x = 1v/cm = 10^-2 v/m, show that by the time it reaches h = 10^-2 m, the electron has reached a velocity of 5.9 x 10³ m/s.

1.1.2

An electron moves in vacuum under the same conditions as in Example 1.1.1 except that the electric field takes the form E = E_x i_x + E_y i_y where E_x and E_y are given constants. When t = 0, the electron is at _x = 0 and _y = 0 and the velocity d_x /dt = v_i and d_y /dt = 0.

(a) Determine _x (t) and _y (t).

(b) For E_x > 0, when and where does the electron return to the plane x = 0?

1.1.3^*

An electron, having velocity v = v_i i_z, experiences the field H = H_o i_y and E = E_o i_x, where H_o and E_o are constants. Show that the electron retains this velocity if E_o = v_i _o H_o.

1.1.4

An electron has the initial position x = 0, y = 0, z = z_o. It has an initial velocity v = v_o i_x and moves in the uniform and constant fields E = E_o i_y, H = H_o i_y.

direction, _y(t).

(a) Determine the position of the electron in the y direction, _y(t).

(b) Describe the trajectory of the electron.

Charge and Current Densities

1.2.1^*

The charge density is _o r/R coulomb/m³ throughout the volume of a spherical region having radius R, with _o a constant and r the distance from the center of the region (the radial coordinate in spherical coordinates). Show that the total charge associated with this charge density is q = _o R³ coulomb.

1.2.2

In terms of given constants _o and a, the net charge density is = (_o /a²) (x² + y² + z²) coulomb/m³. What is the total charge q (coulomb) in the cubical region -a < x < a, -a < y < a, -a < z < a?

1.2.3^*

With J_o and a given constants, the current density is J = (J_o/a²)(y² + z²) [i_x + i_y + i_z ]. Show that the total current i passing through the surface x = 0, -a < y < a, -a < z < a is i = 8J_o a²/3 amp.

1.2.4

In cylindrical coordinates (r, , z) the current density is given in terms of constants J_o and a by J = J_o (r/a)² i_z (amp/m²). What is the net current i (amp) through the surface z = 0, r < a?

1.2.5^*

In cylindrical coordinates, the electric field in the annular region b < r < a is E = i_r E_o (b/r), where E_o is a given negative constant. When t = 0, an electron having mass m and charge q = -e has no velocity and is positioned at r = _r = b.

(a) Show that, in vacuum, the radial motion of the electron is governed by the differential equation mdv_r/dt =-eE_ob/_r, where v_r = d_r/dt. Note that these expressions combine to provide one second-order differential equation governing _r.

(b) By way of providing one integration of this equation, multiply the first of the first-order expressions by v_r and (with the help of the second first-order expression) show that the resulting equation can be written as d[ $fraction GIF #1$ mv_r² + eE_o b ln _r]/dt = 0. That is, the sum of the kinetic and potential energies (the quantity in brackets) remains constant.

(c) Use the result of (b) to find the electron velocity v_r(r).

(d) Assume that this is one of many electrons that flow radially outward from the cathode at r = b to r = a and that the number of electrons passing radially outward at any location r is independent of time. The system is in the steady state so that the net current flowing outward through a surface of radius r and length l, i = 2 rl J_r, is the same at one radius r as at another. Use this fact to determine the charge density (r).

Gauss' Integral Law

1.3.1^*

Consider how Gauss' integral law, (1), is evaluated for a surface that is not naturally symmetric. The charge distribution is the uniform line charge of Fig. 1.3.7 and hence E is given by (13). However, the surface integral on the left in (1) is to be evaluated using a surface that has unit length in the z direction and a square cross-section centered on the z axis. That is, the surface is composed of the planes z = 0, z = 1, x = a, and y = a. Thus, we know from evaluation of the right-hand side of (1) that evaluation of the surface integral on the left should give the line charge density _l.

(a) Show that the area elements da on these respective surfaces are i_z dxdy, i_x dydz, and i_y dxdz.

(b) Starting with (13), show that in Cartesian coordinates, E is equation GIF #1.77 (Standard Cartesian and cylindrical coordinates are defined in Table I at the end of the text.)

(c) Show that integration of _o E da over the part of the surface at x = a leads to the integral equation GIF #1.78

(d) Finally, show that integration over the entire closed surface indeed gives _l.

1.3.2

Using the spherical symmetry and a spherical surface, the electric field associated with the point charge q of Fig. 1.3.6 is found to be given by (12). Evaluation of the left-hand side of (1) over any other surface that encloses the point charge must also give q. Suppose that the closed surface S is composed of a hemisphere of radius a in the upper half-plane, a hemisphere of radius b in the lower half-plane, and a washer-shaped flat surface that joins the two. In spherical coordinates (defined in Table I), these three parts of the closed surface S are defined by (r = a, 0 < < $fraction GIF #2$ , 0 < 2 ), (r = b, $fraction GIF #3$ < < , 0 < 2 ), and ( = $fraction GIF #4$ , b r \leq a, 0 < 2). For this surface, use (12) to evaluate the left-hand side of (1) and show that it results in q.

1.3.3^*

A cylindrically symmetric charge configuration extends to infinity in the z directions and has the same cross-section in any constant z plane. Inside the radius b, the charge density has a parabolic dependence on radius while over the range b < r < a outside that radius, the charge density is zero.

There is no surface charge density at r = b.

(a) Use the axial symmetry and Gauss' integral law to show that E in the two regions is equation GIF #1.80

(b) Outside a shell at r = a, E = 0. Use (17) to show that the surface charge density at r = a is equation GIF #1.81

(c) Integrate this charge per unit area over the surface of the shell and show that the resulting charge per unit length on the shell is the negative of the charge per unit length inside.

(d) Show that, in Cartesian coordinates, E is equation GIF #1.82 Note that (r = x² + y², cos = x/r, sin = y/r, i_r = i_x cos + i_y sin ) and the result takes the form E = E_x (x, y) i_x + E_y (x, y) i_y.

(e) Now, imagine that the circular cylinder of charge in the region r < b is enclosed by a cylindrical surface of square cross-section with the z coordinate as its axis and unit length in the z direction. The walls of this surface are at x = c, y = c and z = 0 and z = 1. (To be sure that the cylinder of the charge distribution is entirely within the surface, b < r < a, b < c < a/2.) Show that the surface integral on the left in (1) is equation GIF #1.83 Without carrying out these integrations, what is the answer?

1.3.4

In a spherically symmetric configuration, the region r < b has the uniform charge density _b and is surrounded by a region b < r < a having the uniform charge density _a. At r = b there is no surface charge density, while at r = a there is that surface charge density that assures E = 0 for a < r.

(a)	Determine E in the two regions.
(b)	What is the surface charge density at r = a?
(c)	Now suppose that there is a surface charge density given at r = b of _s = _o. Determine E in the two regions and _s at r = a.

1.3.5^*

The region between the plane parallel sheets of surface charge density shown in Fig. 1.3.8 is filled with a charge density = 2_o z/s, where _o is a given constant. Again, assume that the electric field below the lower sheet is E_o i_z and show that between the sheets equation GIF #1.84

1.3.6

In a configuration much like that of Fig. 1.3.8, there are three rather than two sheets of charge. One, in the plane z = 0, has the given surface charge density _o. The second and third, respectively located at z = s/2 and z = -s/2, have unknown charge densities _a and _b. The electric field outside the region - $fraction GIF #5$ s < z < s is zero, and _a = 2_b. Determine _a and _b.

1.3.7

Particles having charges of the same sign are constrained in their positions by a plastic tube which is tilted with respect to the horizontal by the angle , as shown in Fig. P1.3.7. Given that the lower particle has charge Q_o and is fixed, while the upper one (which has charge Q and mass M) is free to move without friction, at what relative position, , can the upper particle be in a state of static equilibrium? figure GIF #1

Ampère's Integral Law

1.4.1^*

A static H field is produced by the cylindrically symmetric current density distribution J = J_o \exp (-r/a) i_z, where J_o and a are constants and r is the radial cylindrical coordinate. Use the integral form of Ampère's law to show that equation GIF #1.85

1.4.2^*

In polar coordinates, a uniform current density J_o i_z exists over the cross-section of a wire having radius b. This current is returned in the -z direction as a uniform surface current at the radius r = a > b.

(a)	Show that the surface current density at r = a is
(b)	Use the integral form of Ampère's law to show that H in the regions 0 < r < b and b < r < a is
(c)	Use Ampère's continuity condition, (16), to show that H = 0 for r > a.
(d)	Show that in Cartesian coordinates, H is
(e)	Suppose that the inner cylinder is now enclosed by a contour C that encloses a square surface in a constant z plane with edges at x = c and y = c (so that C is in the region b < r < a, b < c < a/2). Show that the contour integral on the left in (1) is Without carrying out the integrations, use Ampère's integral law to deduce the result of evaluating (d).

1.4.3

In a configuration having axial symmetry about the z axis and extending to infinity in the z directions, a line current I flows in the -z direction along the z axis. This current is returned uniformly in the +z direction in the region b < r < a. There is no current density in the region 0 < r < b and there are no surface current densities.

(a)	In terms of I, what is the current density in the region b < r < a?
(b)	Use the symmetry of the configuration and the integral form of Ampère's law to deduce H in the regions 0 < r < b and b < r < a.
(c)	Express H in each region in Cartesian coordinates.
(d)	Now, consider the evaluation of the left-hand side of (1) for a contour C that encloses a square surface S having sides of length 2c and the z axis as a perpendicular. That is, C lies in a constant z plane and has sides x = c and y = c with c < a/2). In Cartesian coordinates, set up the line integral on the left in (1). Without carrying out the integrations, what must the answer be?

1.4.4^*

In a configuration having axial symmetry about the z axis, a line current I flows in the -z direction along the z axis. This current is returned at the radii a and b, where there are uniform surface current densities K_za and K_zb, respectively. The current density is zero in the regions 0 < r < b, b < r < a and a < r.

(a)	Given that za = 2K_zb, show that K_za = I/ (2a+b).
(b)	Show that H is

1.4.5

Uniform surface current densities K = K_o i_y are in the planes z = $fraction GIF #6$ s, respectively. In the region - s < z < $fraction GIF #7$ s, the current density is J = 2J_o z/si_y. In the region z < - $fraction GIF #8$ s, H = 0. Determine H for - $fraction GIF #9$ s < z.

Charge Conservation in Integral Form

1.5.1^*

In the region of space of interest, the charge density is uniform and a given function of time, = _o (t). Given that the system has spherical symmetry, with r the distance from the center of symmetry, use the integral form of the law of charge conservation to show that the current density is equation GIF #1.91

1.5.2

In the region x > 0, the charge density is known to be uniform and the given function of time = _o (t). In the plane x = 0, the current density is zero. Given that it is x directed and only dependent on x and t, what is J?

1.5.3^*

In the region z > 0, the current density J = 0. In the region z < 0, J = J_o(x, y) cos t i_z, where J_o is a given function of (x, y). Given that when t = 0, the surface charge density _s = 0 over the plane z = 0, show that for t > 0, the surface charge density in the plane z = 0 is _s (x, y, t) = [J_o (x, y)/ ] sin t.

1.5.4

In cylindrical coordinates, the current density J = 0 for r < R, and J = J_o ( , z) sin ti_r for r > R. The surface charge density on the surface at r = R is _s ( , z, t) = 0 when t = 0. What is _s ( , z, t) for t > 0?

Faraday's Integral Law

1.6.1^*

Consider the calculation of the circulation of E, the left-hand side of (1), around a contour consisting of three segments enclosing a surface lying in the x - y plane: from (x, y) = (0, 0) (g, s) along the line y = sx/g; from (x, y) = (g, s) (0, s) along y = s and from (x, y) = (0, s) to (0, 0) along x = 0.

(a)	Show that along the first leg, ds = [i_x + (s/g)i_y]dx.
(b)	Given that E = E_o i_y where E_o is a given constant, show that the line integral along the first leg is sE_o and that the circulation around the closed contour is zero.

1.6.2

The situation is the same as in Prob. 1.6.1 except that the first segment of the closed contour is along the curve y = s(x/g)².

(a)	Once again, show that for a uniform field E = E_o i_y, the circulation of E is zero.
(b)	For E = E_o (x/g)i_y, what is the circulation around this contour?

1.6.3^*

The E field of a line charge density uniformly distributed along the z axis is given in cylindrical coordinates by (1.3.13).

(a)	Show that in Cartesian coordinates, with x = r cos and y = r sin ,
(b)	For the contour shown in Fig. P1.6.3, show that and complete the integrations to prove that the circulation is zero.

1.6.4

A closed contour consisting of six segments is shown in Fig. P1.6.4. For the electric field intensity of Prob. 1.6.3, calculate the line integral of E ds on each of these segments and show that the integral around the closed contour is zero.

1.6.5^*

The experiment in Fig. 1.6.4 is carried out with the coil positioned horizontally, as shown in Fig. 1.7.2. The left edge of the coil is directly below the wire, at a distance d, while the right edge is at the radial distance R from the wire, as shown. The area element da is y directed (the vertical direction).

(a) Show that, in Cartesian coordinates, the magnetic field intensity due to the current i is equation GIF #1.94

(b) Use this field to show that the magnetic flux linking the coil is as given by (1.7.5).

(c) What is the circulation of E around the contour representing the coil?

(d) Given that the coil has N turns, what is the EMF measured at its terminals?

1.6.6

The magnetic field intensity is given to be H = H_o(t) (i_x + i_y ), where H_o(t) is a given function of time. What is the circulation of E around the contour shown in Fig. P1.6.6? figure GIF #2

1.6.7^*

In the plane y = 0, there is a uniform surface charge density _s = _o. In the region y < 0, E = E₁ i_x + E₂ i_y where E₁ and E₂ are given constants. Use the continuity conditions of Gauss and Faraday, (1.3.17) and (12), to show that just above the plane y = 0, where y = 0⁺, the electric field intensity is E = E₁ i_x + [E₂ + (_o /_o)]i_y.

1.6.8

Inside a circular cylindrical region having radius r = R, the electric field intensity is E = E_o i_y, where E_o is a given constant. There is a surface charge density _o cos on the surface at r = R (the polar coordinate is measured relative to the x axis). What is E just outside the surface, where r = R⁺?

Integral Magnetic Flux Continuity Law

1.7.1^*

A region is filled by a uniform magnetic field intensity H_o(t)i_z.

(a) Show that in spherical coordinates (defined in Fig. A.1.3 of Appendix 1), H = H_o (t) (i_r cos - i sin ).

(b) A circular contour lies in the z = 0 plane and is at r = R. Using the enclosed surface in the plane z = 0 as the surface S, show that the circulation of E in the direction around C is - R²_o d H_o/dt.

(c) Now compute the same circulation using as a surface S enclosed by C the hemispherical surface at r = R, 0 < $fraction GIF #10$ .

1.7.2

With H_o(t) a given function of time and d a given constant, three distributions of H are proposed.

Which one of these will not satisfy (1) for a surface S as shown in Fig. 1.5.3?

1.7.3^*

In the plane y = 0, there is a given surface current density K = K_o i_x. In the region y < 0, H = H₁ i_y + H₂ i_z. Use the continuity conditions of (1.4.16) and (6) to show that just above the current sheet, where y = 0⁺, H = (H₁ - K_o)i_y + H_z i_z.

1.7.4

In the circular cylindrical surface r = R, there is a surface current density K = K_o i_z. Just inside this surface, where r = R, H = H₁ i_r. What is H just outside the surface, where r = R⁺?