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