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Transcript
Electric Field Concepts Rules for constructing filed lines • A convenient way to visualize the electric field due to any charge distribution is to draw a field line diagram. At any point the field line has the same direction as the electric field vector • Field lines begin at positive charge and end at negative charge • The number of field lines shown diverging from or converging into a point is proportional to the magnitude of the charge. • Field lines are spherically symmetric near a point charge • If the system has a net charge, the field lines are spherically symmetric at great distances • Field lines never cross each other. ELECTRIC POTENTIAL (The Volt) To develop the concept of electric potential and show its relationship to electric field intensity. In moving the object from point a to b, the work can be expressed by: b W F dL a dL is differential length vector along some portion of the path between a and b ELECTRIC POTENTIAL (cont’d) The work done by the field in moving the charge from a to b is b WE field Q E dL a If an external force moves the charge against the field, the work done is negative: b W Q E dL a ELECTRIC POTENTIAL (cont’d) We can defined the electric potential difference, Vab as the work done by an external source to move a charge from point a to point b as: Vba Where, b W E dL Q a Vab Vb Va ‘a’ is the initial point while ‘b’ is the final point If Vab is negative, there is a loss in potential energy in moving Q from ‘a’ to ‘b’; this implies that the work is being done by the field. However. If Vab is positive, there is a gain in potential energy in the movement, an external agent performs the work Vab is independent of the path taken Vab is measured in joules per coulomb, commonly referred to as volts (V) ELECTRIC POTENTIAL (cont’d) Consider the potential difference between two points in space resulting from the field of a point charge located at origin, where the electric field intensity is radially directed, then move from point a to b to have: b b a a 4 0 r Vba E dL Q 2 a r dra r ELECTRIC POTENTIAL (cont’d) Thus, Vba Q r b 4 0 r r a Q 1 1 Vb Va 4 0 b a The absolute potential at some finite radius from a point charge fixed at the origin: V Q 4 0r ELECTRIC POTENTIAL (cont’d) If the collection of charges becomes a continuous distribution, we could find: V Where, L dL V 4 0 r S dS V 4 0 r V dV V 4 0 r dQ 4 0r Line charge Surface charge Volume charge ELECTRIC POTENTIAL (cont’d) The principle of superposition, where applied to electric field also applies to potential difference. Q1 Q2 V 4 0 r r1 4 0 r r2 QN ... 4 0 r rN Or generally, 1 N Qk V 4 0 k 1 r rk ELECTRIC POTENTIAL (cont’d) Three different paths to calculate work moving from the origin to point P against an electric field. Based on figure, if a closed path is chosen, the integral will return zero potential: E dL 0 EXAMPLE 10 Two point charges -4 μC and 5 μC are located at (2,1-,3) and (0,4,-2) respectively. Find the potential at (1,0,1). SOLUTION - EXAMPLE 10 Let So, Q1 4C and Q2 5C Q1 Q2 V 4 0 r r1 4 0 r r2 Where, r r1 1,0,1 2,1,3 1,1,2 6 r r2 1,0,1 0,4,2 1,4,2 Therefore, V 1,0,1 26 Q1 Q2 4 0 r r1 4 0 r r2 4 10 6 5 10 6 4 0 6 4 0 26 V 1,0,1 5.872 kV ELECTRIC POTENTIAL (cont’d) The electrostatic potential contours from a point charge form equipotential surfaces surrounding the point charge. The surfaces are always orthogonal to the field lines. The electric field can be determined by finding the maximum rate and direction of spatial change of the potential field. ELECTRIC POTENTIAL (cont’d) Therefore, E V The negative sign indicates that the field is pointing in the direction of decreasing potential. By applying to the potential field: Q Q E V ar a 2 r r 4 0 r 4 0 r IMPORTANT !! Three ways to calculate E: If sufficient symmetry, employ Gauss’s Law. Use the Coulomb’s Law approach. Use the gradient equation. EXAMPLE 11 Consider a disk of charge ρS, find the potential at point (0,0,h) on the z-axis and then find E at that point. SOLUTION TO EXAMPLE 11 Find that, dQ S dS and r h2 2 S dd dQ With V then, S a 2 dd V 4 0 0 0 r 4 0r SOLUTION TO EXAMPLE 11 (Cont’d) How to calculate the integral? Let u h2 2 Integral 1 2 u du and then, leads to du 2 d S V h2 2 2 0 S 2 0 h 2 a 0 a2 h To find E, need to know how V is changing with position. In this case E varies along the z-axis, so simply replace h with z in the answer for V, then proceed with the gradient equation. E V V az z S 1 2 z S z 1a z 1 2 a z 2 2 2 2 0 2 z a 2 0 z a Conductors and Insulators • A conductor is a substance that allows current to flow through it :- they transfer charge across them. • In metals, the current is composed of moving electrons. • Electrolytic solutions also conduct current but by the movement of flow of ions. • Insulators have few mobile electrons or ions and the flow of current is inhibited- They keep tight tabs on their electrons. • As fields are increased, dielectric breakdown of insulators occurs and the current is discharged as a surge. • The dielectric strength is the maximum field an insulator can support. Resistance and Ohm’s Law • Resistance is a measure of resistance to flow of electricity. It is defined by Ohm’s Law as follows: V R I (ohm’s Law) Therefore, resistance is in the units of volts per ampere. One volt per ampere is called an ohm (Ω). The reciprocal of resistance is conductance Capacitance The amount of charge that accumulates as a function of potential difference is called the capacitance. Q C V The unit is the farad (F) or coulomb per volt. Capacitance (Cont’d) Two methods for determining capacitance: Q Method Assume a charge +Q on plate ‘a’ and a charge –Q on plate ‘b’. • Solve for E using the appropriate method (Coulomb’s Law, Gauss’s Law, boundary conditions) • Solve for the potential difference Vab between the plates (The assumed Q will divide out) • Capacitance (Cont’d) V Method • Assume Vab between the plates. • Find E , then D using Laplace’s equation. Find ρS, and then Q at each plate using conductor dielectric boundary condition (DN = ρS ) • • C = Q/Vab (the assumed Vab will divide out) Example 12 Use Q method to find the capacitance for the parallel plate capacitor as shown. Solution to Example 12 Place charge +Q on the inner surface of the top plate, and –Q charge on the upper surface of the bottom plate, where the charge density, Q S S from Q S dS Use conductor dielectric boundary, to obtain: Q D az S from DN S Solution to Example 12 We could find the electric field intensity, E Q E az 0 r S The potential difference across the plates is: a Vab E dL b d Q Qd a z dza z 0 r S 0 0 r S Solution to Example 12 Finally, to get the capacitance: Q C Vab C Q 0 r S d Qd 0 r S Bioelectrical Impedance Analysis • Bioelectrical impedance analysis (BIA) is a commonly used method for estimating body composition. • Since the advent of the first commercially available devices in the mid-1980s the method has become popular owing to its ease of use, portability of the equipment and its relatively low cost compared to some of the other methods of body composition analysis. • It is familiar in the consumer market as a simple instrument for estimating body fat. • BIA actually determines the electrical impedance, or opposition to the flow of an electric current, of body tissues, which can be used to calculate an estimate of total body water (TBW). • TBW can be used to estimate fat-free body mass and, by difference with body weight, body fat. Bioelectrical Impedance Theory When constant electric current is applied between two electrodes through a biological medium and the corresponding voltage is measured between the two source poles, the resultant impedance or bioimpedance is determined by Ohm’s law. The recorded voltage is the sum of the potential difference contributions due to the electrical conductivity properties of the tissue medium. The exchange of electrons from source to sink occurs from electrons of the metal electrode (such as platinum or silver-silver chloride) to ions of the tissue medium. The electrode is the site of charge carrier exchange between electrons and ions and thus serves as a transducer of electrical energy. Impedance measurements most commonly use a twoelectrode (bipolar) of four-electrode (tetrapolar) arrangement. Bioelectrical Impedance Theory The Maxwell equation most relevant to bioimpedance is Eq. (1) H D t J D E P Eq. (2) where H - magnetic field strength [A/m], D - electric flux density [coulomb/m2], J - current density [A/m2], E - electric field strength [V/m], - permittivity of vacuum [farad (F)/m], and P - electric polarization dipole moment pr. volume [coulomb/m2]. • If the magnetic component is ignored, Equation 1 is reduced to: D t J Eq. (3) • Equations 1-3 are extremely robust and also valid under nonhomogeneous, nonlinear, and anisotropic conditions. They relate the time and space derivatives at a point to the current density at that point. Bioelectrical Impedance Theory • Impedance and permittivity in their simplest forms are based on a basic capacitor model. • Basic equation of bioimpedance is then (time vectors) Y G jC Typical body segment resistance values SUMMARY (1) The force exerted on a charge Q1 on charge Q2 in a medium of permittivity ε is given by Coulomb’s Law: F12 Where Q1Q2 4 R12 R12 R12a12 2 a12 is a vector from charge Q1 to Q2 Electric field intensity E1 is related to force F12 by: F12 E1 Q2 SUMMARY (2) The Coulomb’s Law can be rewritten as: E Q 4 0 R 2 aR For a continuous charge distribution: dQ E a 2 R 4 0 R For a point charge at origin: E Q 4 0 r 2 ar SUMMARY (3) •For an infinite length line charge ρL on the z axis L E a 2 •For an infinite extent sheet of charge ρS S E aN 2 Electric flux density, D related to field intensity by: D r 0E Where εr is the relative permittivity in a linear, isotropic and homogeneous material. SUMMARY (4) Electric flux passing through a surface is given by: D dS Gauss’s Law states that the net electric flux through any closed surface is equal to the total charge enclosed by that surface: D dS Qenc Point form of Gauss’s Law is D V SUMMARY (5) The electric potential difference Vab between a pair of points a and b in an electric field is given by: b Vab E dL Vb Va a Where Va and Vb are the electrostatics potentials at a and b respectively. For a distribution of charge in the vicinity of the origin, where a zero reference voltage is taken at infinite radius: dQ V 4r SUMMARY (6) E is related to V by the gradient equation: E V Which for Cartesian coordinates is: V V V V ax ay az x y z • The conditions for the fields at the boundary between a pair of dielectrics is given by: ET 1 ET 2 and a 21 D1 D2 S SUMMARY (7) Where ET1 and ET2 are the electric field components tangential to the boundary, a21 is a unit vector from medium 2 to 1 and ρS is the surface charge at the boundary. If no surface charge is present, the components of D normal to the boundary are equal: D N1 D N 2 At the boundary between a conductor and a dielectric, the conditions are: ET 0 and DN S SUMMARY (8) Poisson’s equation is: V V 2 Where the Laplacian of V in Cartesian coordinates is given by: 2 2 2 V V V 2 V 2 2 2 x y z In a charge free medium, Poisson’s equation reduces to Laplace’s equation V 0 2 SUMMARY (9) • Capacitance is a measure of charge storage capability and is given by: Q C V For coaxial cable: L Vab ln b a 2 So, 2L C ln b a For two concentric spheres: Q 1 1 Vab 4 a b So, 4 C 1 a 1 b