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Chapter 4 Steady Electric Currents Electric current, Electromotive force Principle of current continuity, Energy dissipation. 1. 2. 3. 4. 5. 6. Current & Current Density Electromotive Force Principle of Current Continuity Boundary Conditions for Steady Electric Currents Energy Dissipation in Steady Electric Current Fields Electrostatic Simulation 1. Current & Current Density Classification: Conduction current and convection current. The conduction current is formed by the free electrons (or holes) in a conductor or the ions in an electrolyte. The convection current is resulting from the motion of the electron, the ions, or the other charged particles in vacuum, a solid, a liquid or a gas. The amount of charge flowing across a given area per unit time is called the electric current intensity or electric current, and it is denoted by I. The unit of electric current is ampere (A) The relationship between electric current I and electric charge q is I dq dt The current density is a vector, and it is denoted as J. The direction of the current density is the same as the flowing direction of the positive charges, and the magnitude is the amount of charge through unit cross-sectional area per unit time. The relationship between the current element dI across a directed surface element dS and the current density J is dI J dS The electric current across the area S is I J dS S Which states that the electric current across an area is the flux of the current density through the area. In most conducting media, the conduction current density J at a point is proportional to the electric field intensity E at that point so that J E where is called the conductivity, and its unit is S/m. A large means that the conducting ability of the medium is stronger. The above equation is called the differential form of the following Ohm’s law U IR A conductor with infinite is called a perfect electric conductor, or p.e.c.. In a perfect electric conductor, electric current can be produced without the influence of an electric field. There is no steady electric field in a perfect electric conductor. Otherwise, an infinite current will be generated, and it results in an infinite energy. A medium without any conductivity is called a perfect dielectric or an insulator. In nature there exists no any p.e.c. or perfect dielectric. The conductivities of several media unit in S/m Media Conductivities Media Conductivities Silver 6.17 10 7 Sea water 4 Copper 5.80 10 7 Pure water 10 3 Gold 4.10 10 7 Dry soil 10 5 Aluminum 3.54 10 7 Transformer oil 10 11 Brass 1.57 10 7 Glass 10 12 Iron 10 7 Rubber 10 15 The magnitude of the current density of the convection current is not proportional to the electric field intensity, and the direction may be different from that of electric field intensity. If the charge density is , and the moving velocity is v, and then J v As the polarization of dielectrics, the conducting properties of a medium can be homogeneous or inhomogeneous, linear or nonlinear, and isotropic or anisotropic, with same meanings as before. The above equations are valid only for a linear isotropic medium. 2. Electromotive Force We first discuss the chemical action inside the impressed source under open-circuit condition. E Conducting medium P N E E' Impressed source In the impressed source , under the influence of non-electrostatic force the positive charges will be moved continuously to the positive electrode plate P, while the negative charges to the negative electrode plate N. These charges on the plates produce an electric field E, with the direction pointed to the plate N from the plate P, and the electric field E will be stronger with the increase of the charges on the plates. The electric force caused by the charges on the plates will resist the movement of the charges in the source. When the electric force is equal to the non-electrostatic force, the charges are stopped, and the charges on the plates will be constant. Since the non-electrostatic force behaves as the force acting on the charge, the non-electrostatic force is usually considered as that produced by an impressed electric field. P N E E' Impressed source This impressed electric field intensity is still defined as the force acting on unit positive charge, but it is denoted as E'. The impressed electric field E' pushes the positive charges to the positive electrode plate, and the negative charges to the negative electrode plate, and the direction of is opposite to that of the electric field E produced by the charges on the plates. when the impressed electric field is equal but opposite to the electric field produced by the charges on the plates, and the charges will be at rest. P N E E' Impressed source If the conducting medium is connected, the positive charges on the positive electric plate will be moved to the negative electric plate through the conducting medium, while the negative charges on the negative electric plate to the positive electric plate. In this way, the charges on the plates will be decreased, and E < E' . The charges in the source will be moved again. The impressed source will continuously provide the positive charges to the positive electric plate, whereas the negative charges to the negative electric plate, and in view of this a continuous current is formed. Consequently, in order to generate the continuous current in the conducting medium, it must rely on an impressed source. When it is in dynamical balance, the charges on the plates will be constant, and they produce a steady electric field in the impressed source and in the conducting medium. P N E E' Impressed source In the impressed source, E E , and there is a steady current in the circuit consisting of the impressed source and the conducting medium. Although the distribution of the charges on the plates is unchanged, the charges are not at rest. These charges are replaced without interruption. Hence, they are called sustained charges. The steady electric field in conducting medium is produced just by the sustained charge. Once the impressed source is disconnected, the supply of sustained charge to the conducting medium will vanish. P N E E' Impressed source The line integral of the impressed electric field along the path from the negative electric plate N to the positive electric plate P is defined as the electromotive force of the impressed source, and it is denoted as e, i.e. e P N E dl When it is in dynamical balance, E E in the impressed source. Therefore, the above equation can be rewritten as e P N E dl The steady electrode field caused by the sustained charges on the plates is also a conservative field, and the line integral of it around a closed circuit should be zero, i.e. E dl 0 l P Consider that in the conducting medium, J E , we have J dl 0 N E E' Impressed source l For homogeneous media, the above equation becomes J dl 0 l Using Stokes’ theorem, we have J 0 J 0 In homogeneous conducting media the steady electric current field is irrotational. 3. Principle of Current Continuity Assume the density of the sustained charges in the volume V bound by the closed surface S is , then q dV V then S q J dS dV V t t No impressed source P N E E' Impressed source Because the distribution of the charges in the steady electric current field is independent of time, i.e. 0 , we find t J dS 0 S which states that in the steady electric current field the flux of the current density through any closed surface will be zero. If we use a set of curves to describe the current field and let the tangential direction at a point on the curves be the direction of the current density at the point, these curves can then be called the electric current lines. In this way, the electric current lines must be closed, with no beginning or end. This result is called the principle of current continuity. By using the divergence theorem, we obtain J t which is called the charge conservation principle in differential form. Hence, for a steady electric current field, we have J 0 which states that the steady electric current field is solenoidal. 4. Boundary Conditions for Steady Electric Currents The integral forms of the equations for steady electric current field are as follows: J l dl 0 J dS 0 S And the corresponding differential forms are J 0 J 0 From the equations in integral form we can find the boundary condition for the tangential components of the current densities to be J 1t J 2 t 1 2 And the normal components are J1n J 2n J 1t 1 J 2t 2 J1n J 2n The tangential components of the current densities are discontinuous, while the normal components are continuous. Since J E , we find the boundary conditions for the steady electric field can be obtained as follows: E1t E2t 1 E1n 2 E2 n Since there is no the electric field cannot in a perfect electric conductor, the tangential components of a steady electric current cannot exist on the surface. Therefore, when an electric current flows into or out of a perfect electric conductor, the electric current lines are always perpendicular to the surface. 5. Energy Dissipation in Steady Electric Current Fields In a conducting medium, the collision of free electrons with the atomic lattice will generate thermal energy, and this is an irreversible energy conversion process. The impressed source has to compensate the energy dissipation in order to maintain the steady electric current. dl J U dS In a steady electric current field, we construct a small cylinder of length dl and end face area dS, and assume the two end faces of the cylinder are equipotential surfaces. Under the influence of the electric field, electric charge dq is moved to the right end face from the left end face in dt , with the corresponding work done by the electric force as dW dqE dl Edqdl The power dissipation P is P dW dq E dl EIdl EJdSdl dt dt Then the power dissipation per unit volume as pl EJ E 2 J2 If the direction of J is different from that of E, the above equation can be written in the following general form pl E J Which is called the differential form of Joule’s law, and it states that the power dissipation at a point is equal to the product of the electric field intensity and the current density at the point. Suppose the electric potential difference between two end faces U is U, then E . And we know that J I . Hence, the power dl dS dissipation per unit volume can be expressed as pl UI UI dSdl dV The total power dissipation in the cylinder is P pl dV UI which is Joule’s law. Example 1. A parallel plate capacitor consists of two imperfect dielectrics in series. Their permittivities are 1 and 2 , the conductivities are 1 and 2 , and the thickness are d1 and d2, respectively. If the impressed voltage is U, find the electric field intensities, the electric energies per unit volume, and the power dissipations per unit volume in two dielectrics. Solution: Since no current exists outside the capacitor, the electric current lines in the 1 1 d1 capacitor can be considered to be perpendicular U d2 to the boundaries. Then we have 2 2 E11 E2 2 E1d1 E2d2 U In view of this we find E1 2 d1 2 d 2 1 U E2 1 d1 2 d 2 1 U The electric energies per unit volume in two dielectrics, respectively, are 1 1 we1 1 E12 , we 2 2 E22 2 2 The power dissipations per unit volume in two dielectrics, respectively, are pl1 1 E12 , pl 2 2 E22 Two special cases are worth noting: If 1 0 , then E1 U ,E 2 0 , w 0 , pl 2 0 . e2 d1 If 2 0 , then E 0 , we1 0 , pl1 0 , E U . 1 2 d2 U 1= 0 d1 E 2= 0 d2 U E 1= 0 2= 0 Example 2. A quarter of a flat circular conducting washer is shown in the figure. Calculate the resistance between two end faces. Solution: The cylindrical coordinate system should be selected. Assume the electric potential difference between two end faces is U, and let y U t (r,) r 0 a 0 b x The electric potential 1 0 at 0 The electric potential 2 U at π 2 Since the electric potential is related to the angle , it should satisfy the following equation d 2 0 2 d The general solution is C1 C2 Based on the given boundary conditions, we find 2U π The current density J in the conducting medium is y U t J E e (r,) r 0 a 0 b x 2U e r πr Then the current I flowing into the conducting π medium across the end face at is 2 2Ut b dr 2Ut b 2U ln I J dS e ( e t d r ) a S S π r π πr a Consequently, the resistance R between two end faces is R U I π b 2 t ln a 6. Electrostatic Simulation Two fields are found to be very similar in source-free region. Steady Electric Current Field Electrostatic Field ( E 0) ( 0) J dl 0 E dl 0 J dS 0 E dS 0 J 0 E 0 J 0 E 0 l S l S The electric current density J corresponds to the electric field intensity E, and the electric current lines to the electric field lines. If the steady electric current field has the same boundary conditions as that for the electrostatic field, the distribution of the current density will be the same as that of the electric field intensity. Based on this similarity, the solution of the steady electric current field can be found directly from the results of the electrostatic field. In some cases, since the steady electric current field is easy to be constructed and measured, the electrostatic field can be investigated based on the steady electric current field with the same boundary conditions, and this method is called electrostatic simulation. The electrostatic field and the steady electric current field between two electrodes as follows: P N Steady electric current field P N Electrostatic field The calculation of the resistance of conducting media can be determined based on the results of the corresponding electrostatic field. Based on the equations for two fields, we can find the resistance and conductance between two electrodes as R C G C If the capacitance between two electrodes is known, from the above equations the resistance and the conductance between two electrodes can be found out. R C G C The capacitance of a parallel plate capacitor of plate area S and S . If the conductivity of the imperfect dielectric separation d is C d is , the leakage conductance G between two electric plates of the parallel plate capacitor is G S S d d 2π , ln( b / a) where b is the inner radius of the outer conductor, and a is the The capacitance of a coaxial line per unit length is C1 radius of the internal conductor. If the conductivity of the filled dielectric is , the leakage conductance per unit length G1 is G1 2π ln( b / a)