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Chapter 2 - Electrostatics Part A 2.0 Introduction Field theory lays the foundation of electrical, electronics, telecom and computer engineering. It is used in understanding the principle of radar, remote sensing, satellite communication, TV reception, microwave devices, fiber optics, electromagnetic compatibility, etc. 2.0 Introduction A number of phenomena, which cannot be adequately explained by circuit theory, can possibly be explained by field theory. In circuit theory, voltages and currents are the main system variables, and they are treated as scalar quantities. On the other hand, most of the variables in electromagnetic field theory are vectors having magnitude as well as direction. 2.0 Introduction In a simple term, electromagnetic is the study of the effects of electric charges at rest and in motion Both positive and negative charges are sources of electric field Moving charges produce a current, which gives rise to magnetic field A time varying electric field is accompanied by a magnetic field, and vice versa 2.1 Nature of Electric Charge In general, an atom is electrically neutral (total number of positive charge of protons = total number of negative charge of electrons). However, when electrons are removed from atoms of a body, then there will be more positive charges than the negative charges in the atoms and the body is said to be positively charged. On the other hand, if some electrons are added to it, the number of negative charges is more than positive charges, then the body is said to be negatively charged. 2.1 Nature of Electric Charge Electric charge properties: 1) Law of conservation of electric charge The net electric charge can neither be created nor destroyed If a volume contains np protons and ne electrons, then the total charge is q n p e ne e (n p ne )e 2) Principle of linear superposition The total vector electric field at a point in space due to a system of point charges is equal to the vector sum of the electric fields at that point due to the individual charges. 2.1 Nature of Electric Charge First Law of ELECTROSTATICS: “Like charges repel each other, opposite charges attract each other” 2.2 Concept of Electric Field The electromagnetic force consists of an electrical force, Fe and magnetic force, Fm. The electrical force, Fe, is similar to the gravitational force, but with two differences: The source of the gravitational force is mass, whereas the source of the electrical field is electric charge. Electric charge may have positive or negative polarity, whereas mass does not exhibit such a property. 2.2 Concept of Electric Field All matter contains a mixture of neutrons, positively charged protons, and negatively charged electrons. The fundamental quantity of charge is a single electron, usually denoted by the letter e. The unit of electric charge is coulomb (C), named in honor of the 18th century French Scientist Charles Augustin de Coulomb. The magnitude of e is: e 1.6 1019 (C) 2.2 Concept of Electric Field The charge of a single electron is: qe e 1.6 10 19 (C) And that of a proton is equal in magnitude but opposite in polarity: q p e 1.6 1019 (C) Neutrons have no net charge 2.2 Concept of Electric Field Every charged object sets up an electric field in the surrounding space. A second charge "feels" the presence of this field. The second charge is either attracted toward the initial charge or repelled from it, depending on the signs of the charges. Since the second charge also has an electric field, the first charge feels its presence and is either attracted or repelled by the second charge, too. 2.2 Concept of Electric Field If we place two positively charged particles A and B at a distance as shown in figure below, there is a repulsive force between these two particles. This force is of the action-at-a-distance type and can be felt without any intermediate medium between A and B. 2.2 Concept of Electric Field Let us move particle B away. Point P is the point where particle B was placed, and is now in the electric field created by particle A. When particle B is placed back at P, a force F will be exerted on the particle B by the electric field. 2.2 Concept of Electric Field To verify the existence of an electric field at a point, a test charge is placed at that point. If the test charge feels an electric force, then the electric field exists at that point. Thus, an electric field is said to exist at a particular point if an electric force is acting on a charged particle at that point. 2.2 Concept of Electric Field The magnitude and direction of the electric field vector varies from point to point in the field space. Electric field can also vary with time. In this subject we are studying on electric field produced by static electrical charges which is not changing with time. Electrostatic field - field from electric charge at rest. 2.2 Concept of Electric Field Coulomb’s experiments demonstrated that: Two like charges repel one another; whereas two charges of opposite polarity attract The force acts along the line joining the charges, and Its strength is proportional to the product of the magnitudes of the two charges and inversely proportional to the square of the distance between them 2.2 Concept of Electric Field These properties that constitute Coulomb’s Law can be expressed mathematically as: Fe21 q1q2 ˆ Fe21 R (N) (in free space) 12 2 40 R12 where Fe21 is the electrical force acting on +q2 R̂12 +q1 Fe12 R 12 charge q2 due to charge q1 R12 is the distance between th e two charges ˆ is a unit vecto r pointing from charge q to charge q R 12 1 2 0 is a universal constant called the electrical permittivi ty of free space 0 8.854 10 12 farad per meter (F/m) 2.2 Concept of Electric Field Fe21 +q2 R̂12 +q1 Fe12 R 12 The two charges are assumed to be in free space (vacuum) and isolated from all other charges The force Fe12 acting on charge q1 due to charge q2 is equal to force Fe21 in magnitude, but opposite in direction: Fe12 Fe21 2.2 Concept of Electric Field We can express the unit vector R̂12in terms of the position vectors R1 and R2 of the charges q1 and q2 respectively as: z ˆ R 2 R1 R 12 R 2 R1 Fe21 1 40 q1q2 q2 R12 q1 R 2 - R1 R 2 - R1 3 (N) R1 y since R 12 R 2 - R 1 and 1 40 R2 x is called proportion ality constant 2.2 Concept of Electric Field For other medium, q1q2 R 2 - R 1 q1q2 R 2 - R 1 Fe 21 3 4 R 2 - R 1 40 r R 2 - R 1 3 (N) From the expression given above, the force between two-point charge is: directly proportional to the product of the quantities of both charges inversely proportional to the square of the distance between the two charges dependent on the medium in between the two charges 2.3 Electric Field Intensity, E A static electric charge sets up an electric field in the region of space that surrounds it. The quantity we use to measure the strength of the electric field is its intensity. The intensity of an electric field (or electric field strength), E, is the force exerted by the electric field on a unit test charge. 2.3 Electric Field Intensity, E Coulomb’s law states that: An isolated charge, q induces an electric field, E, at every point in space. At any specific point, P, the electric field intensity, E can be expressed as: ˆ ER q 4R 2 (V/m) In the presence of an electric field, E, which may be due to a single charge or a distribution of many charges, the force acting on a test charge, q' when it is placed at that point, is given by F qE (N) 2.4 Streamlines (Flux lines) A convenient method to represent field variations graphically is the use of streamlines, or flux lines. These are directed lines or curves that indicate the direction of the vector field at each point. These continuous lines of force emanate from a positive charge and end on a negative charge. These lines always leave or enter a conducting surface normally. The lines indicate the path that a small positive test charge would take if it were placed in the field. 2.4 Streamlines (Flux lines) A line tangent to a flux line indicates the direction of the electric field at that point. +q Streamlines surrounding a positive charge -q Streamlines surrounding a negative charge 2.4 Streamlines (Flux lines) The arrow shows the direction of the tangential line. The streamlines do not cross among themselves. +q -q Electric field lines between two opposite charges +q +q Electric field lines between two positive charges 2.4 Streamlines (Flux lines) The magnitude of the electric field is represented either by the number of lines (density) passing perpendicularly (normally) through a surface, or by the length of the directed lines. A B Stronger field in region A than B since the density of equal length directed lines is higher q Field strength is strongest at the region closest to q and decreases when it is further away 2.5 Electric Flux In SI units, one line of electric flux is defined as a tube of lines (known as Faraday tube) of force emanates from +1 C and terminates on -1 C charge. Since electric flux is numerically equal to the charge, so unit of flux is measured in coulomb, which is represented by . So = Q coulombs. 2.5 Electric Flux The number of lines of force per unit area is directly proportional to the electric field intensity at that point. They can be related by the equation below: N oE An For free space where N = number of lines of force, An = normal surface area to the field direction at that point, o = the permittivity of the space. 2.5 Electric Flux If A1=A2=A and N1>N2, then N1/A1 > N2/A2 0E1> 0E2, and E1>E2 2.6 Electric Flux Density, D Flux density is given by the normal flux per unit area, denotes as D. If a flux of coulombs passes normally through an area of A m2, then flux density, 2 D A C/m It is related to electric field intensity by the relation, D o r E Therefore D is a vector field similar to which can be represented by lines of force or lines of displacement (or simply "displacement"). 2.6 Electric Flux Density, D D is a vector quantity whose direction at every point is the same as that of E but whose magnitude is o r E One useful property of D is that its surface integral over any closed surface equals the enclosed surface charge. 2.7 Maxwell’s Equations Electromagnetism is based on a set of four fundamental relations known as Maxwell’s equations: D v B E t B 0 D H J t E and D are electric field quantities interrelated by D = εE ε is the electrical permittivity of the material B and H are magnetic field quantities interrelated by B = µH, µ is the magnetic permeability of the material. ρv is the electric charge density per unit volume J is the current density per unit area 2.7 Maxwell’s Equations Maxwell’s equations hold in any material, including free space (vacuum) and at any spatial location All the quantities in Maxwell’s equations may be a function of time, t. These equations are deduced from experimental observations reported by Gauss, Ampere, Faraday and others. Maxwell’s equations form the fundamental tenets of electromagnetic theory. 2.7 Maxwell’s Equations In the static case, none of the quantities appearing in Maxwell’s equations are a function of time ( t 0 ). In this case, all charges are permanently fixed in space, or if they move, they do so at a steady rate so that ρv and J are constant in time. The time derivatives of B and D are zero, and Maxwell’s equations are reduced to: D v E 0 Electrosta tics B 0 H J Magnetosta tics 2.7 Maxwell’s Equations Static and Dynamic Fields Condition Field Quantities Electrostatics Stationary charges Electric Field intensity, E (V/m) Electric Flux density, D (C/m2) D = εE Magnetostatics Steady currents (DC) Magnetic flux density, B (T) Magnetic field intensity, H (A/m) B = μH Dynamics (Time varying fields) Time varying current E, D, B and H 2.7 Maxwell’s Equations Maxwell’s four equations separate into two uncoupled pairs. The first pair involving only the electric field quantities E and D, while the second pair involving only the magnetic field quantities B and H. The electric and magnetic fields are no longer interconnected in the static case This allows the study of electricity and magnetism as two distinct and separate phenomena, as long as the spatial distributions of charge and current flow remain constant in time. 2.8 Charge and Current Distribution There are various forms of electric charge distributions in electromagnetics. If the charges are in motion, they constitute current distribution Charge may be distributed over a volume of space, across a surface, or along a line 2.8 Charge and Current Distribution Charge Densities In electromagnetics, we treat the net charge contained in an elemental volume, ∆v as if it were uniformly distributed within it. Therefore, the volume charge density, ρv can be expressed as v lim v 0 q dq v dv (C/m 3 ) where ∆q is the charge contained in ∆v. In general, ρv is defined at a given point in space (x, y, z), and at a given time, t. 2.8 Charge and Current Distribution Charge Densities Physically, ρv is the average charge per unit volume for a volume ∆v centered at (x, y, z), with ∆v large enough to contain a large number of atoms and at the same time, small enough to be regarded as a point at the macroscopic scale under consideration. The variation of ρv with spatial location is called spatial distribution, or simply distribution. The total charge contained in a given volume, v can be expressed as Q v dv (C) v 2.8 Charge and Current Distribution Charge Densities In conductors, electric charge may be distributed across the surface of a material. The quantity of interest is the surface charge density, ρs which can be defined as q dq s lim ds s 0 s (C/m 2 ) where ∆q is the charge across an elemental surface area, ∆s. 2.8 Charge and Current Distribution Charge Densities Similarly, if the charge is distributed along a line, which need not be straight, the distribution can be characterized as line charge density, ρl , defined as q dq l lim dl l 0 l (C/m) where ∆q is the charge across an elemental length, ∆l. 2.8 Charge and Current Distribution Example 1 – Line charge Distribution Calculate the total charge Q contained in a cylindrical tube of charge oriented along the z-axis as shown. The line charge density is l = 4z, where z is the distance in meters from the bottom end of the tube. The tube length is 10cm. Solution: The total charge Q is 0.1 0.1 0 0 Q l dz 4 zdz 2 z 2 0.1 0 2 102 C 2.8 Charge and Current Distribution Example 2 – Surface Charge Distribution The circular disk of electric charge shown is characterized by a azimuthally symmetric surface charge density that increases linearly with r from zero at the center to 6 C/m2 at r = 3 cm. Find the total charge present on the disk surface. Solution: ρs is symmetrical with respect to the azimuth angle, and can be expressed as a function of r. 2.8 Charge and Current Distribution 6r 2 2 2 10 r (C/m ) 2 3 10 where r is in meters. s In polar coordinates, an element of area is ds = r dr d. For the disk shown, the limits of integration are from 0 to 2 for and from 0 to 3 10-2 m for r. Therefore, Q s ds S 2 3102 0 r 0 2 10 r r drd 2 3 310 r 2 2 10 3 2 0 2 11.31 (mC) 2.8 Charge and Current Distribution Example 3 – Volume Charge Distribution For a spherical shell centered at the origin and extends between R = 3cm and R = 4cm, find the total charge contained in the shell if the volume charge density is 5R×10-4 (C/m3). Solution: v 5R 10 4 (C/m 3 ) Q v dv v 0.04 0.03 2 0 5R 4 10 4 4 5 R 10 4 R 2 sin dR d d 0 0.04 2 2 0.03 5 10 4 0.04 0.03 2.75 (nC) 4 4 2.8 Charge and Current Distribution Current Density Consider a tube of charge with volume charge density ρv moving with a mean velocity, u along the axis of a tube, as shown in the figure below. The charges move a distance ∆l = u ∆t over a period of ∆t. The amount of charge that crosses the tube’s crosssectional surface, ∆s' in time ∆t is therefore 2.8 Charge and Current Distribution Current Density Now consider the more general case where the charges are flowing through a surface ∆s whose surface normal n̂ is not necessarily parallel to u, as shown in figure In this case, the amount of charge ∆q flowing through ∆s is given by q = v u (s cos Ө )t q = vus t 2.8 Charge and Current Distribution Current Density The corresponding current is therefore I = q/t = vus = Js where J = vu (A/m2) J is defined as the current density in ampere per square meter. For an arbitrary surface S, the total current flowing through it is then given by I J ds (A) s 2.8 Charge and Current Distribution Current Density When the current is generated by the actual movement of electrically charged matter, it is called a convection current, and J is called the convection current density. This is distinct from a conduction current, where atoms of the conducting medium do not move. Because the two types of current are generated by different physical mechanism, conduction current obeys Ohm’s law, whereas convection current does not. 2.9 Coulomb’s Law Imagine an electric charge, q located at the origin of a coordinate system. A test charge q’ at a distance R from q experiences a force, which according to Coulomb’s law is: qq ˆ F R (N) 2 4R ˆ is the unit vecto r directing radially from q to q where R Using the above definition the intensity of the electric field surrounding the charge q is therefore: E F ˆ q q ˆ R R (V/m) 2 2 q 4R 40 r R 2.9 Coulomb’s Law This result says that an electric charge located at the origin generates an electric field (E-field) which points radially out from the origin. The strength of the electric field is inverse proportional to the square of the distance from the origin. With F measured in newtons (N) and q' in coulombs (C), the unit of E is (N/C). This is the same as volt per meter (V/m). 2.9 Coulomb’s Law For a material with electrical permittivity ε, the electrical field quantities D and E are related by: D E with r 0 where 0 8.85 10 12 (1 / 36 ) 10 9 (F/m) ε0 is the electrical permittivity of free space, and εr=ε/ε0 is called the relative permittivity (or dielectric constant) of the material. 2.9 Coulomb’s Law For most materials and under most conditions, ε of the material has a constant value independent of both the magnitude and direction of E. If ε is independent of the magnitude of E, then the material is said to be linear because D and E are related linearly, and if it is independent of the direction of E, the material is said to be isotropic. 2.9 Coulomb’s Law Electric field due to multiple point charges The equation for field E due to a single charge can be extended to find the field due to multiple point charges. Consider two point charge, q1 and q2 located at position vector R1 and R2 from the origin of a coordinate system as shown in the Figure x z E E2 E1 R-R1 q1 R1 P R-R2 R q2 R2 y 2.9 Coulomb’s Law Electric field due to multiple point charges At point P with position vector R, the electric field, E1 due to q1 alone is given by E1 q1 R R 1 4 R R 1 3 z E E2 E1 R-R1 (V/m) q1 R1 P R-R2 R q2 Similarly, the electric field due to q2 alone is E2 q2 R R 2 4 R R 2 3 (V/m) x R2 y 2.9 Coulomb’s Law Electric field due to multiple point charges The electric field obeys the principle of linear superposition. The total electric field, E at any point in space is equal to the vector sum of the electric fields induced by all the individual charges. In this case, E E1 E2 1 q1 R R1 q2 R R 2 3 3 4 R R1 R R 2 2.9 Coulomb’s Law Electric field due to multiple point charges Generalizing the preceding result to the case of N point charges, the electric field E at position vector R caused by charges q1, q2, …., qN located at points with position vectors R1, R2, …, RN, is given by E 1 4 N i 1 qi R R i R Ri 3 2.9 Coulomb’s Law Example 4 – Electric Field due to two point charges Two point charges with q1 = 310-5 C and q2 = -510-5 C are located in free space at (2,2,-1) and (0,-2,1), respectively, in a Cartesian coordinate system. Find: (a) the electric field E at (4,2,-1) and (b) the force on a 610-5 charge located at that point. All distances are in meters. Solution: (a) The electric field, E is given by 1 q1 R R1 q2 R R 2 E 3 40 R R1 3 R R 2 2.9 Coulomb’s Law The vectors R1, R2 and R can be R 1 2xˆ 2yˆ zˆ R 2 2yˆ zˆ expressed as: R 4xˆ 2yˆ zˆ Therefore, (b) F q3E 1 32xˆ 54xˆ 4yˆ 2zˆ 5 10 40 8 216 71xˆ 10yˆ 5zˆ 10 -5 (V/m) 4320 E 71xˆ 10yˆ 5zˆ 10 -5 4320 71xˆ 10yˆ 5zˆ 10 -10 (N) 720 6 10 5 2.9 Coulomb’s Law Electric field due to a charge distribution It is possible to have continuous charge distribution along a line, on a surface, or in a volume as shown in the figure below. ρL + + + + + ρs + + + + + + + where + + + Surface charge + + + + + Line charge + + + + + + + ρv + + + + Volume charge L = line charge density, C/m , S = surface charge density,C/m2 , v = volume charge density , C/m3 2.9 Coulomb’s Law Electric field due to a charge distribution Replacing Q with charge element, the electric field intensity : ρL + + + + + + + + ρs + + + + + + Line charge dl ˆ E L 2 R 4o R + + + + + Surface charge ds ˆ E S 2 R 4o R + + + + + + + + ρv + + + + Volume charge E v dv ˆ R 2 4o R where R is the distance of a field point from a differential elements dl, ds or dv contributed by an element dQ of the charge body. 2.9 Coulomb’s Law Example 5 – Line Charge using Coulomb’s law A ring of charge in free space is characterized by a uniform charge density of ρl. The ring has a radius b and positioned in the x-y plane as shown in the figure. Find the electric field intensity, E at a point P(0, 0, h) along the axis of the ring at a distance h from its center. Solution: Consider the electric field generated by a differential segment of the ring, for example, at (b, , 0) as indicated by segment 1 in the figure. This segment has a length, dl = b d and contains charge dq = ρl dl = ρl b d. The vector, R'1 from segment 1 to point P(0, 0, h) is R1 brˆ hzˆ 2.9 Coulomb’s Law R1 R 1 b 2 h 2 Therefore, ˆ R 1 brˆ hzˆ R 1 R 1 b2 h2 The electric field at P(0, 0, h) due to charge of segment 1 is dE1 ˆ l dl l b brˆ hzˆ d R 1 40 R12 40 b2 h2 3 2 1 The field, dE1 has 2 components: (1) dE1r along r̂ and (2) dE1z along ẑ . For symmetry consideration, a field dE2 generated by segment 2 is located diametrically opposite the location of segment 1. This field is identical to dE1, except it has an opposite component r̂ . Hence, the sum of the r̂ components cancelled each other, while the ẑ components add. The sum of the two contributions can be expressed as dE dE1 dE 2 zˆ l bh 1 d 32 2 2 20 b h 2.9 Coulomb’s Law For every ring segment in the semicircle defined over the range 0 ≤ ≤ π, there is a corresponding segment located diametrically opposite at ( + π). The total field generated by the ring can be obtained by integrating the above equation over a semicircle as: l bh 1 d 20 b 2 h 2 3 2 0 bh 1 zˆ l 2 0 b 2 h 2 3 2 E zˆ zˆ h 40 b h 2 2 32 Q where Q 2bl is the total charge contained in the ring. 2.9 Coulomb’s Law Example 6 – Surface Charge using Coulomb’s law z Find the expression of E for an infinite sheet of charge in the x-y plane with uniform charge density, ρs. P (0,0,h) 2 Solution: The charge associated with an elemental area, ds is the total charge is therefore dQ s ds Q s ds x The contribution to E field at point P(0,0,h) by the elemental surface shown in the figure is dQ dE From the figure, we can write h 40 R R ppˆ hzˆ , R R dQ s ds s p d dp 2 ˆ R ˆ R p 2 h2 , R R ρs R p 1 y 2.9 Coulomb’s Law Substituting these equations gives: dE s p d dp ppˆ hzˆ 3 40 p 2 h2 Due to the symmetry of charge distribution, there is a corresponding element 2 whose contribution along p̂ cancels that of element 1, as illustrated in the figure. Therefore, the contribution to E p adds up to zero so that E has only z component. Therefore, E dE z s 40 2 0 p 0 hp d dp p 2 h 2 32 zˆ sh sh s 2 2 3 2 2 2 2 1 / 2 ˆ (2 ) p h pdp zˆ p h z zˆ 0 0 40 2 0 2 0 From this expression, we know that an infinite sheet of charge in the x-y plane has only z component. In other words, the electric field is normal to the sheet and it can be expressed generally as E s nˆ 2 0 Note that the electric field is surprisingly independent of the distance between the sheet and the point of observation, P. In a parallel plate capacitor, the electric field between the two plates having equal and opposite charge is given by E s s nˆ (nˆ ) s nˆ 2 0 2 0 0 2.10 Gauss law Gauss’s law states that: The total electric flux passing through any closed surface is equal to the total charge enclosed by that surface. 2.10 Gauss law The surface, S referred here need not necessarily be a physical surface, but can be any hypothetical surface, which encloses the charge. From the figure, D·ds is the electric flux flowing outward through a differential surface element, ds. The total flux through the surface S is equal to the enclosed charge, Q. The surface, S, is called a Gaussian surface. 2.10 Gauss law From the definition, we can write Ψ = charge enclosed = Q That is, d D ds v dv Or s Q D ds v dv s Applying divergence theorem, D ds D dv s v Comparing the two volume integrals results in D v which is the first of the four Maxwell’s equations. 2.10 Gauss law The first Maxwell’s equation states that the volume charge density is the same as the divergence of the electric flux density. Note that Gauss’s law may be stated in different ways: In the integral form In the differential form or point form Gauss’s law is an alternative statement of Coulomb’s law – proper application of the divergence theorem to Coulomb’s law leads to Gauss’s law. Gauss’s law provides an easy means of finding E or D for symmetrical charge distributions. 2.10 Gauss law A continuous charge distribution has rectangular symmetry if it depends only on x (or y or z), cylindrical symmetry if it depends only on ρ, or spherical symmetry if it depends only on r (independent of θ and ) Gauss’s law always hold no matter the charge distribution is symmetry or not. 2.10 Gauss law For an isolated point charge, it does not matter which law is used to determined the expression for E. However, it does matter when we deal with multiple charges or continuous charge distribution. Coulomb’s law can be used to find E for any specified distribution of charge. However, Gauss’s law is easier to apply than Coulomb’s law, but its usage is limited to symmetrical charge distribution that allows assumption on the variations of magnitude and direction of D as a function of spatial location. 2.10 Gauss law At every point on the surface, only the normal component of D at the surface contribute to the integral form of Gauss’s law. The surface, S, should be chosen so that the magnitude of D is constant and it’s direction is either normal or tangential at every point of each subsurface of S. 2.10 Gauss law Example 7 – Point charge using Gauss’s law For a point charge Q located at the origin, find D on a spherical surface containing a point, P as shown in the figure. z Solution: A spherical surface centered at the origin will satisfy the symmetry condition. This surface is called the Gaussian surface. D is always normal to the Gaussian surface, that is: D D r ˆ D P r r Applying Gauss’s law (Ψ = Qenclosed) gives Q D ds Dr ds , and ds 2 Q 0 0 which is the surface area of the Gaussian surface therefore, Q Dr 4r 2 Thus, D Q r̂ 4r 2 y r 2 sin d d 4r 2 , Gaussian surface x 2.10 Gauss law Example 8 – Infinite line charge using Gauss’s law Consider an infinite line of uniform charge ρl (C/m) which lies along the zaxis. Determine the expression of electric flux density, D at a point, P. l Since D has no z component, surfaces of the cylinder. Hence, D l ρˆ or E 2 D ds 0 l ρˆ 20 on the top and bottom ρ P D y Q l l D ds D ds Q D 2l Line charge ρl C/m Gaussian surface Solution: A cylindrical surface containing P is chosen to satisfy the symmetry condition, as shown in the figure. D is constant and always normal to the cylindrical Gaussian surface: D D ρˆ Applying Gauss’s law to an arbitrary length, l of the line, ds 2l is the surface area of the Gaussian surface where Therefore, z x 2.10 Gauss law Example 9 – Infinite sheet of charge using Gauss’s law Find the electric flux density, D at a point, P for an infinite sheet of uniform charge ρs (C/m2) lying on the x-y plane as shown in the figure. Solution: To determine D, a rectangular box that cut symmetrically by the sheet of charge is chosen. Two of its faces are parallel to the sheet. D will be normal to the sheet: D Dz zˆ z D Infinite sheet of charge Applying Gauss’s law, we obtain: Q s ds D ds Dz ds ds top bottom D has no component along x̂ and ŷ and D ds is zero on the sides of the box. Assuming that the top and bottom area of the box x have an area, A, s A Dz ( A A) Thus, D s 2 zˆ or E s zˆ 2 0 P y Gaussian surface Area A 2.10 Gauss law Gaussian surface Example 10 - Uniformly charged sphere A sphere of radius a has a uniform charge of ρv C/m3. Find the electric flux density, D by constructing Gaussian surfaces for cases r ≤ a and r ≥ a separately. r a a Solution: The charge is spherically symmetrical and a spherical surface is an appropriate surface. For r ≤ a, the total charge enclosed by the surface is Q v dv v dv v 2 0 And Hence, Ψ = Q gives r 2 sin dr d d v 0 r 0 2 D ds Dr ds Dr 0 r D v rˆ , 3 r 0 4 3 r 3 r 2 sin d d Dr 4r 2 for 0 r a For r ≥ a, the charge enclosed by the surface is the entire charge, which is, Q v dv v dv v 2 0 while Therefore, D ds Dr 4r 2 a3 D 2 v rˆ , 3r for r a 0 4 3 2 r sin dr d d a v r 0 3 a r