FIELD LINES. Gauss s Law. Consider an arbitrary volume bounded by a closed surface. The volume contains an assortment of charges.

Transcription

1 FIELD LINES We now consider an alternative method of graphically displaying vectors as directed by arrows where the length of the arrow gives the magnitude of the vector and it s direction gives the direction of the vector. Now we are going to adopt a different convention. At any point in space there will be directed lines. The direction of the lines will give the direction of the vector and the number of lines/area perpendicular to the lines will give the magnitude. For the moment this is just a convenient mathematical device, just as using electric field rather than force directly. However, it will turn out to have a much more profound physical significance as force particles. Graphically it look like this Gauss s Law Consider an arbitrary volume bounded by a closed surface. The volume contains an assortment of charges. Begin by considering a sphere of radius R centered on a single point charge Q. Then if the charge emit N lines they will go radially outward (since they are in the direction of E, and E is radial). Then the # of lines/area is E N 4R But we already know that

2 E kq R Thus N 4 kq We thus have the general result that each charge emits (+) or absorbs (-) 4kQ lines. We now turn to the general case. Because of linearity each charge acts independently. Hence, the total number of lines leaving the volume must be 4kQ IN where Q IN is the total charge in the volume. Now consider an infinitesimal element of surface. We represent it as a vector of magnitude equal to the infinitesimal area, da, and direction perpendicular to the element directed outward. Note that the infinitesimal element will be a plane. Then the number of lines going through the component of area perpendicular to the lines is EdAcosEdA Thus the total number of lines leaving is surface E da The total charge enclosed is vol 'dr'

3 Hence 4k r' d r' EdA vol vol Writing k 4 ε this becomes ε Using the divergence theorem we get ε r' d r' EdA vol vol vol r'dr' Edr' Since the volume is arbitrary we have E ε We should also be able to obtain this results from our earlier result E r' k r' r r' d r' r r' To see how this works we proceed as follows r' E r k r r' d r' r r' We can do this since the integral is over, not. Now

4 r r' ' r r' r r' r r' E r' k r' ' d r' r r' r r' k r' ' r' ' d r' But r' since is derivatives with respect to, not. Hence Er' k r' ' d r' r r' But ' r r' r r' Thus Erkr' ' d r' r r' Now r except at r = since

5 ctn r r r r r r sin r r r rr r r But at r = we have % so we must look further. We use the divergence theorem dr' da r r s ˆr r sin d d rˆ sin d d r s cos 4 Since is for r and yet the integral is finite it must be at r =. But this is the r definition of the Dirac delta function. Hence 4 r r Thus Er4kr' r r' d r' 4kr ε r as above. Next consider

6 r' r r' r r' r r' E r k d r' k r' ' d r' r r' r r' k r' ' r' ' d r' r r' k r' ' ' d r' But the curl of any gradient is. Hence E r From Stokes Theorem E r da Ed Ed E is conservative Thus we can obtain a potential energy in the usual way. Voltage and Potential Energy As usual f PE PE F r dr qedr f i M i i r Define voltage as PE/q. Then f Vf Vi E dr dve dr Vdr Edr EV i

7 Potential Energy and Charge Distribution Consider a collection of point charges. We want the potential energy associated with the collection. There are two energies involved. The first is the creation of the charges. The second is placing them in their final positions. We will ignore the first and assume the charges already exist. We start with them infinitely separated at infinity. Then we move the first charge to its final location. This is free since there is no field to oppose or help the motion. We now move the nd This requires moving it in the field of the first r qq kqq PE k dr r r We now bring in the rd. This time it sees both q and q. Hence Proceeding in this manner we arrive at kq q kq q PE r r N PEtotal k i,j i j qq i r ij j Looked at another way we have Writing out the two sums PE kq N N j total q i ij rij PE qq k qq qq qq qq qq N N total r r rn r r rn

8 Clearly this is not quite right since it has each term twice. Hence we should have qq i r N PEtotal k i,j i j ij j Similarly where N k PE q V total i i i V i N kq r ji ij j is the voltage at i due to all the other charges. We now adapt (unnecessary) the convention that PE = when the charges are at. Then PE qivi i In the continuous case we have PE r V r d r But we know r Er r ε E ε PE ε E Vd r Now

9 EV E V EV EV EV E V EV E ε ε εe PE EVE d r EVdA d r S Now outside the charge distribution E, V EV r r r But da r EVdA If the surface is taken at. Thus PE ε E d r where the integral is over all space. Thus we can associate an energy density εe u with an electric field. Note that this does not necessarily mean that the energy is located at a particular point, rather than if we do the integral over all space we will get the right answer. The same is true with other formulation PE V d r

10 Forces Due to Electric Fields Thus far we have found how to find the electric field for a given charge distribution. Of course what is really observed is force. Hence we need to see how to obtain force from field. This is not quite as trivial as it seems because of the requirement to avoid a charge exerting a force on itself. There are several ways to approach this problem. The most general is through the definition of potential energy. We know that dpe F dr Fdr qedr N Consider an element of surface, d. Suppose this element is displaced a distance d perpendicular to the surface. Then the volume of space external to the surface will decrease by dadr while the volume interior will be increased by the same amount. This will result in a change in PE given by ε ε Ein Eout da dr But this must be equal to Fdr F dr Hence ε F da Eout Ein Thus we need to determine the fields at the surface. This will depend on the charge on the surface and the material. We divide materials into two classes: conductors and insulators. Conductors A conductor is defined as a material in which charges can move over macroscopic distance. It is not necessary that they move freely. All that is required is that they can move are not restricted to a particular atom. Also it is not necessary that all charges can move only that some can. Electrostatics requires that they not be moving. This then requires that there be no

11 electric field in the material or parallel to the surface. In practice when an electric field is applied to the material there will be a brief time (the relaxation time) in which the charges rearrange themselves to cancel the applied field. In most conductors this is ~ - sec or less. We can now deduce several properties of conductors in electrostatics.. The charge density inside a conductor is everywhere zero. To see this use Gauss s Law. Since is zero everywhere inside the conductor, the charge inside any tiny (or otherwise) volume is zero,.. The force on a conductor is always perpendicular to the surface and given by F k A where is the charge/area on the surface. To see this use Gauss s Law to find the field surface and then use the result above to find the force. at the Since the field is perpendicular to the surface and inside it, we have EdA 4k A E 4k Then ε 8k F da 4 k da 6 k k da F k A There is a somewhat more physical way of obtaining this result. Consider the field lines produced by a charge. Topologically there are two possible types of line lines that begin and end, and those that don t. How are each produced? To answer this question we note that can vary in dimensions along the direction of ( dimension) and perpendicular to ( dimensions). The variation along looks like this

12 So that the lines/area changes as we move along the direction. This means that lines are beginning and ending. But this means that Hence produces lines that begin and end. E da E dvol E ε Now consider variations perpendicular to, Consider E d around the dotted path shown Clearly the lower two add and are +, while the upper two add and are -, but E E Ed lower upper But surf Ed E da

13 Thus E and is non-conservative. We already found that is conservative in electrostatics, so we need only consider lines that begin and end We know that the field parallel to the surface is zero for a conductor, and zero inside it. Hence Gauss s Law gives EdA 4kdA E 4k But we know that the lines end. Hence the picture really is Since the charges are equivalent, half of the field is contributed by each sheet of charge. Since a charge does not exert a force on itself, only ½ of the above E exerts a force on the surface F k A as above. Note that the force on a different charge, Q, placed to the right of the conductor would be F 4k Q. Consider a cavity inside a conductor. If there is no charge in the cavity the net charge on the surface of the cavity must be. In fact, the charge density everywhere on the surface of the cavity must be zero as we now show VEd r da VE

14 E V V E d r E d r da4 k V 4 kv da 4 kvqs E everywhere in the cavity. Thus E = and = everywhere on the surface of the cavity. 4. The conductor shields the inside of the cavity from any external fields. 5. Movement of charge inside cavity does nothing outside the conductor.