- 1 -Chapter 7. Electrodynamics7.1. Electromotive ForceAn electric current is flowing when the electric charges are in motion. In order to sustain anelectric current we have to apply a force on these charges. In most materials the current densityJ is proportional to the force per unit charge:J =s f The constant of proportionality s is called the conductivity of the material. Instead ofspecifying the conductivity, it is more common to specify the resistivity r:r=1sFor conductors the resistivity is typically 10-8 W-m; for semiconductor it varies between 0.01 W-m and 1 W-m, and for insulators it varies between 105 W-m and 106 W-m. In most cases theforce on the charges is the electromagnetic force. In that case the current density is equal to:J =s E + v ¥ B ()If the velocity of the charges is small the second term can be ignored, and the equation for Jreduces to Ohm's Law:J =s E Consider a wire of cross-sectional area A and length L. If a potential difference V is appliedbetween the ends of the wire, it will produce an electric field inside the wire of magnitudeE =VLThe current density in the wire is therefore equal toJ =sVLThe total current flowing through the wire is therefore equal to- 2 -I = JA =s AVLThis equation shows that the current flowing from one electrode to the other electrode isproportional to the potential difference between them. This is a rather surprising result since thecharge carriers are constantly accelerating. However, the proportionality between the currentand the potential difference has been found to be correct for most materials. This relation can bewritten asV = IRThe constant of proportionality R is called the resistance of the material. It is in general afunction of the geometry of the system and the conductivity of the materials between theelectrodes. The unit of resistance is the ohm (W). The resistance of the wire is equal toR =VI=Vs AVL=1sLA=rLATo create a current we have to do work. The work required to move a unit of charge across apotential difference V is equal to V. To establish a current I, we need to deliver a power P whereP = VI = I2RThe unit of power is the Watt (1 W = 1 J/s). The work done by the electric force on the chargecarriers is converted into heat (Joule heating).Example: Problem 7.1Two concentric metal spherical shells, of radius a and b, respectively, are separated byweakly conducting material of conductivity s.a) If they are maintained at a potential difference V, what current flows from one to the other?b) What is the resistance between the shells?a) Suppose a charge Q is placed on the inner shell. The electric field in the region between theshells will be equal toE =14pe0Qr2ˆ r The corresponding potential difference between the spheres is equal to- 3 -Va- Vb=- E ∑ dr baÚ=Q4pe01a-1bÊ Ë ˆ ¯ Therefore, in order to maintain a potential difference V between the spheres, we must place acharge Q equal toQ =4pe0V1a-1bÊ Ë ˆ ¯ on the center shell. The total current flowing between the two shells is equal toI =J ∑ da SphereÚ=sE ∑ da SphereÚ=s14pe0Qr24pr2=sQe0= 4psV1a-1bÊ Ë ˆ ¯ b) The resistance between the shells can be obtained from Ohm's law:R =VI=V4psV1a-1bÊ Ë ˆ ¯ =14ps1a-1bÊ Ë ˆ ¯ Example: Problem 7.2a) Two metal objects are embedded in weakly conducting material of conductivity s (see Figure7.1). Show that the resistance between them is related to the capacitance of the arrangementbyR =e0s Cb) Suppose you connected a battery between 1 and 2 and charged them up to a potentialdifference V0. If you then disconnect the battery, the charge will gradually leak off. Showthat V(t) = V0 exp(- t/t), and find the time constant t in terms of e0 and s.a) Suppose a charge Q is placed on the positively charged conductor. The current flowing fromthe positively charged conductor is equal to- 4 -I = J ∑ da SurfaceÚwhere the surface integral is taken over a surface that encloses the positively charged conductor(for example, the dashed surface in Figure 7.1). The expression for I can be rewritten in terms ofthe electric field asI =sE ∑ da SurfaceÚ12sFigure 7.1. Problem 7.2.Using Gauss's law to express the surface integral of E in terms of the total enclosed charge weobtainI =sQe0The charge on the conductor is related to the capacitance of the arrangement and the potentialdifference between the conductors:Q = CVThe current I is therefore equal toI =se0CVThe resistance of the system can be calculated using Ohm's law:- 5 -R =VI=Vse0CV=e0s Cb) The charge Q residing on the positively charged conductor is equal toQ = CV = CRI =-CRdQdtThis equation can be rewritten asdQdt+1CRQ = 0and has the following solution:Qt()= Q0e-t / RCThe potential difference V is equal toVt()=Qt()C=Q0Ce-t / RC= V0e-t / RCThe decay constant t is equal tot= RC =e0sIn any electric circuit a current will only exist if a driving force is available. The mostcommon sources of the driving force are batteries and generators. When a circuit is hooked up toa power source a current will start to flow. In a single-loop circuit the current will be the sameeverywhere. Consider the situation in which the currents are not the same (see Figure 7.2). IfIin > Iout then positive charge will accumulate in the middle. This accumulation of positivecharge will generate an electric field (see Figure 7.2) that slows down the incoming charges andspeeds up the outgoing charges. A reduction in the velocity of the incoming charges will reducethe incoming current. An increase in the velocity of the outgoing charges will increase theoutgoing current. The current will change until Iin = Iout.The total force f on the charge carriers (per unit charge) is equal to the sum of the sourceforce, fs, and the electric force:f = f s+ E- 6 -+++++++++I inI outEEEEFigure 7.2. Current flow.The work required to move one unit of charge once around the circuit is equal tof ∑ dl Ú= f s∑ dl Ú+ E ∑ dl Ú= f s∑ dl Ú=ewhere e is called the electromotive force or emf. The emf determines the current flowingthrough the circuit. This can be most easily seen bye rewriting the force f on the charge carriersin terms of the current density J e= f ∑ dl Ú=J s∑ dl Ú=IasdlÚ= IdlasÚ= IRHere, a is the cross-sectional area of the wire (perpendicular to the direction of the current).Example: Problem 7.5a) Show that electrostatic force alone cannot be used to drive current around a circuit.b) A rectangular loop of wire is situate so