ENE 325 Electromagnetic Fields and Waves Lecture 3 Gauss’s law and applications, Divergence, and Point Form of Gauss’s law 1

Review (1) Coulomb’s law Coulomb’s force electric field intensity or V/m 2

Review (2) Electric field intensity in different charge configurations infinite line charge ring charge surface charge 3

Outline Gauss’s law and applications Divergence and point form of Gauss’s law 4

Gauss’s law and applications “The net electric flux through any closed surface is equal to the total charge enclosed by that surface”. If we completely enclose a charge, then the net flux passing through the enclosing surface must be equal to the charge enclosed, Q enc. 5

Gauss’s law and applications The integral form of Gauss’s law: Gauss’s law is useful in finding the fields for problems that have a high degree of symmetry by following these steps: Determine what variables influence and what components of are present. Select an enclosing surface, Gaussian surface, whose surface vector is directed outward from the enclosed volume and is everywhere either tangential to or normal to 6

Gauss’s law and applications The enclosing surface must be selected in order for to be constant and to be able to pull it out of the integral. 7

Ex1 Determine from a charge Q located at the origin by using Gauss’s law Select a Gaussian surface 3. D r at a fixed distance is constant and normal to a Gaussian surface, can be pulled out from the integral. 8

Ex2 Find at any point P ( , , z) from an infinite length line of charge density  L on the z-axis. 1. From symmetry, 2. Select a Gaussian surface with radius  and length h. 3. D  at a fixed distance is constant and normal to a Gaussian surface, can be pulled out from the integral. ant and normal to a Gaussian surface, can be pulled out from the integral. 9

Ex3 A parallel plate capacitor has surface charge +  S located underneath a top plate and surface charge -  S located on a bottom plate. Use Gauss ’ s law to find and between plates. 10

Ex4 Determine electric flux density for a coaxial cable. 11

Ex5 A point charge of 0.25  C is located at r = 0 and uniform surface charge densities are located as follows: 2 mC/m 2 at r = 1 cm and -0.6 mC/m 2 at r = 1.8 cm. Calculate at a)r = 0.5 cm b)r = 1.5 cm 12

c) r = 2.5 cm 13

Ex6 A uniform volume charge density of 80  C/m 3 is present throughout region 8 mm < r < 10 mm. Let  v = 0 for 0 < r < 8 mm, a)Find the total charge inside the spherical surface r = 10 mm b)Find D r at r = 10 mm. 14

c) If there is no charge for r > 10 mm, find D r at r = 20 mm 15

Divergence and Point form of Gauss’s law (1) Divergence of a vector field at a particular point in space is a spatial derivative of the field indicating to what degree the field emanates from the point. Divergence is a scalar quantity that implies whether the point source contains a source or a sink of field. where = volume differential element 16

Divergence and Point form of Gauss’s law (2) or we can write in derivative form as Del operator: It is apparent that therefore we can write a differential or a point form of Gauss’s law as 17

Divergence and Point form of Gauss’s law (3) For a cylindrical coordinate: For a spherical coordinate: 18

positive indicates a source of flux. (positive charge) negative indicates a sink of flux. (negative charge) Physical example The plunger moves up and down indicating net movement of molecules out and in, respectively. An integral form of Gauss’s law can also be written as 19

Ex7 Let. Determine 20

Ex8 Let in a cylindrical coordinate system. Determine both terms of the divergence theorem for a volume enclosed by r = 1 m, r = 2 m, z = 0 m, and z = 10 m. 21