Institutt for fysikk, NTNU TFY4155/FY1003: Elektrisitet og magnetisme Spring 2005

Summary, week 6 (February 8 and 9)

Meaning of the gradient operator

The vector ∇V points in the direction where V increases most rapidly, i.e., in the direction where the directional derivative of V has its largest value. Since E = −∇V, this means that the electric field points in the direction where V decreases most rapidly. Example: If a point charge q is placed in a position where ∇V = 0, it is not subject to any forces, because F = qE = −q∇V = 0.

Summary so far, and saying hello to Maxwell equation nr 1

The Coulomb law (empirical law for force between two charges q and q′ in mutual distance r):

F = \frac{qq′}{4πε_{0}r^{2}} r̂

Electric field from point charge q (follows from the definition ”force pr unit charge”):

E = \frac{F}{q′} = \frac{q}{4πε_{0}r^{2}} r̂

Conservative force:

∫_{A}^{B} F · dl

is independent of the integration path, i.e., the path between the points A and B. Hence:

∮ F · dl = 0

(i.e., when we integrate around a closed path)

With the definition of E, it follows that the electrostatic field is also conservative, i.e.:

∫_{A}^{B} E · dl

is independent of the integration path, and hence

∮ E · dl = 0

This is one of Maxwell’s equations (for static fields, i.e., fields that do not change in time). A conservative vector field can always be derived from a scalar potential:

E = −∇V

The potential difference between two points A and B can be evaluated if we know the electric field in the space between A and B:

ΔV = V_{B} − V_{A} = − ∫_{A}^{B} E · dl

The superposition principle is valid for the elektric force F (this is an experimental result):

F_{i} = ∑_{j=1}^{n} F_{ij}

= force on charge q_{i} from charges q_{j} (j = 1, 2, … n) Then it follows that the superposition principle is also valid for the electric field E,

E = ∑_{j=1}^{n} E_{j}

and for the electric potential V,

V = ∑_{j=1}^{n} V_{j}

Here, E_{j} and V_{j} are the contributions to the field and the potential, respectively, form charge number j.

Electric flux [FGT 23.1; YF 22.1; TM 22.2; AF 25.3; LHL 19.7; DJG 2.2.1]

ϕ = ∫_{S} E · dA

Sometimes, we write ϕ_{E} to specify that it’s an electric flux we’re talking about. Earlier, we have defined electric field lines so that the electric field strength E = |E| is proportional with the density of field lines, i.e., the number of field lines pr unit area. From the above definition of electric flux ϕ, we may conclude that ϕ simply represents the number of field lines that cross the surface S. The following figure illustrates what this is about:

The surface S is an arbitrary ”thought” or ”chosen” surface in space. The electric field ”exists” in the region where the surface S is ”located”. (E may be zero, or nonzero.) The surface S is then divided into small surface elements dA = n̂dA, with area dA and an orientation in space specified by the surface normal n̂. The flux dϕ through the surface dA is then E · dA. The total flux through the whole surface S is obtained by integrating the contributions dϕ, which is the equation above. Note that the flux is a scalar quantity. However, it may be positive or negative, depending on whether the angle between the vectors E and dA is smaller or larger than 90 degrees.

A closed surface S is a surface which encloses a well defined volume V, e.g., a spherical shell, a peanut shell or similar. The electric flux through a closed surface is written like this (cf. the notation used for path integral around a closed curve):

ϕ_{c} = ∮_{S} E · dA

The index c denotes ”closed”. It is actually not necessary as long as we write down the integral sign with the ring on it. The latter is sufficient to make sure we’re talking about a closed surface. With a closed surface, we may introduce a sign convention for the surface element vector: We choose positive direction for dA when it is directed out of the surface. Then we may conclude that

E · dA > 0 ⇒ flux out through thesurface E · dA < 0 ⇒ flux in through thesurface

Furthermore:

ϕ_{c} > 0 ⇒ net flux out through thesurface ϕ_{c} < 0 ⇒ net flux in through thesurface

For a surface S that is not closed, we don’t have this opportunity to choose the positive and the negative direction for dA. The surface has two sides, and none of these can be claimed to be more ”inside” than the other. However, we may solve the problem by choosing a positive direction of the (closed!) curve that runs around the edge of S. Then, the positive direction of dA is chosen in terms of the right hand rule: Let the four fingers of your right hand point along the positive direction of the curve around S. Then, the remaining finger, the thumb, points in the positive direction of dA.

Gauss’ law [FGT 23.2; YF 22.3; TM 22.2, 22.6; AF 25.4; LHL 19.7; DJG 2.2.1]

Gauss’ law (in socalled integral form; later, if time permits, we shall see that we also have a version of Gauss’ law on socalled differential form):

∮_{S} E · dA = \frac{q_{in}}{ε_{0}}

Here, the integral on the left hand side denotes a surface integral over a closed surface S, while q_{in} is the total net charge inside this closed surface (the ”gaussian surface). Gauss’ law is one of Maxwell’s equations. (For the moment, we are only talking about electrostatics. However, Gauss’ law is also valid when the electric field changes in time.) The content in Gauss’ law may be formulated like this: The net number of field lines out of a volume, i.e., out through the closed surface enclosing this volume, is determined by, and is directly proportional with, the net charge inside this volue, i.e., inside the closed surface. Gauss’ law follows directly from Coulomb’s law, and therefore really represents no new physics.

In connection with the proof of Gauss’ law, we used a thing we called a solid angle Ω. In just the same way as a small sector in a plane spans an angle dϕ, a small sector in space (3D) spans a solid angle dΩ. Furthermore: In the same way as the arc length dl in distance r from the ”origin” then becomes dl = r dϕ, the area dA_{r} of the surface which is perpendicular to r and limited by the small sector then becomes dA_{r} = r^{2} dΩ. Note the analogy between 2D and 3D!

If we let the sector in the plane go once around, this corresponds to the angle

∮ dϕ = ∫_{0}^{2π} dϕ = 2π

Analogously: If we let the sector in space span the whole sphere, this corresponds to a solid angle

∮ dΩ = ∫_{0}^{2π} dϕ ∫_{0}^{π} sin θ dθ = 4π

(Here, we have used spherical coordinates, where dA_{r} = r^{2} sin θ dθ dϕ (see øving 3!), so that dΩ = dA_{r}/r^{2} = sin θ dθ dϕ.)