Engineering Electromagnetics Lecture 3 Dr Ing Erwin Sitompul

Engineering Electromagnetics Lecture 3 Dr. -Ing. Erwin Sitompul President University http: //zitompul. wordpress. com President University Erwin Sitompul EEM 3/1

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density n About 1837, the Director of the Royal Society in London, Michael Faraday, was interested in static electric fields and the effect of various insulating materials on these fields. n This is the lead to his famous invention, the electric motor. n He found that if he moved a magnet through a loop of wire, an electric current flowed in the wire. The current also flowed if the loop was moved over a stationary magnet. ►Changing magnetic field produces an electric field. President University Erwin Sitompul EEM 3/2

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density n In his experiments, Faraday had a pair of concentric metallic spheres constructed, the outer one consisting of two hemispheres that could be firmly clamed together. n He also prepared shells of insulating material (or dielectric material), which would occupy the entire volume between the concentric spheres. President University Erwin Sitompul EEM 3/3

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density n Faraday found out, that there was a sort of “charge displacement” from the inner sphere to the outer sphere, which was independent of the medium. n We refer to this flow as displacement, displacement flux, or simply electric flux. n Where ψ is the electric flux, measured in coulombs, and Q is the total charge on the inner sphere, also in coulombs. President University Erwin Sitompul EEM 3/4

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density n At the surface of the inner sphere, ψ coulombs of electric flux are produced by the given charge Q coulombs, and distributed uniformly over a surface having an area of 4πa 2 m 2. n The density of the flux at this surface is ψ/4πa 2 or Q/4πa 2 C/m 2. n The new quantity, electric flux density, is measured in C/m 2 and denoted with D. n The direction of D at a point is the direction of the flux lines at that point. n The magnitude of D is given by the number of flux lines crossing a surface normal to the lines divided by the surface area. President University Erwin Sitompul EEM 3/5

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density n Referring again to the concentric spheres, the electric flux density is in the radial direction : n At a distance r, where a ≤ r ≤ b, n If we make the inner sphere smaller and smaller, it becomes a point charge while still retaining a charge of Q. The electrix flux density at a point r meters away is still given by: President University Erwin Sitompul EEM 3/6

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density n Comparing with the previous chapter, the radial electric field intensity of a point charge in free space is: n Therefore, in free space, the following relation applies: n For a general volume charge distribution in free space: President University Erwin Sitompul EEM 3/7

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density n Example Find the electric flux density at a point having a distance 3 m from a uniform line charge of 8 n. C/m lying along the z axis in free space. For the value ρ = 3 m, • Can you determine the electric flux density at (1, 7, 7) and (3, 4, 5)? President University Erwin Sitompul EEM 3/8

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Electric Flux Density n Example Calculate D at point P(6, 8, – 10) produced by a uniform surface charge density with ρs = 57. 2 μC/m 2 on the plane x = 9. At P(6, 8, – 10), • Can you determine D at (1, 8, 2) and (12, – 2, 7)? President University Erwin Sitompul EEM 3/9

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law n The results of Faraday’s experiments with the concentric spheres could be summed up as an experimental law by stating that the electric flux passing through any imaginary spherical surface lying between the two conducting spheres is equal to the charge enclosed within that imaginary surface. n Faraday’s experiment can be generalized to the following statement, which is known as Gauss’s Law: “The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. ” President University Erwin Sitompul EEM 3/10

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law n Imagine a distribution of charge, shown as a cloud of point charges, surrounded by a closed surface of any shape. n If the total charge is Q, the Q coulombs of electric flux will pass through the enclosing surface. n At every point on the surface the electric-flux-density vector D will have some value DS (subscript S means that D must be evaluated at the surface). President University Erwin Sitompul EEM 3/11

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law n ΔS defines an incremental element of area with magnitude of ΔS and the direction normal to the plane, or tangent to the surface at the point in question. n At any point P, where DS makes an angle θ with ΔS, then the flux crossing ΔS is the product of the normal components of DS and ΔS. President University Erwin Sitompul EEM 3/12

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law n The resultant integral is a closed surface integral, with d. S always involves the differentials of two coordinates ► The integral is a double integral. n We can formulate the Gauss’s law mathematically as: n The charge enclosed meant by the formula above might be several point charges, a line charge, a surface charge, or a volume charge distribution. President University Erwin Sitompul EEM 3/13

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law n We now take the last form, written in terms of the charge distribution, to represent the other forms: n Illustration. Let a point charge Q be placed at the origin of a spherical coordinate system, and choose a closed surface as a sphere of radius a. n The electric field intensity due to the point charge has been found to be: President University Erwin Sitompul EEM 3/14

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Gauss’s Law n At the surface, r = a, President University Erwin Sitompul EEM 3/15

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions n Let us now consider how to use the Gauss’s law to calculate the electric field intensity DS: n The solution will be easy if we are able to choose a closed surface which satisfies two conditions: 1. DS is everywhere either normal or tangential to the closed surface, so that DS d. S becomes either DSd. S or zero, respectively. 2. On that portion of the closed surface for which DS d. S is not zero, DS is constant. n For point charge ► The surface of a sphere. n For line charge ► The surface of a cylinder. President University Erwin Sitompul EEM 3/16

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions n From the previous discussion of the uniform line charge, only the radial component of D is present: n The choice of a surface that fulfill the requirement is simple: a cylindrical surface. n Dρ is every normal to the surface of a cylinder. It may then be closed by two plane surfaces normal to the z axis. President University Erwin Sitompul EEM 3/17

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions n We know that the charge enclosed is ρLL, President University Erwin Sitompul EEM 3/18

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions n The problem of a coaxial cable is almost identical with that of the line charge. n Suppose that we have two coaxial cylindrical conductors, the inner of radius a and the outer of radius b, both with infinite length. n We shall assume a charge distribution of ρS on the outer surface of the inner conductor. n Choosing a circular cylinder of length L and radius ρ, a < ρ < b, as the gaussian surface, we find: n The total charge on a length L of the inner conductor is: President University Erwin Sitompul EEM 3/19

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions n For one meter length, the inner conductor has 2πaρS coulombs, hence ρL = 2πaρS, n Every line of electric flux starting from the inner cylinder must terminate on the inner surface of the outer cylinder: • Due to simplicity, n If we use a cylinder of radius ρ > b, then the total charge enclosed will be zero. ► There is no external field, President University Erwin Sitompul noise immunity and broad bandwidth, coaxial cable is still the most common means of data transmission over short distances. EEM 3/20

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions n Example A 50 -cm length of coaxial cable has an inner radius of 1 mm and an outer radius of 4 mm. The space between conductors is assumed to be filled with air. The total charge on the inner conductor is 30 n. C. Find the charge density on each conductor and the expressions for E and D fields. President University Erwin Sitompul EEM 3/21

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Application of Gauss’s Law: Some Symmetrical Charge Distributions President University Erwin Sitompul EEM 3/22

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence Homework 3 n D 3. 3. n D 3. 4. n D 3. 5. All homework problems from Hayt and Buck, 7 th Edition. n Deadline: 1 May 2012, at 08: 00. President University Erwin Sitompul EEM 3/23