Chapter2 Static Electric FieldsElectric Field Intensity,ElectricPotentialPolarization ofDielectric,Field EquationsBoundary Conditions.Energy and Force1.Fieldlntensity,Flux,andFieldLines2.Eguations for Electrostatic Fieldsin Free Space3.ElectricPotentialand EguipotentialSurfaces4.Polarizationof Dielectrics5. Equations for Electrostatic Fields in DielectricBoundary ConditionsforDielectricInterfaces6. EBoundary Conditionsfor Dielectric-conductorInterface78. Capacitance9.Energy in Electrostatic Field10.ElectricForcesV
Chapter 2 Static Electric Fields Electric Field Intensity, Electric Potential Polarization of Dielectric, Field Equations Boundary Conditions, Energy and Force 1. Field Intensity, Flux, and Field Lines 2. Equations for Electrostatic Fields in Free Space 3. Electric Potential and Equipotential Surfaces 4. Polarization of Dielectrics 5. Equations for Electrostatic Fields in Dielectric 6. Boundary Conditions for Dielectric Interfaces 7. Boundary Conditions for Dielectric-conductor Interface 8. Capacitance 9. Energy in Electrostatic Field 10. Electric Forces
1.Fieldlntensity,Flux,andFieldLinesThe intensity of the electric field at a point is defined as the forceproduced by the electric field on a unit positive charge at that point,andisdenotedbyEFEqWhere q is the test charge, and F is the force acting on the chargeThe flux of the electric field intensity through a surface is calledelectric flux, and it is denoted as , given by Y, i.eY=[E·dsU7
1. Field Intensity, Flux, and Field Lines The intensity of the electric field at a point is defined as the force produced by the electric field on a unit positive charge at that point, and is denoted by E q F E = Where q is the test charge, and F is the force acting on the charge. The flux of the electric field intensity through a surface is called electric flux, and it is denoted as , given by,i.e. = S E dS
The vector equation for electric field linesExdl=0ElectricfieldtubeDistributionsof electricfield linese6eeFTwo parallelApositiveAnegativechargeplatespoint chargepointchargeUa
The vector equation for electric field lines Edl = 0 Electric field tube Two parallel charge plates A negative point charge A positive point charge Distributions of electric field lines
2.EquationsforElectrostaticFieldsinFreeSpacePhysical experiments show that the intensity of an electrostatic fieldin free space satisfies the following two equations in integral formf, E.ds -1f E·dl =0Where o is the permittivity (or dielectric constant) of free space andits valueis8。 = 8.854187817...×10-12F / m×10-9F/m36元The left equationis called Gauss'slaw, and it shows that the outwardflux ofthe electric field intensity of an electrostatic field over any closedsurface in free space is equal to the ratio of the charge in the closedsurface to the permittivity of free space.The right equation states that the circulation of an electrostaticfield aroundany closed curveis equal to zerou7
2. Equations for Electrostatic Fields in Free Space Physical experiments show that the intensity of an electrostatic field in free space satisfiesthe following two equationsin integral form = S q 0 d E S = l E dl 0 Where 0 is the permittivity (or dielectric constant) of free space and its value is The left equation is called Gauss’s law, and it shows that the outward flux of the electric field intensity of an electrostatic field over any closed surface in free space is equal to the ratio of the charge in the closed surface to the permittivity of free space. 10 F/m 36π 1 8.854187817 10 F/ m 1 2 9 0 − − = The right equation states that the circulation of an electrostatic field around any closed curve is equal to zero
Using the divergence theorem, from Gauss'law we havePV.E-8It is called the differential form of Gauss's law. It shows that thedivergence of the electric field intensity of an electrostatic field at apoint in free space is equal to the ratio of the density of the charge atthe point to the permittivity of free space.From Stokes'theorem and the above equation, we haveVxE=0which shows that the curl ofan electrostaticfield in free space is zeroeverywhereThe electrostatic field infree spaceis a lamellarorirrotationaloneU>
0 E = E = 0 which shows that the curl of an electrostatic field in free space is zero everywhere. Using the divergence theorem, from Gauss’law we have It is called the differential form of Gauss’s law. It shows that the divergence of the electric field intensity of an electrostatic field at a point in free space is equal to the ratio of the density of the charge at the point to the permittivity of free space. From Stokes’theorem and the above equation, we have The electrostatic field in free space is a lamellaror irrotational one
After knowing the divergence and the rotation of the electricfield intensity, one may write, with the aid of the Helmholtz'stheoremE=-VΦ+V×AwheredvprV'.E(r)dvd(r)4元r-rV'xE(r)dyA(r) =rITU7
E = − + A − = − = V V V V d ( ) 4π 1 ( ) d ( ) 4π 1 ( ) |r r | E r A r |r r | E r r where After knowing the divergence and the rotation of the electric field intensity, one may write, with the aid of the Helmholtz’s theorem x P z y r O dV (r) r − r r V
Substituting the electric field eguationsinto the above resultsgivesp(rdvΦ(r) :A(r)= 04元80HenceE=-VdΦThe scalar function @is called the electric potential, and the electricfield intensity at a point in free space is equal to the negative gradientof the electric potential at that point.According to the National Standard of China, the electricpotentialis denoted by the Greek small character , i.e.E=-VpU7
− = V V 0 d ( ) 4π 1 ( ) |r r | r r A(r) = 0 Substituting the electric field equations into the above results gives Hence E = − The scalar function is called the electric potential, and the electric field intensity at a point in free space is equal to the negative gradient of the electric potential at that point. E = − According to the National Standard of China, the electric potentialis denoted by the Greek small character , i.e
Substituting the electric potential expression into this equation,we haveE(n)- er)r-r)duJr 4元s0r-rlIf the electric charge is distributed on a surface S or on a curve l,we havePs(r')(r-r)Ps(r')dsdsp(r) :E(r)4元80slr-4元80Ir-rprp,(r)(r-r0dlE(r)Ir-r4元0It is easy to see that the electric field intensity can bedetermined directly from the above equations if the distribution ofthe chargeis known.U7
If the electric charge is distributed on a surface S or on a curve l, we have − = S S S 0 d | ( ) 4π 1 ( ) r r | r r − − = S S S 3 0 d | ( )( ) 4π 1 ( ) r r | r r r E r − = l dl ( ) 4π 1 ( ) 0 |r r | r r l − − = l l l 3 0 d | ( )( ) 4π 1 ( ) r r | r r r E r Substituting the electric potential expression into this equation, we have V V − − = d 4π ( )( ) ( ) 3 0 r r r r r E r It is easy to see that the electric field intensity can be determined directly from the above equations if the distribution of the charge is known
Summary( a ) The charge q in the Gauss'law should be the sum of all positiveand negative charges in the closed surface S.(b ) The electric field lines cannot be closed and intersecteach other.( c ) The line integral of the electric field intensity along a pathbetween any two points is independent of the path, and electrostaticfield is a conservative field as the gravitational field.(d ) If the distribution ofthe charge is known, the electric fieldintensity can be found based on Gauss'law, the electric potential, orthe distribution of the charge
（a）The charge q in the Gauss’ law should be the sum of all positive and negative charges in the closed surface S. Summary （b）The electric field lines cannot be closed and intersect each other. （c）The line integral of the electric field intensity along a path between any two points is independent of the path, and electrostatic field is a conservative field as the gravitational field. （d）If the distribution of the charge is known, the electric field intensity can be found based on Gauss’ law, the electric potential, or the distribution of the charge
Example 1 Calculate the electric field intensity produced bya point charge.Solution: The point charge is the charge whose volume is zero.Because of the symmetry of the point charge, if the point charge isplaced at the origin of a spherical coordinate system, the electricfieldintensity must be independent of the angles Oand oConstruct a sphere of radius r and let the point charge be at thecenter of the sphere, then the magnitude of electric field intensity atall of the points on the surface of the sphere will be equal. If the pointcharge is positive, the direction of the electric field intensity is thesame as that of the outward normal to the surface of the sphere.f,E·ds - 1Applying Gauss'law60
Example 1 Calculate the electric field intensity produced by a point charge. Solution: The point charge is the charge whose volume is zero. Because of the symmetry of the point charge, if the point charge is placed at the origin of a spherical coordinate system, the electric field intensity must be independent of the angles and . Construct a sphere of radiusr and let the point charge be at the center of the sphere, then the magnitude of electric field intensity at all of the points on the surface of the sphere will be equal. If the point charge is positive, the direction of the electric field intensity is the same as that of the outward normal to the surface of the sphere. = S q 0 d Applying Gauss’ law E S