Institutt for fysikk, NTNU TFY4155/FY1303: Elektrisitet og magnetisme Vår 2004

Øving 6

Guidance: Monday February 16 To be delivered by: Thursday February 19

Exercise 1

The largest electric field that can be maintained in air is about 3 MV/m. Larger fields result in a discharge (socalled corona discharge). In the lectures, we have shown that a metal sphere will have all its net charge on the surface. We have also shown that the electric field at the surface is E = σ/ε_{0}, where σ is the surface charge density.

a) What is then the maximum surface charge density that a metal surface can maintain without resulting in a discharge?

b) What is the minimum radius of a metal sphere if it is supposed to maintain a charge 1C? [Correct answer is either 25 nm, 1.5 mm, 6.6 cm, 54 m or 2.3 km]

c) A typical metal consists of atoms arranged in a crystal structure, with a nearest neighbour distance of about 0.3 nm. What is then the number of surface atoms pr m^{2}? You may assume that the surface atoms are arranged in a regular quadratic lattice. [Extra: Will your answer be the same if the surface atoms are arranged in a regular triangular lattice?]

d) The surface charge in a) is located on the metal described in c). Assuming all the net charge is distributed only in the outermost atomic layer at the surface, how big fraction of the atoms in this layer has got an extra electron? [Correct answer is either 3.3 · 10^{−9}, 1.5 · 10^{−5}, 4.3 · 10^{−3} or 0.17]

Exercise 2

The figure below shows a cross section through the center of a metal sphere with a spherical cavity inside. (The cavity is not concentric with the metal sphere.) Inside the cavity, there is a positive point charge q (located in the cross section that goes through the center of the two spheres, but not located in the center of the cavity). The metal sphere is otherwise electrically neutral so that the net charge of the whole system is q. The point charge is fixed in the given position.

What can you then say about the distribution of (free) charge in the metal sphere, given that we have electrostatic (equilibrium) conditions? [Hint 1: What is the electric field inside the metal? Hint 2: Use Gauss’ law to support your arguments.] Sketch field lines for the electric field E. Are you able to figure out what E looks like outside the sphere? [The solution is simple. However, the explanation behind the answer is perhaps not so simple?]

Exercise 3 (≃ exercise 3, exam 15. August 2003)

An electric dipole consists of two point charges q and −q with a (fixed) distance a between them. The dipole is placed in a homogeneous “external” electrostatic field E = Ex̂. Assume that the dipole is located in the xy plane, and in such a way that the vector a from −q to q, and therefore also the dipole moment p = qa, makes an angle θ with E. We take θ to be positive when going counterclockwise, from E to p, as shown in the figure.

a) What is the total force (from the external field E) on the dipole?

b) From the mechanics course, we remember that the torque τ around a given axis is defined as τ = ∑_{i} r_{i} × F_{i}, where r_{i} is the “arm” from the axis to the position where the force F_{i} acts. Show that for the electric dipole in the homogeneous field, the torque around the axis that is perpendicular to the plane containing p and E, and going through the center of the dipole, is

τ = p × E = −E × p = −pE sin θ ẑ

c) Finally, you are supposed to find an expression for the potential energy U(θ) of the electric dipole described above. Sketch U(θ). Which orientation of the dipole (relative to E) represents a stable equilibrium?

Some helpful information concerning c): Let us for simplicity stay within the xy plane only. A force F = F_{x} x̂ + F_{y} ŷ = F_{r} r̂ + F_{α} α̂ acting in a position r = r cos α x̂ + r sin α ŷ then yields a torque τ = r × F around the z axis:

Further, we know that the force F may be derived from the potential energy U using the gradient operator: F = −∇U. In polar coordinates (r, α), we have

∇ = r̂\frac{∂}{∂r} + α̂\frac{1}{r} \frac{∂}{∂α}

With this information, you should be able to show that

τ = −∂U/∂α,

and next, how U in our problem depends on the angle between E and p.

Comments: An electric insulator, i.e., a dielectric, typically consists of molecules with zero net charge. However, the internal charge distribution (i.e., the positions of the atomic nuclei and the electrons) may be such that the molecule has a nonzero dipole moment. Such polar molecules may be described in terms of simple electric dipoles, like the one we have discussed above. An example: Water, H_{2}O.

Since oxygen is more electronegative than hydrogen (i.e., it “wants” an extra electron more than hydrogen does), the electron distribution will be somewhat distorted in the direction of the oxygen atom within a water molecule. This means that close to the O atom, we have a small net negative charge, e.g. −2δ. Since the molecule is overall electrically neutral (and given the symmetry of the water molecule), there must be a small positive net charge +δ near each H atom. A dielectric may also consist of atoms or molecules without such a polar charge distribution, i.e., with zero (permanent) electric dipole moment. However, if such a material is placed in an external electric field, the electrons and the nuclei will be pulled in separate directions, resulting in an induced electric dipole moment p_{ind} directed along E. The magnitude of such induced dipole moments is typically small compared to permanent dipole moments, like the one we have in water. However, qualitatively the behaviour will be the same. Therefore: If you have understood this exercise, you have essentially understood how a dielectric behaves in an external electric field.

The cross product between two vectors The cross product between two vectors is a third vector with direction perpendicular to both, and with a magnitude given by the product of the magnitudes of the two, multiplied with the sine of the angle between them. Thus:

c = a × b

is a vector with magnitude

c = |c| = |a| · |b| · sin θ = a · b · sin θ

Example: The cross product between a = 10x̂ and b = 5ŷ is

c = a × b = 50ẑ