Phys 742 - EM I In-class test 1 - 29 February 1996

1. Two conducting spheres of radius a carry opposite charges.

(a) Find the capacitance C of this system when the spheres are separated by a distance b >> a (One can obtain an expansion in a/b. The first term of this expansion is enough for full credit on this problem, but you can get the next term for extra credit if time permits). Give a number (in cm and in μF) when a = 1 cm and b =10 cm.

(b) What is the force between the spheres? Obtain the answer directly and check that it is also given by a general formula involving C with the C of part (a)

2. Consider the problem of finding the potential in a charge-free region surrounded by a

surface S when (a) the potential is assigned on S

(b) the field is assigned on S

(c) both the potential and the field have assigned values on S Discuss these three cases. Can the potential be assigned arbitrarily on the boundary

and is then the solution unique? Can the field components be assigned arbitrarily on the boundary, or which components can be assigned arbitrarily, and is then the solution unique? And what about case (c)? You are not expected to prove your answers, but they must be precise answers.

3.

(a) We know that the real part of the complex potential i arcsin(z/a) represents the electrostatic potential of a charged strip. What is the charge per unit length on the strip? Sketch the field lines and the equipotentials.

(b) What electrostatic potential problem is solved by the real part of arcsin(z/a)? (If you can follow the branches of the complex arcsin you will think that the answer is obvious; in any case do the following). Find the electric field all along the x axis. From its behavior for x > a argue that you have there a conducting plate. Similarly for x < a. By integrating E_{x} from −a to a, find the potential difference between the plates.

4. A grounded conducting sphere of radius a is immersed in a uniform field E_{0}.

(a) What is charge density on the surface of the sphere?

(b) What is the induced dipole moment? What is the induced quadrupole moment?

5. Sketch the field lines and equipotentials of a dipole.

Solution

1.

(a) This is very similar to Jackson’s 1.7. Let the spheres have charges ±q. The potential on the sphere with charge +q is approximately q/a − q/b. On the other sphere, it is −q/a + q/b. Taking the difference and dividing by q,

\frac{1}{C} ≃ 2 (\frac{1}{a} − \frac{1}{b}) or C ≃ \frac{1}{2} \frac{ab}{b − a} ≃ \frac{1}{2} (a + \frac{a^{2}}{b} )

For a = 1 cm and b = 10 cm, C = (5/9) cm = 0.556 cm = 0.617 pF. (The other formula gives 0.55 cm). See also the Comments.

(b) The force is clearly F = −q^{2}/b^{2}. This is consistent with

F = −\frac{∂}{∂b} (\frac{q^{2}}{2C} )|_{q=const} = −\frac{q^{2}}{2} \frac{∂}{∂b} (\frac{1}{C} )

2.

When the potential is assigned on S, there is always a unique solution (Dirichlet problem). When the normal component of the field is assigned on S, there is always a solution that is unique up to an additive constant (Neumann problem). In part (c) the problem is overspecified and has no solution in general.

3.

(a) If F = i arcsin(z/a), the complex field is

−\frac{dF}{dz} = \frac{−i/a}{\sqrt{1 − z^{2}/a^{2}}} = \frac{−1}{\sqrt{z^{2} − a^{2}}}

The complex potential of a unit line charge is −2 ln z and the complex field is 2/z. Comparing the two fields for large z, we see that the charge per unit strip length is −1/2.

(b) Recall that \frac{dF}{dz} = −E_{x} + iE_{y}. For F = arcsin(z/a),

\frac{dF}{dz} = \frac{1}{\sqrt{a^{2} − z^{2}}}

It follows that E_{x} vanishes for |x| > a, meaning that the halfplane for x > a is at a constant potential, and the halfplane for x < −a is too. The potential difference is

V_{right} − V_{left} = − ∫_{−a/2}^{a/2} E_{x}dx = ∫_{−a/2}^{a/2} \frac{dx}{\sqrt{a^{2} − x^{2}}} = arcsin \frac{x}{a}|_{−a}^{a} = π

One can get this directly by examining Re F, which is just arcsin(x/a) for |x| < a and y = 0.

4.

The charge density is (3/4π)E_{0} cos θ, the dipole moment is E_{0}a^{3}, the quadrupole moment is zero.

Comments

1.

(a) (Alternative method) Using C_{21} = C_{12} (always true) and C_{22} = C_{11} (true in this case),

C = \frac{C_{11}C_{22} − C_{12}C_{21}}{C_{11} + C_{22} + C_{12} + C_{21}} = \frac{C_{11} − C_{12}}{2}

For b >> a we expect C_{11} = a + O(a^{2}/b^{2}). To find C_{11} and C_{12}, we must put sphere 1 at potential 1 and sphere 2 at potential 0. To leading order, the charge on sphere 1 must be q = a (so that q/a = 1) and we must put on sphere 2 an image charge q′ = −qa/b = −a^{2}/b. By definition, C_{11} = q and C_{12} = q′. This gives back the result

C = \frac{1}{2} (a + \frac{a^{2}}{b} )

The disadvantage of this method is that it is limited to spheres, because it uses explicitly the method of images. The other method is more general. Suppose that the spheres are replaced by two bodies of capacitances C_{1} and C_{2}. (C_{1} is the capacitance of body 1 all by itself and is only approximately equal to C_{11}. Similarly for C_{2}.) Then the potential on body 1 is approximately q/C_{1} − q/b and the potential on body 2 is approximately −q/C_{2} + q/b. Taking the difference and dividing by q,

\frac{1}{C} ≃ \frac{1}{C_{1}} + \frac{1}{C_{2}} − \frac{2}{b}

C ≃ \frac{C_{1}C_{2}}{C_{1} + C_{2} − 2C_{1}C_{2}/b} ≃ \frac{C_{1}C_{2}}{C_{1} + C_{2}} (1 + \frac{2}{b} \frac{C_{1}C_{2}}{C_{1} + C_{2}} )

Comparing with the general formula

C = \frac{C_{11}C_{22} − C_{12}^{2}}{C_{11} + C_{22} + 2C_{12}}

we see that, to leading order, C_{12} = −C_{1}C_{2}/b.