1. Example 2.4 shows that a uniform plane of charge, with charge density , has electric field

   

   where points normal to, and away from the plane. Thus, for a horizontal plane, points up (in the direction) above the plane, and down (in the direction) below the plane.

   We check that this obeys the electrostatic boundary condition

   

   where is the normal vector above the plane.

   Here

   

   obeying the boundary condition.

   

   Here, from example 2.5, the field is zero everywhere but between the plates, where it is .

   Now we check our electrostatic boundary condition across the two charged planes. For the bottom plane at z=0,

   

   for the bottom plane has charge density and . For the top plane at z=d,

   

   for the top plane has charge density , with again equal to .

2. 
   

   

   1. We find the surface charge densities , , and by noting that inside the conductor. Thus the interior surface a must totally screen inside (by carrying total surface charge ). This surface is spherically symmetric about , so we expect a uniform charge distribution:

      

      The exact same argument holds for surface b:

      

      The conducting sphere is uncharged, so the migration of charges to the surfaces a and b is accompanied by the migration of compensating charge to the outer surface of the conductor at R. Since everywhere inside the conductor, this charge distributes itself uniformly over the surface R:

      

   2. Outside the conductor, the field is spherically symmetric due to the only uncanceled charge distribution, the spherically symmetric one at r= R. We can find it by Gauss' law

      

      Here points radially outward, perpendicular to a spherical Gaussian surface at r, so = for surface area element da on the spherical Gaussian surface. also depends only on r, remaining constant on the fixed r Gaussian surface, and so factors out of the flux integral:

      

      The total charge enclosed by the Gaussian spherical surface, outside the conductor, is . Thus

      

      the same field induced by a charge at the center of the sphere.

   3. We find the field within each cavity by Gauss' law, for a spherical Gaussian surface at r within the cavity. As before, the spherical symmetry of the charge distribution (here a point charge at the origin) means that E is radial and constant on a spherical Gaussian surface centered on the charge, so

      

      Applied to cavities a and b, then,

      

   4. and each experience no force. Let's consider : 4 charge distributions exert forces on : at a, at b, at , and at R. The force each would each exert on is , where is the electric field each distribution generates at .

      The forces due to at a and at R are both zero by Gauss' law. Calculating an electric field anywhere inside a charged spherical shell always gives vanishing electric field, as no charge is enclosed by the spherical Gaussian surface inside the shell.

      The forces due to at b and at cancel. See this by considering as a Gaussian surface a spherical shell centered on with radius greater than b. The charge enclosed is exactly zero, at r=0 plus on the shell at r=b. Thus this configuration also creates zero electric field anywhere outside the sphere centered on ; and in particular, at .

      A parallel argument leads to zero force on .

3. 
   

   

   

   Setting this equal to by Gauss' law gives

   

   Of course, for s ;SPMlt;a or s;SPMgt;b, vanishes, as a cylindrical Gaussian surface encloses zero net charge (for s;SPMgt;b, because +Q on the cylindrical shell at s=a is canceled by -Q on the cylindrical shell at s=b).

   To find the capacitance of this coaxial capacitor we need the potential difference between the +Q and -Q conductors at s=a and s=b. This is

   

   the unusual sign because we are looking for minus the usual potential difference V(b) - V(a), since by convention we need V(+Q) - V(-Q). This gives

   

   since is radial. Plugging in our result for between a and b,

   

   The capacitance is defined as

   

   The capacitance per unit length is thus

   

4. 
   

   

   

   which holds because the slot contains no charges.

   Here,

   

   As discussed in Example 3.3, separation of variables leads to eigenfunction solutions

   

   the choice of exponentials for x and sinusoids for y being dictated by boundary condition (4-4), which cannot be satisfied by a linear combination of sinusoids. Boundary condition (4-4) requires A=0, and boundary condition (4-1) requires D=0. Boundary condition (4-2) requires that

   

   We thus have eigenfunctions

   

   satisfying boundary conditions (4-1), (4-2), and (4-4). The only remaining boundary condition is (4-3), which cannot be obeyed by a single eigenfunction. We thus construct the general solution as a superposition of all allowed eigenfunctions, with coefficients :

   

   We impose boundary condition (4-3) on this sum, which fixes the coefficients :

   

   As explained in example 3.3, this sum is a Fourier sine expansion for the function , whose coefficients are given by

   

   Here, plugging in

   

   gives

   

   Since cosine is -periodic, and we need only work out for n=1,2,3, 4. The term in parentheses gives

   

   so we have

   

   This gives, for our solution to the full stated problem,

   

5. We are asked to solve the boundary value problem

   

   where units of volts are assumed. As Boas mentions, it is much easier to solve two simpler problems:

   

   and

   

   The sum then solves the original boundary value problem.

   For problem 1: the 2-dimensional Laplace's equation has a basis of eigensolutions

   

   We have chosen this basis because only for sinusoidal eigensolutions X(x) can we force X(x) to vanish at two distinct points in x, as required. To make vanish at x=0, for all y, we restrict ourselves to the solutions,

   

   To make vanish at y=0, where , we must have . We then have

   

   To make vanish at x=10, we must have nonzero only when , that is, . For our last boundary condition, setting when y=30, we have

   

   Thus are the Fourier coefficients in a Fourier expansion of f = 100, 0;SPMlt;x;SPMlt;10. These are

   

   Thus our solution for is

   

   For problem 2: we take as our basis of eigensolutions for the 2-dimensional Laplace's equation

   

   again because we need sinusoidal eigensolutions Y(y) to make Y vanish at two distinct points y. To make vanish at y=0, for all x, we restrict ourselves to the solutions,

   

   To make vanish at x=0, where , we must have . We then have

   

   To make vanish at y=30, we must have nonzero only when , that is, . For our last boundary condition, setting when x=10, we have

   

   Thus are the Fourier coefficients in a Fourier expansion of f = 100, 0;SPMlt;y;SPMlt;30. These are

   

   Thus our solution for is

   

   Reassembling, our full solution is

   

6. Here we have the boundary value problem

   

   with boundary conditions

   

   As discussed in Example 3.5, separation of variables V(x,y,z) = X(x)Y(y)Z(z) leads to

   

   Here, we need to force X(0) = X(a) = 0 and Y(0) = Y(a) = 0; as noted before, the only eigensolutions we can force to zero at two distinct points are the sinusoidal ones (and not the exponential ones), so we must choose and , giving us eigensolutions

   

   where

   

   Boundary conditions (6-1) and (6-3) force us to discard the and eigensolutions. Boundary conditions (6-2) and (6-4) require that

   

   Boundary condition (6-5) implies that

   

   We thus have eigenfunctions

   

   where

   

   satisfying boundary conditions (6-1) through (6-5). The only remaining boundary condition is (6-6), which cannot be obeyed by a single eigenfunction. We thus construct the general solution as a superposition of all allowed eigenfunctions, with coefficients :

   

   We impose boundary condition (6-6) on this sum, which fixes the coefficients :

   

   As explained in example 3.5, this sum is a two-dimensional Fourier sine expansion for the function , whose coefficients are given by

   

   Here, plugging in gives

   

   This gives, for our solution to the full stated problem,

   

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