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## Electric Fields

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**Electrostatic Phenomena**Coulomb’s Law r 1 q q F1 F2 F = ˆ 1 2 F r 12 12 pe 2 4 r 12 o q1 Superposition Ftotal = F1+ F2 + ... q q2 Text Reference: Chapter 22.1 through 22.3 Examples: 22.1 – 22.5 Last Time…**Define Electric Field in terms of force**on a test charge How to think about fields Electric Field Lines Example Calculation: Electric Dipoles Text Reference: Chapter 22.4 through 22.6 Examples: 22.5, 6, 7, 8, 9, 10, and 11 Today...**Lecture 2, Act 1**Two balls of equal mass are suspended from the ceiling with nonconducting wire. One ball is given a charge +3q and the other is given a charge +q. +q +3q +3q +q (a) (b) g +3q +q Which of the following best represents the equilibrium positions? +3q +q (c)**Lecture 2, Act 1**+q +3q +q +3q +q (a) (b) (c) The force on the +3q charge due to the +qcharge must be equal and opposite to the force of the +3q charge on the +qcharge Amount of charge on each ball determines the magnitude of the force, but each ball experiences the same magnitude of force Symmetry, therefore, demands (c) P.S. Knowing the form of Coulomb’s law you can write two equations with two unknowns (T andq) Which best represents the equilibrium position? +3q Remember Newton’s Third Law!**What is a Field?**(from Lect. 1 notes) A FIELD is something that can be defined anywhere in space It can be a scalar field (e.g., Temperature field) It can be a vector field (e.g., Electric field) Fields represent physical quantities**A Scalar Field (from Lect. 1 notes)**72 73 77 75 71 82 77 84 68 80 64 73 83 82 88 55 66 88 80 75 88 90 83 92 91 These isolated temperatures sample the scalar field (you only learn the temperature at the point you choose, but T is defined everywhere (x, y) )**A Vector Field (from Lect. 1 notes)**72 73 77 75 71 82 77 84 68 80 64 73 83 56 55 57 66 88 80 75 88 90 83 92 91 It may be more interesting to know which way the wind is blowing... That would require a vector field (you learn both wind speed and direction)**The Electric Field**A simple, yet profound observation - The net Coulomb force on a given charge is always proportional to the strength of that charge. q1 = F F1 + F2 F1 F æ ö r ˆ ˆ q q r q r q ç ÷ = + 1 1 2 2 F ç ÷ pe 2 2 4 r r è ø 0 1 2 F2 q2 test charge - Wecan now define a quantity, the electric field, which is independent of the test charge, q, and depends only on position in space: r r F The qiare the sources of the electric field º E q**The Electric Field**r r F º E q With this concept, we can “map” the electric field anywhere in space produced by any arbitrary: F Bunch of Charges Charge Distribution + + - + + + + - + + + - + + - + + + + + + + “Net” E at origin - These charges or this charge distribution “source” the electric field throughout space**Example: Electric Field**y 1) Notice that the fields from the top-right and bottom left cancel at the origin? a a +q +q 2) The electric field, then, is just the field from the top -left charge. It points away from the top-left charge as shown. 3) Magnitude of E-field at the origin is: a a Q x a +q kq = E 2a2 kq cosq = Ex 2a2 kq - sinq = Ey kq 1 2a2 = 2a2 2 kq 1 - = 2a2 2 What is the electric field at the origin due to this set of charges? The x and y components of the field at (0,0) are:**Example: Electric Field**y a a q 1 +q = +q k E x 2 2 a 2 a a x Q a +q Note: If the charge Q is positive, the force will be in the direction of the electric field If the charge Q is negative, the force will be against the direction of the electric field F is F is Now, a charge, Q, is placed at the origin. What is the net force on that charge?**Let’s Try Some Numbers...**If q = 5mC, a = 5cm, and Q = 15mC. y Then Ex= 6.364 106 N/C and Ey= -6.364 106 N/C a a +q +q a a x Fx = QEx and Fy = QEy a Fx=95.5 N Fy=-95.5 N So... +q and We also know that the magnitude of E = 9.00 106 N/C We can, therefore, calculate the magnitude of F F = |Q| E = 135N**Lecture 2, Act 2**y E Q1 d Q2 x Two charges, Q1 and Q2, fixed along the x-axis as shown produce an electric field, E, at a point (x,y)=(0,d)which is directed along the negative y-axis. - Which of the following is true? (a) Both chargesQ1 and Q2are positive (b) Both chargesQ1andQ2are negative (c) The chargesQ1andQ2have opposite signs**Lecture 2, Act 2**y E Q1 d Q2 x Q1 Q1 Q2 Q2 Q1 Q2 Two charges, Q1 and Q2, fixed along the x-axis as shown produce an electric field, E, at a point (x,y)=(0,d)which is directed along the negative y-axis. - Which of the following is true? (a) Both chargesQ1 and Q2are positive (b) Both chargesQ1andQ2are negative (c) The chargesQ1andQ2have opposite signs E E E (a) (b) (c)**Reality of Electric Fields**• The electric field has been introduced as a mathematical convenience, just as the gravitational field of Physics 111 There is MUCH MORE to electric fields than this! IMPORTANT FEATURE:E field propagates at speed of light • NO instantaneous action at a distance (we will explain this when we discuss electromagnetic waves) • i.e., as charge moves, resultant E-field at time tdepends upon where charge was at timet - dt For now, we avoid these complications by restricting ourselves to situations in which the source of the E-field is at rest. (electrostatics)**Mathematics of Fields**Scalar Fields • Number associated with each point in space • May be time dependent (future) • Expressed as a functiong(x, y, z, t) Vector Fields • Vector associated with each point in space • May be time dependent (future) • Expressed as a functionG(x, y, z, t) Physical Fields • Obey a simple rule • Created by sources • Continuous and well behaved • What field looks like depends on rule and sources z y x Pick a coordinate system… This specifies the formula: g(x, y, z, t) G(x, y, z, t)**Ways to Visualize the E Field**vector map field lines + chg + chg + + Consider the E-field of a positive point charge at the origin**Rules for Vector Maps**+ chg + Direction of arrow indicates direction of field Length of arrowsµlocal magnitude of E**Rules for Field Lines**- + Field at two white dots differs by a factor of 4 sincer differs by a factor of 2 Local density of field lines also differs by a factor of 4 (in 3D) • Lines leave (+) charges and return to (-) charges • Number of lines leaving/entering chargeµamount of charge • Tangent of line = direction of E • Local density of field linesµlocal magnitude of E**y**2a +Q a a x a -Q (b) Ex(2a,a) = 0 (c) Ex(2a,a) > 0 (a) Ex(2a,a) < 0 Ex Solution: Draw some field lines according to our rules. NOTE– from Lecture 3, ACT 1 • Consider a dipole with the y-axis as shown. • Which of the following statements aboutEx(2a,a)is true?**Field Lines From Two Opposite Charges**Dipole Dipoles are central to our existence!**The Electric Dipole**y +Q a x q a E E r -Q Ex = ?? Ey = ?? Symmetry Ex(x,0) = 0 see the appendix for further information What is the E-field generated by this arrangement of charges? Calculate for a point along x-axis: (x, 0)**Electric Dipole Field Lines**What can we observe about E? Ex(x,0) = 0 Ex(0,y) = 0 We derived: ... for r >> a, Lines leave positive chargeand return to negative charge Field largest in space between two charges**Field Lines From Two Like Charges**• There is a zero halfway between the two charges • r >> a: looks like the field of point charge (+2q) at origin**y**+ + + + + + + + + R + + + + x + + + + + + + + + The key thing to remember here is that the total field at the origin is given by the VECTOR SUM of the contributions from all bits of charge. If the total field were given by the ALGEBRAIC SUM, then (b) would be correct. (exercise for the student). Note that the electric field at the origin produced by one bit of charge is exactly cancelled by that produced by the bit of charge diametrically opposite!! Therefore, the VECTOR SUM of all these contributions is ZERO!! Lecture 2, ACT 3 Consider a circular ring with total charge +Q. The charge is spread uniformly around the ring, as shown, so there is λ = Q/2pR charge per unit length. The electric field at the origin is (b) (a) zero (c)**Electric Field inside a Conductor**2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ There is never a net electric field inside a conductor – the free charges always move to exactly cancel it out. • A two electron atom, e.g., Ca • heavy ion core • two valence electrons • An array of these atoms • microscopically crystalline • ions are immobile • electrons can move easily • Viewed macroscopically: • neutral**Summary**Define E-Field in terms of force on “test charge” How to think about fields Electric Field Lines Example Calculation: Electric Dipole Reading Assignment: Chapter 23.1 through 23.2 Examples: 23.1 and 23.4**Appendix A:**Other ways to Visualize the E Field Graphs Ex, Ey, Ezas a function of(x, y, z) Er, Eq, Efas a function of(r, q, f) Field Lines + chg Ex(x,0,0) + x Consider a point charge at the origin**Appendix A**y Consider a point charge fixed at the origin of a coordinate system as shown. –Which of the following graphs best represent the functional dependence of the Electric Field for fixed radius r? r f x Q Er Er Er 3A Fixed r > 0 2p 0 0 f 2p 0 f f 2p (a) (b) (c) Ex Ex Ex 3B Fixed r > 0 0 0 0 f f 2p f 2p 2p**Appendix “ACT”**y Consider a point charge fixed at the origin of a coordinate system as shown. – Which of the following graphs best represent the functional dependence of the Electric Field for fixed radiusr? r f x Q Er Er Er 3A Fixed r > 0 2p 0 0 f 2p 0 f f 2p (a) (b) (c) • At fixed r, the radial component of the field is a constant, independent off!! • For r>0, this constant is > 0. (note: the azimuthal component Ef is, however, zero)**Appendix “ACT”**Ex Ex Ex At fixed r, the horizontal component of the field Ex is given by: 3B Fixed r > 0 0 0 0 f f 2p f 2p 2p y Consider a point charge fixed at the origin of a coordinate system as shown. –Which of the following graphs best represent the functional dependence of the Electric Field for fixed radius r? r f x Q (a) (b) (c)**Appendix B: Electric Dipole**y +Q a x q a E E r -Q What is the E-field generated by this arrangement of charges? Calculate for a point along x-axis: (x, 0) Ex = ?? Ey = ?? Symmetry Ex(x,0) = 0**Coulomb Force**Radial E y Electric Dipole +Q What is the Electric Field generated by this charge arrangement? a x a -Q Now calculate for a pt along y-axis: (0,y) Ey = ?? Ex = ??**y**Electric Dipole r +Q a x a Case of special interest: (antennas, molecules) r > > a -Q For pts along x-axis: For pts along y-axis: For r >>a For r >>a,**Field Lines**• Consider the E-field produced by a single point charge • The field is spherically symmetric + chg 1) magnitude of E-field depends only on R 2) at radius R, magnitude = E = kQ/R2 Suppose N field lines at surface of sphere of radius R Then this is the density of the field lines at the surface: + Note: density of field lines is only proportional to the magnitude of the field as a result of E being inversely proportional to R2 This concept will become more useful (and physical) when we discuss Gauss’ Law in terms of electric flux! Since Coulomb’s Law is an inverse square law, we can use the density of the field lines to represent the magnitude of the field!