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Transcript
ELECTROSTATICS COULUMB’S LAW ELECTRIC FIELD INTENSITY LINE, SURFACE & VOLUME CHARGES ELECTRIC FLUX DENSITY GAUSS’S LAW ELECTRIC POTENTIAL BOUNDARY CONDITIONS CAPACITANCE 1 Electrostatic Field Problem – Example: Parallel Plate Capacitor Scalar Field: Electrostatic Potential Vector Field: Electrostatic Field Strength 2 Bear in mind!!!! • Electromagnetics is the study of the effect of charges at rest and charges in motion. • Some special cases of electromagnetics: Electrostatics: charges at rest Magnetostatics: charges in steady motion (DC) Electromagnetic waves: waves excited by charges in time-varying motion 3 Electrostatics • Electrostatics is the physics term for static charge. • Electro means charge, and of course static means stationary or not • moving. Charge • 2 Types of Charge: Positive (+) and Negative (-) 4 Electrostatics Attraction and Repulsion Unlike a gravitational force which always attracts, electrostatic force may repel or attract depending on the type charge. Ben's Rule and Paula Abdul - Opposites attract and likes repel. (+) (-) = attract (+) (+) = repel (-) (-) = repel 5 Electrostatics: Microscopic View Electrons have a negative charge and protons have a positive charge. Neutrons and most materials have a net charge of zero or neutral. When certain types of objects are rubbed together, electrons from one object may be transferred to an object with a greater affinity for the electrons. When this happens, the object that gave up the electrons is positive, whereas the object that collected the electrons is negative. Key: Only electrons move. They exist in the outer shells of 6 the atom. • Electric charge is inherently quantized such that the charge on any object is an integer multiple of the smallest unit of charge which is the magnitude of the electron charge e = 1.602 10-19 C. • On the macroscopic level, we can assume that charge is “continuous.” • The following list shows part of the triboelectric sequence. When any two substances shown in this list are rubbed together, the top one will become positively charged while the lower one will become negatively charged. The further apart the two substances are in the list, the greater the electrification. ++ Asbestos Fur (rabbit) Glass Mica Wool Quartz Fur (cat) Lead Silk Human skin, Aluminum Cotton Wood Amber Copper, Brass Rubber Sulfur Celluloid India rubber -7 Static Charges: Coulomb’s Law All bodies are able to take a charge of electricity and this is termed static electricity. The charge on a body is measure by means of the force between the charges. The Coulomb force law, which only applies to charged points, is stated below.. The force of attraction or repulsion between two charged points is directly proportional to the charges and inversely proportional to the square of the distance between them. In vector form, it is stated thus, F12 Q1Q2 40 R 12 2 a12 8 Coulomb’s Law Where; F = Force between points (N) Q1, Q2 = Charges on point 1 and point 2 (Coulomb) R = radial separation on points/distance (m) a12 = the unit vector in the direction from Q1 to 2 o =QPermittivity of the free space (vacuum) with a value given by: 0 8.85 10 12 F 10 9 F m 36 m 9 Coulomb’s Law Charge Q1 exerts a vector force F12 in Newton's (N) on charge Q2, F12 Q1Q2 40 R 12 2 a 12 Note: a negative force results if the points have opposite charges and a positive force results if the points have the same polarity 10 Coulomb’s Law The above equation can be simplified as follows Q1Q2 F k R2 The proportional constant, k is: k 1 40 1 4 , r 0 is constant= 8.99x109 11 Coulomb’s Law • If the space between the charges is another material or air, the law may be written F12 Q1Q2 40 r R 12 2 a12 where r relative permittivity of material. 12 Example 1 Find the force on charge Q1, 20 C, due to charge Q2, -300 C, where Q1 is at (0,1,2)m and Q2 at (2,0,0)m Solution 1 z F1 Q1 R21 (0,1,2) y Q2 x (2,0,0) 13 Solution 1 (cont’d) • Because 1C is a rather large unit, charges are often given in microcoulombs (C), nanocoulombs (nC) and picocoulombs (pC). Referring to figure, R21=-2ax+ay+2az R=3 a21=1/3(-2ax+ay+2az) using Coulmb' s Law equation; F12 40 R 12 20 10 300 10 2a 4 8.854 10 3 6 F21 Q1Q2 6 12 2a x a y 2a z N 6 3 2 x 2 a12 a y 2a z 3 The force magnitude is 6N and the direction is such that Q1 is attracted to Q2 (unlike charges attract) 14 Electric Field Intensity If Q1 is fixed to be at origin, a second charge Q2 will have force acting on Q1 and can be calculated using Coulomb’s Law. We also could calculate the force vector that would act on Q2 at every point in space to generate a field of such predicted force values. 15 Electric Field Intensity (Cont’d) It becomes convenient to define electric field intensity E1 or force per unit charge as: F12 E1 Q2 This field from charge Q1 fixed at origin results from the force vector F12 for any arbitrarily chosen value of Q2 16 Electric Field Intensity (Cont’d) Coulomb’s law can be rewritten as E Q 40 R 2 aR to find the electric field intensity at any point in space resulting from a fixed charge Q. 17 Example 2 Find E at (0,3,4) m in cartesian coordinates due to a point charge Q =0.5C at the origin. Solution to Example 2 a R (3a y 4a z ) / 5 R 3a y 4a z 0.6a y 0.8a z R 5 0.5 106 0.6a y 0.8a z E 2 9 4 10 / 36 5 Thus E 180V / m in the direction a R 0.6a y 0.8a z 18 Example 3 Let a point charge Q1 = 25nC be located at P1 (4,-2,7). If ε = ε0, find electric field intensity at P2 (1,2,3). Solution to Example 3 By using the electric field intensity, E Q 4o R 2 aR This field will be: 25 10 9 E aR 4o R 19 Solution to Example 3 (Cont’d) Where, R12 r2 r1 3a x 4a y 4a z and R 41 E Q 4o R 2 aR Q 4o R 25 10 9 4 8.854 10 ?? 12 41 3 3 R 3a 2 x 4a y 4a z 20 Electric Field Intensity (Cont’d) If there are N charges, Q1,Q2...QN located respectively at point with position vectors r1,r2...rN the electric field intensity at point r is: E Q1 4 0 r r1 E 2 r r1 .. r r1 1 40 N k 1 r rN QN 4 0 r rN 2 r rN Qk r rk r rk 3 21 Field Lines The behavior of the fields can be visualized using field lines: Field vectors plotted within a regular grid in 2D space surrounding a point charge. 22 Field Lines (Cont’d) Some of these field vectors can easily be joined by field lines that emanate from the positive point charge. The direction of the arrow indicates the direction of electric fields The magnitude is given by density of the lines 23 Field Lines (Cont’d) The field lines terminated at a negative point charge The field lines for a pair of opposite charges 24