Problem 3.1

  Two spherical conducting shells of radii ra and rb are arranged concentrically and are charged to the

potentials and , respectively. If rb>ra, find the potential at points between the shells and at points r>rb.

By symmetry, we note that is independent of and . This simplifies the Laplacian. We know that the bounding surfaces are at constant potential and that the

region between is charge-free, so we can use .

We can then use the BC to find the constants. First, and second, . Two equations and two unknowns. Solving for b in each equation,

Solving for a,

Plugging back into b,

And finally, the potential: (I will use the first form of b, although you can use the second form)

Problem 3.5

  Expand the function

F(u) = (1-2xu+u2)-1/2

in a Taylor series up to the . Note that the coefficients are the first four Legendre polynomials Pn(x). In fact, F(u) is a generating function for all of the Legendre Polynomials:

Recall that a Taylor series expansion is

where F(n)(u0) is the nth derivative of F(u) evaluated at u0. So, for the Taylor series, we need the derivatives.

Evaluating these at u0=0, we have:

which gives F(n)(0) = [n! Pn(x)], the first four Legendre Polynomials. Indeed, a Taylor expansion with u0=0 gives

Problem 3.7

 

Obtain in cylindrical coordinates, Eq. (3-8), from the rectangular coordinates form, Eq. (3-6),

by direct substitution: and .

Start with equation (3-6).

Since the z component is going to come along for the ride, I will ignore it and tack it back on at the end. We need to consider the chain rule on the x and the y coordinates.

where xi is either x or y. We will need , , , and . The first two are straightforward

since . (In order to do this correctly, you must write as a function of the rectangular coordinates exclusively.)

In order to find , we need to solve for (and write it in terms of rectangular coordinates only).

So, . Notice on the other hand that . It is not always true that the derivative is the reciprocal of the "inverted" derivative. Therefore,

Similarly, to find , we need to solve for . This

follows similarly to , but differs by the minus sign and the trig function:

Therefore,

So, putting these together and using the property , I find that the two-dimensional Laplacian gives

Problem 3.12

  A long cylindrical conductor of radius a bearing no net charge is placed in an initially uniform electric

field . The direction of is perpendicular to the cylinder axis. (a) Find the potential at the points exterior to the cylinder. (b) Find the charge density on the cylindrical surface.

Aside on the Electric Field:

The external electric field will induce the charges inside the conducting cylinder to move until there is no E-field inside the cylinder (i.e., constant potential). This effect will produce a surface charge density on the cylinder. If there is a tangential component to the E-field, then the surface charges will move along the surface. Since everything eventually equilibrializes, the charges must have moved into an organization such that there is no tangential E-field component on the surface.

This is expressed by the BC that the tangential component is continuous across the boundary coupled with the zero E-field inside the conductor. So that the tangential component of the E-field is zero outside the conductor as well.

That the normal component of the E-field is discontinuous implies that there is a normal component outside the conductor and that it terminates on the surface charges of the conductor.

For an image of the net field after the conductor has come to equilibrium, see page 64 Figure 3-2. This figure is actually of a sphere in an E-field, but if we consider the circle to be a cross section of the cylinder, the picture still looks correct.

If we choose cylindrical coordinates with along the cylinder. Our potential, then, is independent of

z, . On the other hand, the external E-field direction, which I set as , is a little awkward to express.

So, .

The potential inside the conductor is a constant by the properties of conductors (as discussed

above). To find the potential outside (in the charge-free region) we must solve Laplace's Equation. (

) I will use cylindrical coordinates with along the cylinder. Our potential, then, is

, independent of z.

The physical restrictions are

the surface is at constant potential, , we only see the external field infinitely far away:

Since is -dependant, it must have the property that . (Notice that this is not true of polynomials.)

So, Laplace's equation in cylindrical coordinates and independent of z is

We assume separability, which means that I can write . The potential may not actually be separable, but we assume it for convenience. The uniqueness theorem guarantees that if we assume it and can find a solution, then it must be The Solution. If we cannot find a solution, then we can't assume separability. So,

Each side must equal a constant since I can vary and independently without losing the inequality. Let's call the constant N2. The right-hand side (RHS) simplifies easily:

The solutions to this are trig functions: . We recall

the physics restriction that . Since

is only equal to if is times an integer, there is a restriction on our original

constant: N must be an integer. This gives the -dependence.

The LHS gives

If N=0, this is as we have seen in class with solution . If ,

then the solution is polynomial. Let , then find the characteristic equation:

which implies that N=n (and must be the same when we combine the and solutions). We put this all together, and notice that when N is negative, we can use the properties that the cosine is even and the sine is odd to simply modify the coefficients. We therefore can restrict ourselves to non-negative integers and absorb the negative into the coefficients Ac and As.

The general solution is

We have two plus four times an infinite number of unknowns! That's quite a few. Let's see if we can get rid of a few with the Boundary conditions.

The first BC is that

Since in this limit, , we can say nothing about the primed coefficients. However, all

and all Bn = 0. Furthermore, A0 is undetermined and A1=-E0. Finally, since

, A'0=0. This simplifies our solution to

Now, we only have one plus two times an infinite number of unknowns! That's not so bad. Let's see if we can get rid of a few more.

The second BC is that the surface is at constant potential, . Our solution yields

This must be true for any value of . That can only happen if B'n = 0, , and a-1 A'1 =

E0a. Again, this is a significant simplification. In fact, it leaves us with . This simplifies our solution to

newline Now we need to check our math by considering various limits.

which implies newline Finally, we can find the surface charge density from the BC on the electric field:

Problem 3.14

  A point charge q is situated a distance d from a grounded conducting plane of infinite extent. Obtain the total charge induced on the plane by direct integration of the surface charge density.

Recall that there are two boundary conditions:

Since the first equation relates the surface charge density to the electric field,

Since the tangential component of the E-field is continuous across the boundary and the E-field is zero inside the conductor (i.e., on the other side) the electric field must be normal to the conducting surface (i.e., must have no tangential component) on the charge side. That is to say, the conducting surface affects the net E-field; it is not merely a radial field due to the point charge.

We can use the method of images to find the E-field at the surface of the conductor. I will imagine a charge -q a distance d on the other side of the conductor. This will produce an E-field normal to the plane which is half-way between the two charges. The potential is fixed up to an arbitrary constant, which I set when I decide to call the plane zero potential.

Set the origin at the plane. Using cylindrical coordinates, I will place the charges so that q is at z=d and -q is at z=-d. The net electric field then is

Therefore, finding the total charge is straightforward; note that is the outward normal ( ):

Let and .

This is exactly the size of the image charge! Consider the possibility of this being a coincidence and ask during office hours.

Problem 3.16

 

Two grounded conducting planes intersect at and a point charge q lies between them. Determine the positions of the image charges that will give the electric field between the planes.

See the Aside on Problem 12 for some comments about the electric field on a conductor.

When a charge is brought near a conducting surface, a surface charge is induced until the net electric field near the conductor is normal to the surface. By imagining an image charge on the other side of the conducting surface, on can find the field due to the charges and more easily calculate the field which actually lies outside the conductor.

Notice that this method will make it appear as though there is an electric field inside the conductor

as well. Although we know that there is no E-field inside the conductor. So, while we are doing these problems, we must bear in mind that the solutions that we find only allow us to find the E-field in the specified region. We have to rework the problem to find the E-field in a different region.

So, before we work this problem, let's notice that when you have a charge q next to a conducting plane (say a distance d), then an image charge -q placed a distance d on the other side will produce a field which is everywhere normal to the plane where we originally had the conducting surface. When you have multiple surfaces, you have to consider one at a time and you have to be aware that each image charge will affect not only the surface across which it mirrors, but also the other surfaces. (If that doesn't make sense, continue reading the solution.)

I am going to use cylindrical coordinates with z along the axis of intersection of the planes. So, the

charge is a distance from the axis. I will further set along one of the planes and

as the other plane. The charge is at which is less than . Notice that the problem does not specify the location of the charge between the planes.

So consider the plane first. In order to set the field right for it, we need an image charge

at and at . In fact, it turns out that in this problem, all

image charges will have . Notice Table 3.1 lists the value for the two extreme limits of q moving between the conducting planes and the value for q half-way between the planes at

.

Consider the plane next (ignore the plane). In order to set the field right for it, we need to

provide an image charge to balance the original charge q as well as

(so q3 = +q) to balance the other image charge q1. Since the plane is at and the distance

of the charge from the plane can be written as an angle , the image charge needs to

be that same angle beyond the plane: . Relative to the

plane, the first image charge q1 is (because it is from the plane). So, the q3

image needs to be this same angle beyond the -plane . Again refer to Table 3.1.

Our two new charges now need to be balanced for the plane. So we ignore the plane and

consider the extra unbalanced image charges q2 and q3. The next image charge will

balance charge q2. The location of q2 relative to the plane can be written

. To balance this, I place q4 at .

Further, we will need to balance q3. The location of q3 relative to the plane can

be written . To balance this, I place q5 at

.

Finally, we must verify that q4 and q5 are balanced on the plane. In fact, across , q4

balances q5, q3 balances q1, and q2 balances q. Across , q1 balances q, q2 balances q4, and q3 balances q5. Everything is balanced and we are done.

  Table: Distribution of image charges. Notice the limits. As the original charge approaches the

conducting surface, an equal and opposite charge will be locally induced on the conducting surface. When very close, this will produce a small dipole field. When in contact, this will negate any E-field.

In the limit, there are always a q and a -q which overlap. This is also true for the

limit. In the half-way between case, the images are symmetrically distributed about the circle.

image charge angle

original q 0

1 q1 = -q 0

2 q2 = -q

3 q3 = +q

4 q4 = +q

5 q5 = -q

Problem 2.8

 

Show that the equipotential surface of the preceeding problem is spherical in shape. What are the coordinates of the center of this sphere?

Since the electric field points towards decreasing electrical potential, the positive charge must be at high potential, and the negative charge must be at low potential. The point

x=1.18 a has with and , so the potential

must have a local minimum in the -direction at x=1.18a (although it may be a saddle-point if it is

not in minimum in the -direction). Very far away ( ) the charges look like a single charge

of , so overall, the E-field points inwards (radially) which implies that is spherical far

away. Since we want the equipotential, let's consider :

This is zero if

We collect terms and look for the equation of a circle (since that is what we are asked to look for)

This is the equation of a circle centered about .

Problem 2.12

  A conducting object has an arbitrarily shaped hollow cavity in its interior. If a point charge q is introduced into the cavity, prove that the charge -q is induced on the surface of the cavity. [ Hint: Use Gauss' Law.]

Gauss' Law says: We know that there is no E-field inside a perfect conductor (or the internal charges would move). We know that immediately surrounding a charge there is an E-field. Since there is an E-field in the cavity but not in the conductor, something must occur near the interior boundary of the conductor. Let us set a "Gaussian surface" at the boundary. (Recall the example, from class about the "Gaussian pillbox" at the surface of the conductor.) Consider a surface just inside the cavity: Qenc = q. Consider a surface just inside the

conductor. Since , from Gauss' Law, Qenc = 0. Therefore there must be Qsurface such that q + Qsurface = 0, which implies that Qsurface = -q.

Problem 2.15

  A spherical charge distribution has a volume charge density that is a function only of r, the distance

from the center of the charge distribution. In other words, . If is as given below, determine the electric field as a function of r. Integrate the result to obtain an expression for

the electrostatic potential , subject to the restriction that . Let A and be

scalar constants. (Note that has units of a volume charge density, Coulombs-per-meter3, but A has units of a surface charge density, Coulombs-per-meter2.)

To calculate the E-field in general, we can use

but can be awfully complicated (as you may see in graduate school). This is the hard way to solve the problem - See below for the easy way. This denominator has the following forms in recatangular, cylindrical, and spherical coordinates.

And then you must integrate this with respect to the primed coordinates. It is crucial to simplify this denominator if at all possible or it will easily lead to something which cannot be integrated analytically and which is a pain in the rear to integrate numerically. To simplify this, use the fact that

is a predetermined given vector. Set your coordinate axis . The advantage to this is that

(using ) in rectangular coordinates x = y = 0 and z=r0,

in cylindrical coordinates and z=r0, and

in spherical coordinates .

Furthermore, we find:

These are simpler, but spherical coordinates still has a to be integrated over. We can simplify this further, but I'll save that for a different example. In cylindrical coordinates, the integrand is the simplest but the limits of integration become awkward:

In this case, since the limits of integration for involve z', we must integrate first and z' second. In spherical coordinates, the limits of integration are easy but the integrand is hard:

  In general, if you can use Gauss' Law to find the E-field, do so. It is significantly easier.

 

a)

b)

a)

Find the E-field from the charge density. Since , there is spherical

symmetry. That is to say, as well. Furthermore, . For r>R, create a Gaussian sphere with radius r. Since all of the charge is enclosed, Gauss' Law says:

This has the correct dimensions and is the correct functional form for r>R. Now consider r<R. For r<R, again create a Gaussian sphere with radius r. Since not all of the charge is enclosed we must consider the fraction enclosed. Gauss' Law says:

This has the correct dimensions and is the correct functional form for r>R. So, the electric field is

It may seem peculiar for the electric field to remain constant inside the sphere, but you'll notice that E falls off as r-2 outside the sphere as expected and that the E-field at the boundary is continuous. We expect the E-field to fall off as r-2, as you get out further and further, and we might expect the E-field to grow as we increase r inside the charge density (as it does in part (b)), but in this particular case, the amount of charge we are enclosing is only just enough to compensate for increasing the size of the Gaussian sphere. The corresponding potential must be done for r>R separately from r<R: For r>R,

For r<R,

So, the electric potential is

This is consistent with . Notice also that this is continuous at the point r=R.

b)

Find the E-field from the charge density. Since , there is spherical

symmetry. That is to say, as well. Furthermore, . For r>R, create a Gaussian sphere with radius r. Since all of the charge is enclosed, Gauss' Law says:

This has the correct dimensions and is the correct functional form for r>R. For r<R, again create a Gaussian sphere with radius r. Since not all of the charge is enclosed we must consider the fraction enclosed. Gauss' Law says:

This has the correct dimensions and is the correct functional form for r<R. So, the electric field is

Notice that, since , we can write the r<R version as

The corresponding potential must be done for r>R separately from r<R: For r>R,

(as expected.) For r<R,

So, the electric potential is

This is consistent with . Notice also that this is continuous at the point r=R.

Problem 2.19

  The screened Coulomb potential

is appropriate for a charge q in a semiconducting medium. Calculate the corresponding electric field and charge density.

To find the electric field from the potential, we simply differentiate with a gradiant:

To find the charge density from the electric field, we simply use Gauss' law

and differentiate with a divergence:

Problem 2.24

  Three charges are arranged in a linear array. The charge -2q is placed at the origin, and two

charges, each of +q are placed at and , respectively. (a) Find a relatively

simple expression for the potential which is valid for distances . (b) Make a plot of the equipotential surfaces in the x-z plane.

This charge distribution is a double-dipole, generally known as the linear quadrupole. This is a multipole expansion. First write down the potential (Eqn 2-15 with N=3):

where I have used . Then I expand using the binomial expansion

which is valid for |x|<1. I'll define and observe that

implies that terms of order and higher are negligible.

The important observations to make are that

the result involved all terms of order , some of these came from the x term and some came from the x2 term,

the non-l/r terms cancelled out - these are the monopole terms - this is not a monopole,

the order terms cancelled out - these were the dipole terms - this is not a dipole the terms that were left are the quadrupole terms - this is primarily a quadrupole.

Problem 2.3

 

Point charges of are situated at each of three corners of a square whose side is

. Find the magnitude and direction of the electric field at the vacant corner point of the square.

There is enough symmetry in the problem that it doesn't matter which corner is vacant; but we must choose one. For convenience, I am going to choose a coordinate system that lies along the diagonals of the square. Note that the diagonal of the square is

So, the charges and the point of interest are at

Since (it should be obvious that) the -components of field due to charges q1 and q2 will cancel, I will not bother to calculate them.

The difference with the book is rounding.

Problem 2.4

 

An infinitely lone line charge has uniform charge density . Using direct integration, find the electric field at a distance r from the line.

Let us do a harder problem and consider a finite rod of length L and place the origin somewhere between the endpoints. There is cylindrical symmetry. Consider the E-field at a point

from the origin in the - plane. (i.e., Set the origin such that z0 = 0.) We

can consider an infinitesimal charge at the point z along the rod and integrate (add-up) this affect across the rod (from z=-l1 to z=l2, where L=l1+l2). [SEE CLASS FOR SET-UP.]

Use a trig-substitution, such that . Notice that the

definition implies that there is a triangle with one leg z and the other leg . This

implies that and .

Notice the following: 1.

The -term only goes away half-way between the end-points (l1 = l2). 2.

The -term does not go away even half-way between the endpoints.

3. In the limit as l1 and l2 go to infinity, this is the infinite rod case.

Aside: Finding the E-field around a finite rod

Remember that defines the point of interest. Once we have the E-field in these terms, we can then consider these to be the variables which tell me where the E-field is being found, even though we considered them to be constants during this calculation. The z-dependence of the E-field comes in through the l1 and the l2. We chose to set the origin such that z0=0. If we want a general form for the E-field surrounding a finite rod, we need to be more general. To be general, we need to convert the l1 and the l2 back into terms of L and z0. This is managable, but the explicit final answer will look different for different choices of origin. It will have the form of the E-field found above in terms of l1 and l2, but it should be written in terms of L and z0. It is this conversion that is dependant on the choice of origin. I can explain this better in person

Problem 2.7

 

(a) Two point charges, -q and , are situated at the origin and at the point (a, 0, 0) respectively.

At what point along the -axis does the electric field vanish? (b) In the - plane, make a plot of the equipotential surfacewhich goes through the point just referenced in the previous part. Is this point a true minimum (global as opposed to local) in the potential?

I will leave part (b) for you to answer. We can ignore the component in this case. Since the charges are opposite sign, the field cannot be zero in the region between them. Both the + charge

at the point x=a (I assume a>0) and the - charge at the origin create an E-field in the direction. So, the zero point must either be x<0 or x>a. In the region x<0, we have a large charge close by and a small charge further away -- these cannot cancel out. So, the zero point must be x>a, where we have a close small charge and a large far charge. So, consider the net E-field.

This is zero when

Solving for x:

The minus solution gives x = .317 a < a, which can't be true. (This gives a false answer because the math does not see that the field due to the positive charge has switched sign.) The plus solution

gives (x >a), which is consistent with our expectations.