Presentation on theme: "EE3321 ELECTROMAGENTIC FIELD THEORY"— Presentation transcript:
1EE3321 ELECTROMAGENTIC FIELD THEORY Week 4Gauss’ Law, Flux DensityGaussian SurfaceElectric PotentialPoisson’s Equation
2Gauss’ LawAs it turns out, Coulomb's law is actually a special case of Gauss' Law, a more fundamental description of the relationship between the distribution of electric charge in space and the resulting electric field.Gauss's law is one of Maxwell's equations, a set of four laws governing electromagnetics.The law bears the name of Karl Friedrich Gauss ( ), one of the greatest mathematicians of all time who also made significant contributions to theoretical physics.
3Differential form of Gauss’ Law In differential form, Gauss' law states that the divergence of the E field is proportional to the charge density that produces it:whereρ is the total electric charge density (in units of C/m³)єo is the electric constant (8.854 x 10-12 F/m)
4ExerciseSuppose that a charge is distributed in a sphere of radius R =1 m. In the region 0 < R < 1 m the electric field is given by E = 1 x 10 – 6 R aR. Find the charge density in this region.Recall the expression for the divergence in spherical coordinates.Find the charge density for R< 1 mPlot the charge density as a function of R.
5Integral form of Gauss’ Law We can make use of the divergence theorem to obtain the integral form of Gauss’s law:whereQ is the charge enclosed by the surface of integration and is equal to the volume integral of the charge density.
6Flux Density We define the electric flux density as Then, the electric flux is given by
7Electric DipoleConsider the case of an electric dipole. What is the total flux coming out of the rectangular surface?
8Gaussian SurfaceA Gaussian surface is a closed two-dimensional surface through which a flux or electric field is to be calculated.The surface is used in conjunction with Gauss's law, allowing one to calculate the total enclosed electric charge by performing a surface integral.Gaussian surfaces are usually carefully chosen to exploit symmetries of a situation to simplify the calculation of the surface integral.If the Gaussian surface is chosen such that for every point on the surface the component of the electric field along the normal vector is constant, then the calculation will not require difficult integration as the constant can be pulled out of the integration sign.
9Spherical SurfaceA spherical Gaussian surface is used when finding the electric field or the flux produced by any of the following:a point chargea uniformly distributed spherical shell of chargeany other charge distribution with spherical symmetryThe spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
10Point Electric ChargeWhat is the total flux out of a spherical surface around a point charge?What is the divergence of D?
11ExerciseLet D = Do R aR for R≤ a. Determine the amount of charge enclosed by a spherical surface of radius b for:b<ab=ab>a
12ExerciseConsider a charged spherical shell of negligible thickness, with a uniformly charge density ρs and radius a.Use Gauss's law to find the magnitude of the resultant electric field E inside the charged shell.
13Tesla CageIn a Tesla cage, the net flux is zero and the magnitude of the electric field is also zero.
14Cylindrical Gaussian Surface A cylindrical Gaussian surface is used when finding the electric field or the flux produced by an infinitely long line of uniform charge.
15The PillboxThis Gaussian surface is used to find the electric field due to an infinite plane of uniform charge.
16Electric PotentialThe electrical potential difference is defined as the amount of work needed to move a unit electric charge from the second point to the first, or equivalently, the amount of work that a unit charge flowing from the first point to the second can perform.The potential difference between two points A and B is the line integral of the electric field E
17Line IntegralThe integration path is an arbitrary path connecting point A of known potential to the observation point B as shown below.The value of V is independent of the integration path.
18Line IntegralThe value of V is independent of the integration path.
19ExerciseLet E = Eo az (V/m) for z>0. Find the potential between the points A and B:A = (0, 0, 0) and B = (1, 1, 1)Steps: set up the integral following a simple pathA to (1, 0, 0)(1, 0, 0) to (1, 1, 0)(1, 1, 0) to B
20Potential at Infinity The potential at infinity is said to be zero. If a test charge +q moves toward a charge +Q starting at an infinite range it will gain potential at point P.
21Gradient of VWhen the magnetic field is constant in time, it is possible to express the electric field as the gradient of the electrostatic potential.
22ExerciseA spherically charged shell of radius a, centered at the origin, creates the potential V = – Voa/R (Volts) for R>a.Determine the corresponding electric field E.Recall the expression of the gradient operator in spherical coordinates.Notice that the potential V is independent of the azimuth and elevation angles.
23ExerciseConsider the potential V = 4 ln ( a/r ) in cylindrical coordinates. Find E.Recall the expression for the gradient operator in cylindrical coordinates.Notice that the potential V is independent of the azimuth angle and the elevation z.
24Equipotential Lines and Surfaces An equipotential line is a line in space where the potential is constant.An equipotential surface is a surface in space where the potential is constant.No work is required for a charge to move along an equipotential surface. Work is required for a charge to move to a different equipotential surface.
25Point ChargeFor a point charge, an equipotential line takes the shape of a circle, and an equipotential surface is a sphere centered on the charge.
26CapacitorConsider two large parallel plates separated by a short distance L.Assuming that the plates have a uniform charge distribution, the electric field lines are perpendicular to the plates and the equipotential lines are parallel to the plates.
27Potential WellIf a test charge approaches the charged sphere it will experience a decreasing potential.The potential is symmetric and decreases as the observation gets closer to the sphere.The concentric orbits in the figure represent lines of equal potential.
28Poisson’s EquationPoisson's equation relates the potential V to the charge density ρ.To determine V one needs to make use of boundary conditions.If the charge density is zero, then Laplace's equation results.Cathode Tube
29ExerciseLet d2V/dz2 = 0 with boundary conditions V(0) = 0 and V(z=d)=100 V. Find for V(z).Integrate Laplace’s Equation over z.Integrate again.Apply boundary conditions.
30Homework Read sections 4-4 and 4-5 Solve the following end-of-chapter problems:4.22, 4.24, 4.28, 4.30, 4.34