ENERGY IN ELECTROSTATICS

Transcription

1 ENERGY IN ELECTROSTATICS We now turn to the question of energy in electrostatics. The first question to consider is whether or not the force is conservative. You will recall from last semester that a conservative force is one for which the work done by the force is independent of the path taken. Gravity was a conservative force, while friction was not. Hence we need to determine whether the work done by following path (1) from A to B is the same as by following path (2). Recalling the definition of work: dfdr B AB Fdr A We note that changing the direction of motion changes the sign of the work done, and thus: AB BA Hence if the work done on path (1) is the same as that done on path (2) we can go out on (1) and back on (2) and the total work done must be zero. We conclude that the work done around any closed path must be zero for a conservative force. The question then is what condition must be met by the force for this to happen. To answer this requires a little math diversion into vector calculus. I realize this will be new to many of you, so for the moment regard it as cultural. We will not need it until later when we come to Maxwell s equations, but it is useful to see it beforehand. DIVERGENCE OF A VECTOR A vector can change either along its direction or perpendicular to its direction. In other words, if the vector points in the ˆx direction it can have:

2 da dx But it could also have: da da dy dz Since the former would require only one number to describe it we expect it to be a scalar, whereas the latter requires two numbers and is a vector. If we knew both of these at all points plus the value of the vector at one point we would completely describe the vector. For this reason the fundamental equations of electricity and magnetism simply give these derivatives. We define the average value of the divergence of a vector as shown below. Consider a region of space of ume V bounded by a closed surface: We now define the average value of the divergence of B to be: We then define the divergence of B as: Av Div B Div B lim 1 BdA 1 BdA This definition is independent of the particular coordinate system used, but for present purposes we will use Cartesian coordinates to get an idea of how to actually calculate the divergence. Consider a rectangular box with one corner at the point (x,y,z) and sides parallel to the coordinate planes. Now consider the faces at x and x + dx which are parallel to the yz plane.

3 Recalling that da points outward from the ume we find the integral over these faces to be: Bx dxb(x) B B x x(x) dydz Bx x dx dydz dxdydz dxdydz dx x Proceeding similarly on the other two pairs of faces we find: We then get: B B x y B B da z dxdydz x y z 1 Div B lim B da v V 1 B B x y B lim z dxdydz dxdydz dxdydz x y z B B x y B z x y z We now define a vector differential operator as follows. is to operate on any type of entity (vectors, scalars, tensors). It is the ratio of the change in the entity to the physical distance moved when the coordinate changes by an infinitesimal amount. For example, in Cartesian coordinates the physical distance moved when the coordinate changes by dx is dx. But in polar coordinates the physical distance moved when θ changes by dθ is Rdθ. Thus: V V V V V 1 V V xˆ yˆ zˆ ˆ zˆ ˆ x y z z Since is a vector we can multiply it into a scalar and get a vector. We can dot it into a vector and get a scalar, or we can cross it into a vector and get a vector. All three will be important.

4 From the result above we see that the divergence of a vector is simply: Div B B We can now use these results to get what is known as the divergence theorem: VAv B BdA 1 V B d BdA V Bd BdA This is an extremely useful result which we will use repeatedly this semester. It is important to understand its physical meaning. Since gives the number of lines leaving the ume, BdA B d must give the number of lines produced in the ume. Thus B must give the number of lines produced per unit ume. We can now use this to derive a differential form of Gauss s Law: 1 EdA r' d 3 r' g 1 E d r' r' d r' 3 3 g

5 Since the ume is arbitrary we must have: 1 st Maxwell Equation. E g