1 November 2002 Physics 217, Fall 2002 1Today in Physics 217: charges in motion Forces on dielectrics: an electrostatic pump Currents Magnetic forces on point charges and currents- - - - - - - - - - -+ + + + + + + + + + +E+Q-QFx1 November 2002 Physics 217, Fall 2002 2Force and work on dielectricsA real parallel-plate capacitor – one with finite-size plates –has an electric field that extends past its boundaries. The field past the edges is called the fringing field. E+Q-Q1 November 2002 Physics 217, Fall 2002 3Force and work on dielectrics (continued)A dielectric slab inserted part-way into the capacitor is polarized by the field, and the bound charge experiences forces exerted by the fringing fields that push the dielectric further into the capacitor. What is the force?- - - - - - - - - - -+ + + + + + + + + + +E+Q-QFx1 November 2002 Physics 217, Fall 2002 4Force and work on dielectrics (continued)There are two ways we could go about the calculation: Calculate the fringing electric displacement, D, by solving Laplace’s equation for the boundary conditions presented by the capacitor plates. (If the plates are circles or squares, this would be straightforwardly solved by separation of variables.) Then put the dielectric slab back, and calculate the field and bound charge distribution. Then calculate the total force on the bound charges. Believe it or not, you now know how to do it this way, and I have half a mind to make this a homework problem. But I won’t, because there’s an easier way to do it that ignores the fringing field completely, even though that’s what does the work.1 November 2002 Physics 217, Fall 2002 5Force and work on dielectrics (continued) We know the potential energy (work done by an external agent) for empty and dielectric-filled capacitors, so calculate Wand use to compute the force.Let’s begin by assuming that the charge on the capacitor is constant, and that the plate dimensions are a×A, with separation d. ThenBreak the capacitor into two parts, one with dielectric and one without. These two capacitances are in parallel (have the same potential difference), soW=−F—2211, constant.22QWCV QC==()emptydielectric44axaxCC Cddεππ−=+=+A1 November 2002 Physics 217, Fall 2002 6Force and work on dielectrics (continued)Then the force is Sure enough, the dielectric is pulled into the capacitor. Here the charge is constant as this happens, so the potential difference Vbetween the plates is less than is was when the capacitor was empty. If the potential were fixed in the problem – if, say, a battery were present, then2222222211ˆˆ221111ˆˆˆ.244 2 4 2edQ QdCWdx C dxCQaa Qa Vadd dCCdεεχππ π=− =− =−=−= =Fx xxxx—1 November 2002 Physics 217, Fall 2002 7Force and work on dielectrics (continued)This time, the voltage is constant and the charge on the capacitor plates increases as the dielectric is pulled in. But either way, the dielectric is pulled in. You will explore this some more on the homework in the classic “capacitor oil pump” problem, 4.28.us2222121 Same as before...21ˆ.2xedW F dx VdQ Fdx VdQdW dQ dC dCFVVVdx dx dx dxdCVdxVadχ=+=−+=− + =− +==Fx1 November 2002 Physics 217, Fall 2002 8CurrentsConsider a very long conducting wire, part of which is shown below. It is neutral overall but contains charges that can move under the influence of an externally-applied electric field, and charges that can’t:The motion of the +λdistribution is a steady current:0λ==+λλ−vEIvλ=Units: esu/sec in cgs, coul/sec = amp in MKS.1 November 2002 Physics 217, Fall 2002 9Currents (continued)During a time a charge passes a given point on the wire; Iis the rate at which charge passes that point.  Though in general v is of course a vector, we don’t usually write Ias a vector, because we usually run currents through wires (one dimension).  It is, however, useful to think of the surface and volume analogues of this line-charge current as vectors, since they will come into any discussion of currents within a wire.Suppose we have a conducting sheet, with surface density σin mobile charges and -σin fixed charges. Then the surface current density ist∆vtλ∆σ=KvUnits: esu/(cm sec) in cgs, amp/m in MKS.1 November 2002 Physics 217, Fall 2002 10Currents (continued)The total current from thesheet iswhere is an infinitesimallength element perpendicular to thedirection the current flows.And, suppose you have a conducting volume, with charge density ρof mobile charges and -ρof fixed charges. Then we get to define the most useful of current densities:⊥∆Addd⊥⊥=⇔=∫IIK KAAd⊥AK1 November 2002 Physics 217, Fall 2002 11Currents (continued)The units of current density J areIntegrating over the whole surface area of a conductor, one gets the current flowing out of the volume it encloses:daAdddaρ⊥==⋅=∫JvIJaIJ2-1 -2esu cm sec or amp m .AId dτ=⋅=⋅∫∫Ja JvV—1 November 2002 Physics 217, Fall 2002 12Currents (continued)Because electric charge is conserved, the only way for a current to flow out of a volume is for the charge inside it to decrease:We can equate the integrands of the last two volume integrals to produce.AdQ dddddt dt tρρτ τ∂⋅=− =− =−∂∫∫∫JavVV.tρ∂⋅=−∂J—Continuity equation1 November 2002 Physics 217, Fall 2002 13Magnetic force on point chargesThe reason we consider currents, of course, is that they exert forces on each other that cannot be explained by electrostatics: parallel currents attract each other, and opposite currents repel each other, for a force that is perpendicular to v. As you are already aware, this is because of a new force that has nothing obvious (at the moment) to do with Coulomb’s law, the magnetic force, and an associated magnetic field, B. This force is empirically determined for a point charge to bemagmagelec in MKS unitsqqcqqc=× ×=+ =+×vFBvBvFF F E BLorentz force lawc= speed of light1 November 2002 Physics 217, Fall 2002 14Magnetic unitsIn cgs, E and B have the same units, usually expressed as statvolt/cm for E or gauss (= statvolt/cm) for B.In MKS, E and B have different units:[E] = V/m or Nt/coulqqc=+×vFE Bqqc=+×vFE B[]-1 -14Tesla T Nt sec coul m10 gauss .====B1 November 2002 Physics 217, Fall 2002 15Notes about B The electrostatic force can do