Presentation on theme: "Week 3 Electric Charge, Force, and Field Charge Density, Current Density Continuity Equation."— Presentation transcript:
Week 3 Electric Charge, Force, and Field Charge Density, Current Density Continuity Equation
Electric Charge Around 600 BC, Thales of Miletus reported that charge could be accumulated by rubbing fur on various substances such as amber. He noted that the charged amber buttons could attract light objects such as hair. He also noted that if he rubbed the amber for long enough, he could even get a spark to jump. Amber is “electron” in Greek, which is spelled ήλεκτρον
. Electric Charge Convention The convention of positive and negative charges is accredited to Benjamin Franklin. The electron charge turned out to be negative under this convention. The electron has a negative charge of 1.6 x 10 -19 As.
Electric Charge Convention The force of a positive charge is radially outward. The force of a negative charge is radially inward. If the charges are opposite, the charges will attract each other. If the charges are similar, the charges will repel each other.
The Electroscope It consists of a vertical metal rod from the end of which hang two parallel strips of thin flexible gold leaf. A metal ball is attached to the other end where the charge to be tested is applied. When the metal sphere is touched with a charged object the gold leaves spread apart in a 'V'. Since the leave they receive the same sign charge they repel each other.
Electronic Electroscope Build the circuit below to detect the electric field E surrounding an electrified object Sensor built atop a 9v battery connector
Coulomb’s Law The law of electrostatic attraction and repulsion, developed in the 1780s by French physicist Charles Augustine de Coulomb, may be stated in scalar form as follows: The magnitude of the electrostatic force between two point electric charges is directly proportional to the product of the magnitudes of each charge and inversely proportional to the square of the distance between the charges.
Inverse-Square Law The magnitude of the force exerted by two charges over each other is given by F = k Q 1 Q 2 Newtons (N= kg m s -2 ) R 2 k = __1___ = 9 x 10 9 N m 2 / (As) 2 4πε o ε o = 8.85 x 10 -12 F/m electric constant or permittivity of free space
Exercise Find the magnitude of the force F for Q 1 = 5 nAs (or nC in old convention), Q 2 = -3 nAs, and R = 2 m.
Inverse-Square Law Lines represent the “flux” emanating from the source. The total number of flux lines depends on the strength of a charge and is constant with increasing distance. A greater density of flux lines (lines per unit area) means a stronger force field.
Inverse-Square Law The density of flux lines is inversely proportional to the square of the distance from the source because the surface area of a sphere increases with the square of the radius. Thus the strength of the field is inversely proportional to the square of the distance from the source.
Electric Force Vector The force exerted by Q1 on a positive unit charge located at a distance R is given by F 12 = k Q 1 Q 2 a 12 Newtons ( N = kg m s -2 ) R 2 The direction of the force due to a positive charge Q 1 (x 1, y 1, z 1 ) at an observation point P (x 2, y 2, z 2 ) is given by: a 12 = x (x 2 – x 1 ) + y (y 2 – y 1 ) + z (z 2 – z 1 ) [(x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 + (z 2 – z 1 ) 2 ] 1/2
Exercise Determine the force F that Q 1 = 1 x 10 -9 (C) exerts on Q 2 = 2 x 10 -9 (C). Assume x and y coordinates are given in meters.
Electric Field The space surrounding an electric charge has a property called an electric field. This electric field exerts a force on other electrically charged objects. The electric field (force per unit charge) at a distance R is given by E = k Q 1 a 12 Newtons/meter (N/m) R 2
Exercise Find the force exerted on a charge Q 1 = + 10 nAs by a field E = 3 N/m a y.
Electric Dipole The simplest example of a pair of electric charges of equal magnitude but opposite sign, separated by some, usually small, distance. The E field of an electric dipole behaves as shown here.
Electric Dipole In the far field (R>>L), the electric field due to two point charges on the z axis separated by a short distance L is given by: E = QL (2cosθ a R + sinθ a θ ) 4πε o R 2 Dipoles can be characterized by their dipole moment p = QL z For the simple electric dipole given above, the electric dipole moment would point from the negative charge towards the positive charge, and have a magnitude equal to the strength of each charge times the separation between the charges.
Exercise Calculate the divergence of the far field produced by an electric dipole E = QL (2cosθ a R + sinθ a θ ) 4πε o R 2
Institute of Electrical and Electronics Engineers The logo of the or IEEE (read eye- triple-e) depicts Franklin’s kite, the electric dipole symbol, and the magnetic dipole symbol. IEEE is an international non-profit, professional organization for the advancement of technology related to electricity. It has the most members of any technical professional organization in the world, with more than 365,000 members in around 150 countries.
Charge Distributions Electric field intensity away from an infinite line charge distribution ρ along z axis E = ____ρ_____ a r 2πε o r Electric field intensity away from an infinite surface charge distribution ρ s on x-y plane E = ____ρ s _____ a z z >0 2ε o
Charge Density The linear, surface, or volume charge density is the amount of electric charge in a line, surface, or volume. It is measured in Coulombs per meter (As/m), square meter (C/m²), or cubic meter (C/m³), respectively. Since there are positive as well as negative charges, the charge density can take on negative values. Like any density it can depend on position.
Notation Different variables are used to denote the various dimensions of charge density. Commonly used variables: λ or ρ l for line charge density (C/m) σ or ρ s for surface charge density(C/m²) ρ or ρ v for volume charge density (C/m³)
Differentials Length dx, dy, or dz r dφ, dz(angular aperture on x-y plane) R sinθ dφ (angular aperture) Surface dx dy, dx dz, or dy dz r dφ dz(cylindrical surface) R 2 sinθ dθ dφ(spherical surface) Volume dx dy dz r dr dφ dz(cylindrical volume) R 2 sinθ dR dθ dφ(spherical volume)
Line Charge Density Consider a line density λ on a thing wire along the y axis as shown here.
Surface Charge Density Consider a surface density σ on a thing wire along the y axis as shown here.
Volume Charge Density Consider a volume density ρ in the spherical configuration shown here.
Getting E from a Charge Distribution Use Coulomb’s Law: The field differential dE at an observation point P(x, y, z) is proportional to the charge differential dQ The field differential dE is inversely proportional to the square of the distance R’ between the observation point P and where dQ is located The total field E is the integral of dE Take advantage of geometrical symmetry whenever possible
Field Produced by Line Charge The field differential dE z is: proportional to the charge differential dQ = λdr Inversely proportional to R’ = (r 2 + z 2 ) ½ The projection of dE along the z axis Due to symmetry: The r component of E will be zero on the z-axis E will point in the z- direction
Field Produced by Ring Charge The field differential dE z is: proportional to the charge differential dQ = λ rdφ Inversely proportional to R’ = (r 2 + z 2 ) ½ The projection of dE along the z axis Due to symmetry: The r component of E will be zero E will point in the z- direction
Exercise Assume that σ is constant. Determine the differential dE produced by a ring Determine the projection of dE along the x axis
Current Density J It’s a measure of the density of flow of the electric charge or electric current per unit area of cross section. In SI units, the electric current density is measured in amperes per square meter. Current density J defined as a vector whose magnitude is the current per cross-sectional area.
Current Density J A common approximation to the current density assumes the current simply is proportional to the electric field, as expressed by J = σ E where E is the electric field strength and σ is the electrical conductivity. Conductivity σ is the reciprocal (inverse) of electrical resistivity and has the SI units of Siemens per meter (S m -1 )
Current I The current through a surface S can be calculated using a surface integral: The current is the net flux of the current density vector field flowing through the surface S.
Practical Issues In the domain of electrical wiring (isolated copper), maximum current density can vary from 4A/mm 2 for a wire isolated from free air to 6A/mm 2 for a wire at free air. If the wire is carrying high frequency currents (above 100kHz) the skin effect may affect the distribution of the current across the section by concentrating the current on the surface of the conductor. In the domain of printed board, for top and bottom layers, maximum current density can be as high as 35A/mm 2 with a copper thickness of 35 µm. Inner layers cannot dissipate as much power as outer layers, thus it is not a good idea to put high power lines in inner layers. In the domain of semiconductor, the maximum current density is given by the manufacturer. But a common average is 1mA/µm (180 nm technology) (Ampere per width of the line).
Exercise The maximum current density of an insulated copper wire is 4 A/mm 2. The conductivity of the metal is 59.6 × 10 6 S/m. Determine the maximum electric field that the wire can sustain.
It’s the principle that electric charge can neither be created nor destroyed. The quantity of electric charge is always conserved. The net flow out of a chosen volume must equal the net change in charge held inside the volume: The surface integral on the left expresses the current outflow from the volume, and the negatively signed volume integral on the right expresses the decrease in the total charge inside the volume.
Continuity Equation From the divergence theorem Hence Because this relation is valid for any volume, no matter how small, no matter where located, we have
Exercise Suppose that a charge of electrons is suddenly injected in a block of silver. As time goes by, the electrons would separate due to Coulomb forces. Let the charge density be given by ρ = ρ o e –αt where ρ o is the initial charge density and α is a material constant. Determine the initial rate of change of this charge density.