Physics 121 - Electricity and Magnetism Lecture 3 - Electric Field Y&F Chapter 21 Sec. 4 – 7 • • • • • • • • • Recap & Definition of Electric Field Electric Field Lines Charges in External Electric Fields Field due to a Point Charge Field Lines for Superpositions of Charges Field of an Electric Dipole Electric Dipole in an External Field: Torque and Potential Energy Method for Finding Field due to Charge Distributions – Infinite Line of Charge – Arc of Charge – Ring of Charge – Disc of Charge and Infinite Sheet Motion of a charged paricle in an Electric Field - CRT example 1 Copyright R. Janow – Fall 2015 Recap: Electric charge Basics: • • • • Positive and negative flavors. Like charges repel, opposites attract Charge is conserved and quantized. e = 1.6 x 10-19 Coulombs Ordinary matter becomes electrically neutral – screening In conductors, charges are free to move around • screening and induction • In insulators, charges are not free to move around • but materials polarize Coulombs Law: forces at a distance enabled by a field Constant k=8.89x109 Nm2/coul2 3rd law pair of forces F12 q1 repulsion shown r12 q2 F 21 Force on q1 due to q2 (magnitude) F12 k q1q2 2 r12 Superposition of Forces or Fields n Fnet on 1 F1,i F1,2 F1,3 F1, 4 ..... i 2 Copyright R. Janow – Fall 2015 Fields Scalar Field Examples: Vector Field Examples: • Temperature - T(r) • Pressure - P(r) • Gravitational Potential energy – U(r) • Electrostatic potential – V(r) • Electrostatic potential energy – U(r) • Velocity - v(r) • Gravitational field/acceleration - g(r) • Electric field – E(r) • Magnetic field– B(r) Fields “explain” forces at a distance – space altered by source Gravitational Field force/unitmass g(r ) Lim ( m0 0 Fg (r ) m0 m0 is a “test mass” ) versus Electrostatic Field force/unit charge Fe (r ) E(r ) Lim ( ) q0 0 q0 q0 is a positive “test charge” “Test” masses or charges map the direction and magnitudes of fields Copyright R. Janow – Fall 2015 Field due to a charge distribution TEST CHARGE q0 z ARBITRARY CHARGE DISTRIBUTION (PRODUCES E) Fnet Test charge q0: • • small and positive does not affect the charge distribution that produces E. A charge distribution creates a field • r • y x E(r ) Lim ( q0 0 F (r ) q0 most often… F (r ) E(r ) q0 ) • • Map E field by moving q0 around and measuring the force F at each point E field exists whether or not the test charge is present E(r) is a vector parallel to F(r) E varies in direction and magnitude F q0E F = Force on test charge q0 at point r due to the charge distribution E = External electric field at point r = Force/unit charge SI Units: Newtons / Coulomb Copyright R. Janow – Fall 2015 later: V/m Electrostatic Field Examples Field Location Value Inside copper wires in household circuits 10-2 N/C Near a charged comb 103 N/C Inside a TV picture tube 105 N/C Near the charged drum of a photocopier 105 N/C Breakdown voltage across an air gap (arcing) 3×106 N/C E-field at the electron’s orbit in a hydrogen atom 5×1011 N/C E-field on the surface of a Uranium nucleus 3×1021 N/C F E q0 • • • Magnitude: E=F/q0 Direction: same as the force that acts on the positive test charge SI unit: N/C Copyright R. Janow – Fall 2015 Electric Field 3-1 A test charge of +3 µC is at a point P. An external electric field points to the right. Its magnitude is 4×106 N/C. The test charge is then replaced with another test charge of –3 µC. What happens to the external electric field at P and the force on the test charge when the change happens? A. The field and force both reverse direction B. The force reverses direction, the field is unaffected. C. The force is unaffected, the field is reversed D. Both the field and the force are unaffected E. The changes cannot be determined from the information given Copyright R. Janow – Fall 2015 Electric Field due to a point charge Q Coulombs Law test charge q0 1 F Qq 0 4 0 r 2 Find the field E due to point charge Q as a function over all of space r̂ E E F q0E r F E q0 Q 1 4 0 Q r̂ r2 r r̂ r E • Magnitude E = KQ/r2 is constant on any spherical shell (spherical symmetry) • Visualize: E field lines are radially out for +|Q|, in for -|Q| • Flux through any closed shell enclosing Q is the same: F = EA = Q.4r2/40r2 = Q/0 Radius cancels The closed (Gaussian) surface intercepts all the field lines leaving Q Copyright R. Janow – Fall 2015 Use superposition to calculate net electric field at each point due to a group of individual charges Example: for point charges + + Fnet F1 F2 ... Fn F F F F Enet tot 1 2 ... n q0 q0 q 0 q0 E1 E2 ... En Enet at i 1 qi r̂ 2 ij 4 0 j rij Do this sum for every test point i Copyright R. Janow – Fall 2015 Visualization: electric field lines (lines of force) • • • • • Map direction of an electric field line by moving a positive test charge around. The tangent to a field line at a point shows the field direction there. The density of lines crossing a unit area perpendicular to the lines measures the strength of the field. Where lines are dense the field is strong. Lines begin on positive charges (or infinity and end on negative charges (or infinity). Field lines cannot cross other field lines Copyright R. Janow – Fall 2015 DETAIL NEAR A POINT CHARGE STRONG WEAK NEAR A LARGE, UNIFORM SHEET OF + CHARGE • No conductor - just an infinitely large charge sheet • E approximately constant in the “near field” region (d << L) L + + + + + + d q + + + + TWO EQUAL + to + + + + + + + + + + F CHARGES (REPEL) to EQUAL The field has uniform intensity & direction everywhere except on sheet + AND – CHARGES ATTRACT Copyright R. Janow – Fall 2015 Field lines for a spherical shell or solid sphere of charge Shell Theorem Outside: Same field as point charge Inside spherical distribution at distance r from center: • E = 0 for hollow shell; • E = kQinside/r2 for solid sphere Copyright R. Janow – Fall 2015 Example: Find Enet at point “O” on the axis of a dipole • Use superposition • Symmetry Enet parallel to z-axis r z d/ 2 r- z d / 2 kq kq Eat O E E 2 2 r r • Limitation: z > d/2 or z < - d/2 1 1 Eat O kq 2 2 (z d / 2) (z d / 2) z E+ O Er+ Eat O r- z 2kqd 2 2 2 ( z d / 4 ) DIPOLE MOMENT q d points from – to + q p qd Exact For z >> d : point “O” is “far” from center of dipole +q d and 1 z3 since d 1 z z=0 -q Eat O Exercise: Do these formulas describe E at the point midway between the charges Ans: E = -4p/20d3 • • • • qd 1 p 1 2 0 z 3 2 0 z 3 Fields cancel as d 0 so E 0 E falls off as 1/z3 not 1/z2 E is negative when z is negative Does “far field” E look like point charge? Copyright R. Janow – Fall 2015 Electric Field 3-2: Put the magnitudes of the electric field values at points A, B, and C shown in the figure in decreasing order. A) EC>EB>EA B) EB>EC>EA C) EA>EC>EB D) EB>EA>EC E) EA>EB>EC .B .C .A Copyright R. Janow – Fall 2015 A Dipole in a Uniform EXTERNAL Electric Field Feels torque - Stores potential energy (See Sec 21.7) ASSSUME THE DIPOLE IS RIGID p qd Dipole Moment Vector px E Torque = Force x moment arm = - 2 q E x (d/2) sin(q) = - p E sin(q) (CW, into paper as shown) Potential Energy U = -W U dq pE sin(q)dq pE cos(q) UE p E • |torque| = 0 at q = 0 or q = • |torque| = pE at q = +/- /2 • RESTORING TORQUE: (q) (q) OSCILLATOR • U = - pE for q = 0 minimum • U = 0 for q = +/- /2 • U = + pE for q = maximum Copyright R. Janow – Fall 2015 3-3: In the sketch, a dipole is free to rotate in a uniform external electric field. Which configuration has the smallest potential energy? A C B E D E 3-4: Which configuration has the largest potential energy? Copyright R. Janow – Fall 2015 Method for finding the electric field at point P - given a known continuous charge distribution EP This process is just superposition 1 q 1 lim 2 i r̂i 4 0 q 0 i ri 4 0 dq r 2 r̂ 1. Find an expression for dq, the “point charge” within a differentially “small” chunk of the distribution dl for a linear distributi on dq dA for a surface distributi on dV for a volume distributi on 2. Represent field contributions at P due to point charges dq located in the distribution. Use symmetry E 1 q r̂ 4 0 r 2 dE dq 4 0r 2 r̂ 3. Add up (integrate) the contributions dE over the whole distribution, varying the displacement and direction as needed EP dE (line, surface, or volume integral) dist Copyright R. Janow – Fall 2015 Example: Find electric field on the axis of a charged rod • • Rod has length L, uniform positive charge per unit length λ, total charge Q. Q/L. Calculate electric field at point P on the axis of the rod a distance a from one end. Field points along x-axis. dq dx 1 dq 1 dx dE 4 0 x 2 4 0 x 2 • Add up contributions to the field from all locations of dq along the rod (x [a, L + a]). L a E a dx 4 0 x 2 4 0 E L a a Q 4 0a(L a) dx x 2 4 0 L a 1 x a • Interpret • L => • L << • L >> 1 Q 1 1 4 0 L a L a Limiting cases: 0 ? (rod is point charge a ? (same) a ? Copyright R. Janow – Fall 2015 Electric field due to an ARC of charge, at center of arc L • Uniform linear charge density = Q/L dq dq = ds = Rdq • P on symmetry axis at center of arc Net E is along y axis need Ey only kdq k dq cos(q) dEP 2 r̂ dEP, y ĵ 2 R R • Angle q is between –q0 and +q0 R q q0 dE q0 P • Integrate: q0 1 1 q0 EP, y R cos( q ) d q sin( q ) | q0 4 0 R 2 q0 4 0 R note : cos(q)dq sin(q) EP,y 2k sin(q0 ) ĵ R • For a semi-circle, q0 = /2 EP,y 2k ĵ R and sin( q) sin(q) In the plane of the arc • For a full circle, q0 = EP, y 0 Copyright R. Janow – Fall 2015 Electric field due to a straight LINE of charge Point P on symmetry axis, a distance y off the line L dq = dx x q0 q0 y r q dEP ( q) P dEP ( q) solve for line segment, then let y << L point “P” is at y on symmetry axis uniform linear charge density: = Q/L by symmetry, total E is along y-axis x-components of dE pairs cancel kdq dEP 2 r̂ r dEP, y k dq cos( q) r 2 ĵ • “1 + tan2(q)” cancels in numerator and denominator k cos(q)dq dE y,P ĵ y • Integrate from –q0 to +q0 k q0 k q E y,P ĵ cos(q)dq ĵ sin(q) q0 0 y q0 y x y tan(q) r 2 x2 y2 r 2 y2 [1 tan2 (q)] Find dx in terms of • • • • SEE Y&F Example 21.10 q: dx d [tan(q)] d sin(q) y y dq dq dq cos(q) 2 y [1 tan (q)] dx y [1 tan (q)] dq 2 dq dx y [1 tan2 (q)] dq 2k EP ĵ sin(q0 ) y Finite length wire • As q0 /2 (y<<L) LONG wire looks infinite 2k Falls off as 1/y EP ĵ y Copyright Along –y direction R. Janow – Fall 2015 Electric field due to a RING of charge at point P on the symmetry (z) axis • • • • SEE Y&F Example 21.9 Uniform linear charge density along circumference: = Q/2R dq = charge on arc segment of length ds = Rdf P on symmetry axis xy components of E cancel Net E field is along z only, normal to plane of ring kdq dEP 2 r̂ r dEz,P dq ds R df k dq cos(q) r 2 k̂ cos(q) z/r r 2 R2 z 2 k Rzdf dEP,z k̂ 3 r • Integrate on azimuthal angle f from 0 to 2 EP,z k Rz 2 k df 2 2 3 / 2 0 [R z ] 2R Q total charge on disk EP,z kQz [ R2 z2 ] k̂ 3/2 integral = 2 Ez 0 as z 0 (see result for arc) ds R df df • Limit: For P “far away” use z >> R EP,z kQ z2 Ring looks like a point charge if point P is very far away! Exercise: Where is Ez a maximum? Set dEz/dz = 0 Ans:R. Janow z = R/sqrt(2) Copyright – Fall 2015 Infinite (i.e.”large”) uniformly charged sheet Non-conductor, fixed surface charge density Infinite sheet d<<L “near field” uniform field E 2 0 for infinite, non - conducting charged sheet d L Method: solve non-conducting disc of charge Copyright for point on z-axis then approximate z << R R. Janow – Fall 2015 Electric field due to a DISK of charge for point P on z (symmetry) axis See Y&F Ex 21.11 dE • Uniform surface charge density on disc in x-y plane = Q/R2 • Disc is a set of rings, each of them dr wide in radius • P on symmetry axis net E field only along z • dq = charge on arc segment rdf with radial extent dr q s z dq dA r dr df k dq 1 z r dr df dEP,z cos( q ) k̂ k̂ 2 3/2 2 2 4 s 0 r z dA=rdfdr R cos(q) z/s s2 r 2 z 2 P f r x • Integrate twice: first on azimuthal angle f from 0 to 2 which yields a factor of 2 then on ring radius r from 0 to R R 2 r dr EP,z z 2 k 2 3 / 2 4 0 0 [r z ] Note Antiderivative r [r 2 z 2 ]3 / 2 1 Edisk 2 0 d 1 2 dr [r z 2 ]1 / 2 k̂ 1/2 z2 R 2 z For z<< R: disc looks infinitely large P is close to the surface Edisk k̂ 2 0 “Near field” is constant – disc approximates an infinite sheet of Copyright chargeR. Janow – Fall 2015 Motion of a Charged Particle in a Uniform Electric Field F qE Fnet ma •Stationary charges produce E field at location of charge q •Acceleration a is parallel or anti-parallel to E. •Acceleration is F/m not F/q = E •Acceleration is the same everywhere in uniform field Example: Early CRT tube with electron gun and electrostatic deflector ELECTROSTATIC ACCELERATOR PLATES (electron gun controls intensity) FACE OF CRT TUBE heated cathode (- pole) “boils off” electrons from the metal (thermionic emission) ELECTROSTATIC DEFLECTOR PLATES electrons are negative so acceleration a and electric force F are in the direction opposite the electric Copyright R. Janow – Fallfield 2015 E. Motion of a Charged Particle in a Uniform Electric Field F qE Fnet ma Kinematics: ballistic trajectory y is the DEFLECTION of the electron as it crosses the field Acceleration has only a constant y component. vx is constant, ax=0 L vx ay x y eE m Time of flight t yields vx L vx t electrons are negative so acceleration a and electric force F are in the direction opposite the electric field E. Measure deflection, find t via kinematics. Evaluate vy & vx eE 2 y t t m v y a y t v v 2x v 2y 1 a 2 y 2 1 2 ( ) 1/ 2 Copyright R. Janow – Fall 2015