Electricity and Magnetism PHYSICS (ELECTRICITY AND MAGNETISM) By Assoc Prof Dr Dương Hoài Nghĩa Email : dhnghia@hcmut.edu.vn - dhoainghia@yahoo.com Web site : www4.hcmut.edu.vn/~dhnghia phone: 0918 416 425 Code of module: PH015IU Level: Credits: Credits – hours per week of lectures Prerequisites: Basic Mathematical Analytics, High School Physics Evaluation & grading: 2-hour examination Objectives of module: • Know and understand basic physical processes and phenomena • Solve basic physics problem by applying both theoretical and experimental techniques • Understand and acquire skills needed to use physical laws governing real process and to solve them in the engineering environment Synopsis of module: Chapter 1: Electric Fields • Properties of Electric Charges • Insulators and Conductors • Coulomb’s Law • The Electric Field Electric Field Lines • Electric Field of a Continuous Charge Distribution • Motion of Charged Particles in a Uniform Electric Field • Electric Flux Gauss’s Law • Conductors in Electrostatic Equilibrium • Insulator with Uniform Charge Density Chapter 2: Electric Energy and Capacitance • Potential Difference and Electric Potential • Potential Differences in a Uniform Electric Field • Electric Potential and Potential Energy Due to Point Charges • Electric Potential Due to Continuous Charge Distributions • Electric Potential Due to a Charged Conductor • Capacitance • Combinations of Capacitors • Energy Stored in a Charged Capacitor • Capacitors with Dielectrics Chapter 3: Current and Resistance, Direct Current Circuits • Electric Current • A Model for Electrical Conduction • Resistance and Ohm’s Law • Electrical Energy and Power • Electromotive Force • Kirchoff’s Rules • Resistors in Series and in Parallel • RC Circuits Electricity and Magnetism Chapter 4: Magnetism • The Magnetic Field • Magnetic Force Acting on a Current-Carrying Conductor • Torque on a Current Loop in a Uniform Magnetic Field • Motion of a Charged Particle in a Uniform Magnetic Field • The Hall Effect • The Biot–Savart Law • Ampere’s Law • The Magnetic Field of a Solenoid • Magnetic Flux Gauss’s Law in Magnetism • Displacement Current and the General Form of Ampère’s Law • Magnetism in Matter • The Magnetic Field of the Earth Chapter 5: Electromagnetic Induction • Faraday’s Law of Induction • Motional emf • Lenz’s Law • Induced emf and Electric Fields • Self-Inductance • RL Circuits • Energy in a Magnetic Field • Mutual Inductance Chapter 6: Alternating-Current Circuits • AC Sources and Phasors • Resistors in an ac Circuit • Inductors in an ac Circuit • Capacitors in an ac Circuit • The RLC Series Circuit • Power in an ac Circuit • Resonance in a Series RLC Circuit • The Transformer and Power Transmission Chapter 7: Electromagnetic Waves • Maxwell’s Equations and Hertz’s Discoveries • Plane Electromagnetic Waves • Energy Carried by Electromagnetic Waves • Momentum and Radiation Pressure • Production of Electromagnetic Waves by an Antenna • The Spectrum of Electromagnetic Waves References: Halliday D., Resnick R and Merrill, J (1988) Fundamentals of Physics Extended third edition John Willey and Sons, Inc Alonso M and Finn E.J (1992) Physics Addison-Wesley Publishing Company Hecht, E (2000) Physics Calculus Second Edition Brooks/Cole Faughn/Serway (2006) Serway’s College Physics Thomson Brooks/Cole Electricity and Magnetism Chapter 1.1 ELECTRIC FIELDS Properties of Electric Charges Every object contains a vast amount of electric charge Object which contains equal amounts of the two kinds of charge is call electrically neutral One with an imbalance is electrically charged The net charge of an object is the difference between the amount of positive charge and negative charge of the object Experiment (Fig 1.1 and Fig 1.2): Rub one end of a glass rod with silk → electrons are transfered to silk → the glass rod contains a positive net charge Rub one end of a plastic rod with fur → electrons are transfered to the plastic rod → the plastic rod contains a negative net charge Fig 1.1 : Charges with the same electrical sign repel each other Fig 1.2 : Charges with opposite electrical signs attract each other Electric charge is conserved The net charge of any isolated system can not change Electric charge is quantized Elementary charge is e = 1.602 x 10-19 C The charge of an electron is –e The charge of a proton is +e Electric current dq i= dt [A] (1.1) Electricity and Magnetism 1.2 Insulators and Conductors Conductors : materials through which charge can move rather freely (metal, tap water, human body, …) Insulators (nonconductor) : materials through which charge can not move freely (plastic, glass, chemically pure water, … ) Superconductors : materials that are perfect conductors, allowing charge to move without any hindrance Experiment (Fig 1.3) : Put a plastic rod with negative net charge near a neutral copper rod Conduction electrons on the copper rod are repelled to the far end of the copper rod by the negative charge on the plastic rod Then the negative charge on the plastic rod attracts the remaining positive charge on the near end of the copper rod Fig 1.3 1.3 Coulomb’s Law 1) The electrostatic force of attraction or repulsion between two charged particles (point charges) which are at rest and in vacuum F=k |q1|.|q | r2 [N] (1.2) = 8.99 x 109 Nm2/C2 4πε o εo = 8.85 x 10-12 C2/Nm2 : the permittivity constant where k = and Fig 1.4 Fig 1.5 Fig 1.6 Fig 1.7 2) The force on any one charge due to a collection of other charges is the vector sum of the forces due to each individual charge (Fig 1.7) r r (1.3) F= ∑ Fi i Electricity and Magnetism Experiment (Fig 1.8) a) An aluminum ball with zero net charge assumes a vertical position at the end of a thread or string b) A negatively charged ball is brought close to the neutral ball which becomes polarized c) The positive pole of the aluminum ball is attracted to the negatively charged ball up to contact d) After contact the aluminum ball becomes negatively charged by charge transfer through the point of contact and is repelled by the negatively charged ball e) The aluminum ball will stop at equilibrium in a position deviated from vertical at an angle determined by the charges of the balls Fig 1.8 3) Properties A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at its center If a charged particle is located inside a shell of uniform charge, there is no net electrostatic force on the particle from the shell 1.4 The Electric Field Electric Field Lines 1) Electric field is defined as the electric force per unit charge The direction of the field is taken to be the direction of the force it would exert on a positive test charge The electric field is radially outward from a positive charge and radially in toward a negative point charge 2) To find the electric field at point P near a charged object : Place a positive charge qo (called test charge) at r P Measure the electrostatic force F that acts on the test charge The electric field at point P due to the charged object r r F [V/m, N/C] (1.4) E= qo Electricity and Magnetism 3) Field line diagrams : 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 Electric field lines diverge from positive charges and converge into negative charges Rules for constructing filed lines: a) Field lines begin at positive charge and end at negative charge b) The number of field lines shown diverging from or converging into a point is proportional to the magnitude of the charge c) Field lines are spherically symmetric near a point charge d) If the system has a net charge, the field lines are spherically symmetric at great distances e) Field lines never cross each other 4) The electric field of a point charge E= |q| 4πε o r [N/C] Fig 1.9 : Electric field lines of a point charge Fig 1.10 : Electric field of charges (1.5) Electricity and Magnetism 5) The electric field of an electric dipole : an electric dipole consists of two charges + q and –q, of equal magnitude but opposite sign, that are separated by a distance d (Fig 1.11) E = E+ - E- = E≈ qd 2πεo z3 q q - 2 4πεo r+ 4πεo r− p = = q q - 4πεo (z − d / 2) 4πεo (z + d / 2)2 (1.6) 2πε o z p = qd : dipole moment [Cm] The vector p points from the negative charge to the positive charge (Fig 1.12) Fig 1.11 (1.7) Fig 1.12 1.5 Electric Field of a Continuous Charge Distribution 1) The electric field of a charged ring (Fig 1.13) λ : linear charge density [C/m] ⇒ dq = λds dE = E= λds 4πεo r λds = (1.8) (1.9) 4πεo (z + R ) λzds ∫ dE cos(θ) = ∫ 4πεo (z2 + R )3 / = 2πRλz E= ( 4πεo z + R ) 3/ = ( qz 4πεo z + R ( λz 4πεo z + R 2πR ∫ ds ) 3/ ) 3/ (q = 2πRλ : the total charge on the ring) if z >> R then E = if z = then E = q 4πεo z : from a large distance, the ring looks like a point charge (1.10) Electricity and Magnetism Fig 1.13 : A ring of uniform positive charge Fig 1.14 : A disk of uniform positive charge 2) The electric field of a charged disk (Fig 1.14) σ : surface charge density [C/m2] dq = σdA = σ2πrdr : the charge on the ring with radius r zσ2πrdr dE = ( 4πεo z + R E= ∫ dE = As R → ∞ : E → (1.11) (1.12) ) 3/ zσrdr ∫ 2εo (z2 + R )3 / = σ  z 1 −  2ε o  z2 + R     (1.13) σ : electric field produced by an infinite sheet of uniform charge 2εo 1.6 Motion of charged particles in a uniform electric field 1) Point charge in an electric field : The electrostatic force on a point charge q r r F = qE [N] (1.14) Fig 1.15 2) Fig 1.16 describes the essential features of an ink-jet printer Drops are shot out from generator G and receive a (negative) charge in a charging unit C An input signal from a computer controls the charge given to each drop and thus the effect of field E on the drop and the position on the paper at which the drop lands About 100 tiny drops are needed to form a single character Electricity and Magnetism Let m be the mass of the drop The acceleration of the drop along the vertical axis is ay = qE/m → y = qEt 2m The speed of the drop along the horizontal axis vx = constant → x = vxt Let L be the length of the deflecting plate, the vertical deflection of the drop is y = Fig 1.16 qEL2 2mv x Fig 1.17 3) A dipole in an electric field : Fig 1.17 shows an electric dipole in a uniform external electric field E Two centers of equal but opposite charge are separated by distance d The line between them represents rigid connection The magnitude of the net torque τ = -Eqsin(θ)d = -pEsin(θ) [Nm] (1.15) (by convention, τ < because it tends to rotate the dipole in the clockwise direction) The torque acting on a dipole tends to rotate it into the direction of the field E If we choose the potential energy to be zero when θ = 90o then the potential energy U at any angle θ is θ U= − ∫ rr τdθ = -pEcos(θ) = − pE [J] (1.16) 90 o 4) In a water molecule, the two hydrogen atoms and the oxygen atom not lie on a straight line but form an angle of about 105o Moreover the 10 electrons of the molecule tend to remain closer to the oxygen nucleus than to the hydrogen nuclei This makes the oxygen side of the molecule slightly more negative than the hydrogen side and creates an electric dipole moment p that points along the symmetry axis of the molecule If the water molecule is placed in an external electric field, it is rotated into the direction of the electric field (as shown in Fig 1.17) Fig 1.18: a H2O molecule Electricity and Magnetism 10 Example : A neutral water molecule H2O in its vapor state has an electric dipole moment of magnitude 6.2x10-30 Cm How far apart are the molecule’s centers of positive and negative charge ? If the molecule is placed in an electric field of 1.5x104 N/C, what maximum torque can the field exert on it ? How much work much an external agent to rotate this molecule by 180o in this field, starting from its fully aligned position ? Since there are 10 electrons and 10 protons in a neutral water molecule, the magnitude of its dipole moment is p = qd = 10ed ⇒ d = 3.9 pm The torque on a dipole is maximum when the angle between E and p is 90o τmax = pEsin(90o) = 9.3x10-26 Nm The work done by an external agent -pEcos(180o) - [ -pEcos(0o) ] = 1.9x10-25 J 1.7 Electric flux Gauss’ law 1) Flux of an Electric Field (Fig 1.19) The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field r r Φ = E.dA = E.dA.cos(θ) Fig 1.19 [Nm2/C] (1.17) Fig 1.20 Fig 1.21 It is often simpler to find the flux through one surface of an object than through another In the case of the cone the flux through the base (Area = R2) is the same as the flux through the lateral surface, but it is much easier to calculate the flux through the base = E Alateralcos( ) = E ( R2) 2) Gauss’ Law The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity Φ= Q εo (1.18) Electricity and Magnetism 11 Fig 1.22 Example : A cylindrical Gaussian surface, closed by end caps, is immersed in a uniform electric field The cylinder axis is parallel to the field direction (Fig 1.23) Φ= ∫ r r E.dA = r r E.dA + ∫a r r E.dA + ∫b r r E.dA = ∫c (a : left cap, b : right cap, c : cylinder surface) Fig 1.23 Fig 1.24 Example : A spherical Gaussian surface (of radius r) centered on a point charge q (Fig 1.24) q E= 4πεo r r r Φ = E.dA = ∫ [N/C] q ∫ 4πεor dA = (1.19) π r 2q 4πεo r = q εo [Nm2/C] (1.20) 1.8 Conductors in electrostatic equilibrium 1) The net electric charge of an isolated conductor is located entirely on the outer surface of the conductor Because the mutual repulsion of like charges from Coulomb's law demands that the charges are as far apart as possible, hence on the outer surface of the conductor Using Gauss’ law and this fact, we deduce that the electric field inside the conductor is zero Any net electric field inside the conductor would cause charge to move since it is abundant and mobile 2) The external electric field near the surface of a charged conductor is perpendicular to the surface Because if there were a field component parallel to the surface, it would cause mobile charge to move along the surface This violates the assumption of equilibrium Using Gauss’ law and the fact that the electric field inside the conductor is zero, we deduce the external electric field E= σ εo where σ : surface charge density (1.21) Electricity and Magnetism 12 1.9 Insulator with uniform charge density 1) Infinite nonconducting line of charge : the electric field at any point due to an infinite line of charge with uniform linear charge density λ is perpendicular to the line of charge and has magnitude E= λ 2πε o r (1.22) where r is the perpendicular distance from the line of charge to the point 2) Infinite nonconducting sheet of charge : the electric field due to an infinite nonconducting sheet with uniform surface charge density σ is perpendicular to the plane of the sheet and has magnitude E= σ 2εo (1.23) 3) Nonconducting spherical shell : The electric field outside a spherical shell of charge with radius R, uniform volume charge density and total charge q, is directed radially and has magnitude E= q 4πεo r for r ≥ R , (1.24) The charge behaves, for external points, as if it were all located at the center of the sphere The field inside the shell is E= qr 4πε o R , for r < R (1.25) r is the distance from the center of the shell to the point at which E is measured Problems Electric charges 1.1) What is the magnitude and direction of the electrostatic force on each charge in Fig P1.1 ? The charges are q1 = 10e, q2 = -20e, where e = 1.602 x 10-19 C is the elementary charge, and r = 1mm Fig P1.1 Fig P1.2 1.2) In Fig P1.2, q1 = 10e, q2 = -20e, q3 = -10e, where e = 1.602 x 10-19 C, r = 0.1mm What is the magnitude and direction of the electrostatic force on each charge ? 1.3) A proton and two electrons form three corners of an equilateral triangle with sides of length 3x10-6 m What is the magnitude of the net electrostatic force at each corner? 1.4) Two equally charged particles are held 3.2x10-3m apart and then released from rest The initial acceleration of the first particle is 7m/s2 and that of the second is 9m/s2 The mass of the first particle is 6.3x10-7kg Find the mass of the second particle and the magnitude of the charge of each particle 1.5) The magnitude of the electrostatic force between the two identical ions that are separated by a distance of 5x10-10 m is 3.7x10-9 N What is the charge of each ion ? How many electrons are missing from each ion 1.6) In Fig P1.3, q1 = 10e, q2 = -20e, q3 = -10e, q4 = 20e, where e = 1.602 x 10-19 C, r = 0.1mm What is the magnitude and direction electrostatic force on each charge ? Electricity and Magnetism 13 1.7) In Fig P1.4, two tiny conducting balls of identical mass m and identical charge q hang from nonconducting threads of length L Assume that θ is so small that tan(θ) ≈ sin(θ) a) Find the equilibrium separation x of the balls b) Explain what happens to the balls if one of them is discharged Fig P1.3 Fig P1.4 Fig P1.5 1.8) In crystals of the salt cesium chloride, cesium ions Cs+ form the eight corners of a cube and a chlorine ion Cl- is at the cube’s center (Fig P1.5) The edge length of the cube is r = 0.40 nm The Cs+ ions are each deficient by one electron (and thus each has a charge of +e) a) What is the magnitude of the net electrostatic force exerted on the Cl- ion by the eight Cs+ ions at the corners of the cube ? b) If one of the Cs+ ions is missing, the crystal is said to have a defect What is the magnitude of the net electrostatic force exerted on the Cl- ion by the seven remaining Cs+ ions ? Electric fields 1.9) In Fig P1.6, q1 = q, q2 = -2q, q3 = -q, q4 = 2q, where q is the elementary charge, r = 0.1mm What is the electric field at the center of the square ? Fig P1.6 Fig P1.7 : Plastic rod of charge –Q Fig P1.8 1.10) What is the electric field due to the plastic rod with uniformly distributed charge -Q at point P (Fig P1.7)? 1.11) A thin glass is bent into a semi-circle of radius a as shown below A charge +q is uniformly distributed along one half of the glass, and a charge -q is uniformly distributed along the other half of the ring Use Coulomb’s law to determine the magnitude and direction of the electric field strength at the point P (Fig P1.8) 1.12) Find the electric field a distance z above the midpoint of a straight line segment of length 2L which carries a uniform linear charge density λ 1.13) Find the electric field a distance z above one end of a straight line segment of length L, which carries a uniform linear charge density λ Electricity and Magnetism 14 1.14) Find the electric field of a dipole at B, C, D in Fig P1.9 Fig P1.9 Fig P1.10 1.15) An electric dipole consists of two charges q1 = +2e and q2 = -2e separated by a distance d = 10-9m The electric charges are placed along the y-axis as shown in Fig P1.10 Suppose a constant external electric r r r field E ext = i + j N/C is applied (a) What is the magnitude and direction of the dipole moment? (b) What is the magnitude and direction of the torque on the dipole? (c) Do the electric fields of the charges q1 and q2 contribute to the torque on the dipole? Briefly explain your answer r r r 1.16) An electric dipole with dipole moment p = (3 i + j ) x1.24 x10 -30 Cm is in an electric field r r E = 4000 i N/C (a) What is the potential energy of the electric dipole ? (b) What is the torque acting on it ? (c) If an external agent turns the dipole until its electric r r r p = (-4 i + j ) x1.24 x10 -30 Cm How much work is done by the agent dipole moment is 1.17) The electric field at point P (x,y) of an electric dipole is Ex = x and Ey = -y where Ex and Ey are the r components of the electric field vector E in x and y axis respectively (Fig P1.11) Find and draw the electric field lines (the curves of electric force) Hint : dx / Ex = dy / Ey Fig P1.11 Fig P1.12 r 1.18) In fig P1.12, a uniform, upward electric field E of magnitude E = 2,000 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively The plate has length L = 10 cm and separation d = cm An electron is then shot between the plates from the left edge r of the lower plate The initial velocity vo of the electron makes an angle α = 45° with the lower plate and has magnitude x 106 m/s Will the electron strike one of the plates ? If so which plate and how far horizontally from the left edge will the electron strike ? Electricity and Magnetism 15 Gauss’ law 1.19) Find the electric field at all points due to a long, solid cylinder of radius R and uniform linear charge density λ 1.20) A solid non-conducting sphere of radius R has a uniform charge distribution of volume density ρs Cm-3 Determine an expression for the electric field inside and outside the sphere as a function of the distance from the center of the sphere 1.21) Consider an uncharged metal shell of inner radius a and outer radius b If a charge +Q is placed within the center of the shell, draw a diagram of the electric field around the charge +Q and within the shell Using Gauss' law, determine the strength of the electric field inside, within and outside the shell 1.22) Consider a metal shell of inner radius a and outer radius b What is the charge distribution on the inner surface and outer surface of the shell a) if negative charge is added to the outer surface of the shell from an external source b) if electrons are extracted from outer surface of the shell Determine the strength of the electric field inside, within and outside the shell 1.23) Find the electric field of a long, nonconducting, solid cylinder of radius cm which has a nonuniform volume charge density ρ = Ar2 where A = 2.5µC/m5 and r is the radial distance from the cylinder axis 1.24) A charge distribution that is spherically symmetric but not uniform radially produces an electric field of magnitude E = Kr4, directed radially outward from the center of the sphere Here r is the radial distance from that center and K is a constant What is the volume density ρ of the charge distribution ? 1.25) Use Gauss's Law to find the electric field everywhere due to a uniformly charged insulator shell (Fig P1.13) The shell has a total charge Q, which is uniformly distributed throughout its volume Fig P1.13 Fig P1.14 1.26) In Fig P1.14, a solid nonconducting sphere of radius a = cm is concentric with a spherical conducting shell of inner radius b = 1.5a and outer radius c = 1.7a The sphere has a net uniform charge q1 = +5 fC Determine an expression for the electric field as a function of the distance from the center of the sphere 1.27) In Fig P1.14, a conducting sphere of radius a = mm is concentric with a spherical conducting shell of inner radius b = mm and outer radius c = 3.5 mm The sphere has a net charge q1 = +5 pC Initially the net charge of the conducting shell is zero a) Determine the charge distribution on the conducting shell b) Determine the electric field as function of the distance r from the center of the sphere c) Determine the electric potential as function of the distance r from the center of the sphere Electricity and Magnetism 16 Homeworks H1.1 Four charges q1, q2, q3, q4 form four corners of a square with side r [mm] a) What is the magnitude of the net electrostatic force at each corner ? b) What is the magnitude and direction of the electric field at the center of the square ? Fig H1.1 Fig H1.3 n q1 q2 q3 q4 r 2e -2e 4e -4e 0.2 2e -4e 4e -2e 0.4 4e -2e -4e 2e 0.6 -2e 2e -4e 4e 0.1 2e 4e -4e -2e 0.3 -4e -2e 4e 2e 0.5 6e -6e 8e -8e 0.7 6e -8e 8e -6e 0.8 6e -8e -6e 8e 0.9 10 -6e 6e -8e 8e 0.1 11 6e 8e -8e -6e 0.2 12 -8e -6e 8e 6e 0.3 13 e -e 3e -3e 0.4 14 e -3e 3e -e 0.4 15 3e -e -3e e 0.6 16 -e e -3e 3e 0.7 n q1 q2 q3 q4 r 17 3e -3e 5e -5e 0.2 18 3e -5e 5e -3e 0.3 19 5e -3e -5e 3e 0.4 20 -3e 3e -5e 5e 0.5 21 3e 5e -5e -3e 0.6 22 -5e -3e 5e 3e 0.7 23 7e -7e 9e -9e 0.8 24 7e -9e 9e -7e 0.9 25 7e -9e -7e 9e 0.2 26 -7e 7e -9e 9e 0.3 27 7e 9e -9e -7e 0.4 28 -9e -7e 9e 7e 0.5 29 e -e 3e -3e 0.6 30 e -3e 3e -e 0.7 31 3e -e -3e e 0.8 32 -e e -3e 3e 0.9 n q1 q2 q3 q4 r 33 2e -2e 8e -8e 0.2 34 2e -8e 8e -2e 0.3 35 8e -2e -8e 2e 0.4 36 -2e 2e -8e 8e 0.5 37 2e 8e -8e -2e 0.6 38 -8e -2e 8e 2e 0.7 39 4e -4e 6e -6e 0.8 40 4e -6e 6e -4e 0.9 41 4e -6e -4e 6e 0.2 42 -4e 4e -6e 6e 0.3 43 4e 6e -6e -4e 0.4 44 -6e -4e 6e 4e 0.5 45 e -e 3e -3e 0.6 46 e -3e 3e -e 0.7 47 3e -e -3e e 0.8 48 -e e -3e 3e 0.9 n q1 q2 q3 q4 r 49 4e -4e 3e -3e 0.2 50 4e -2e 3e -5e 0.4 51 -4e 2e 5e -3e 0.6 52 -4e 4e -3e 3e 0.1 53 -4e -2e 3e 5e 0.3 54 4e 2e -5e -3e 0.5 55 8e -8e 7e -7e 0.7 56 8e -6e 7e -9e 0.8 57 -6e 8e 7e -9e 0.9 58 -8e 8e -7e 7e 0.1 59 -8e -6e 7e 9e 0.2 60 8e 6e -9e -7e 0.3 61 3e -3e e -e 0.4 62 3e -e e -3e 0.4 63 -3e e 3e -e 0.6 64 -3e 3e -e e 0.7 Electricity and Magnetism 17 H1.2 Find the electric field a distance z [mm] above one end of a straight line segment of length L [mm], which carries a uniform linear charge density λ [µC/m] n 10 11 12 13 14 15 16 L 10 11 12 13 14 10 11 12 z 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 5 λ n L z λ 17 13 0.2 18 14 0.3 19 0.4 20 0.5 21 0.6 22 0.7 23 10 0.8 24 11 0.9 25 12 0.2 26 13 0.3 27 14 0.4 28 0.5 29 0.6 30 0.7 31 0.8 32 10 0.9 n L z λ 33 11 0.2 34 12 0.3 35 13 0.4 36 14 0.5 37 0.6 38 0.7 39 0.8 40 0.9 41 10 0.2 42 11 0.3 43 12 0.4 44 13 0.5 45 14 0.6 46 0.7 47 0.8 48 0.9 n L z λ 49 0.2 50 10 0.3 51 11 0.4 52 12 0.5 53 13 0.6 54 14 0.7 55 0.8 56 0.9 57 0.2 58 0.3 59 10 0.4 60 0.5 61 10 0.6 62 11 0.7 63 12 0.8 64 13 0.9 H1.3 In Fig H1.3, a solid nonconducting sphere of radius a [mm] is concentric with a spherical conducting shell of inner radius b [mm] and outer radius c [mm] The sphere has a net uniform charge q [fC] Determine an expression for the electric field as a function of the distance from the center of the sphere n a b c q 15 20 22 16 21 23 17 22 24 10 18 23 25 11 19 24 26 12 20 25 27 13 21 26 28 14 22 27 29 15 23 28 30 16 10 24 29 31 17 11 25 30 32 18 12 15 19 22 19 13 16 20 23 20 14 17 21 24 21 15 18 22 25 22 16 19 23 26 23 n a b c q 17 20 24 27 24 18 21 25 28 25 19 22 26 29 26 20 23 27 30 27 21 24 28 31 28 22 25 29 32 29 23 15 21 23 30 24 16 22 24 31 25 17 23 25 32 26 18 24 26 33 27 19 25 27 34 28 20 26 28 35 29 21 27 29 36 30 22 28 30 37 31 23 29 32 38 32 24 30 33 39 n a b c q 33 25 31 34 40 34 15 22 26 41 35 16 23 27 42 36 17 24 28 43 37 18 25 29 44 38 19 26 30 45 39 20 27 31 46 40 21 28 32 47 41 22 29 33 48 42 23 30 34 49 43 24 31 35 50 44 25 32 36 51 45 15 23 26 52 46 16 24 27 53 47 17 25 28 54 48 18 26 29 55 n a b c q 49 19 27 30 24 50 20 28 31 25 51 21 29 32 26 52 22 30 33 27 53 23 31 34 28 54 24 32 35 29 55 25 23 36 30 56 15 24 26 31 57 16 25 27 32 58 17 26 28 33 59 18 27 29 34 60 19 28 30 35 61 20 29 31 36 62 21 30 32 37 63 22 31 33 38 64 23 32 34 39 ... 16 21 23 17 22 24 10 18 23 25 11 19 24 26 12 20 25 27 13 21 26 28 14 22 27 29 15 23 28 30 16 10 24 29 31 17 11 25 30 32 18 12 15 19 22 19 13 16 20 23 20 14 17 21 24 21 15 18 22 25 22 16 19 23. .. 29 21 27 29 36 30 22 28 30 37 31 23 29 32 38 32 24 30 33 39 n a b c q 33 25 31 34 40 34 15 22 26 41 35 16 23 27 42 36 17 24 28 43 37 18 25 29 44 38 19 26 30 45 39 20 27 31 46 40 21 28 32 47 41. .. 16 -e e -3 e 3e 0.7 n q1 q2 q3 q4 r 17 3e -3 e 5e -5 e 0.2 18 3e -5 e 5e -3 e 0 .3 19 5e -3 e -5 e 3e 0.4 20 -3 e 3e -5 e 5e 0.5 21 3e 5e -5 e -3 e 0.6 22 -5 e -3 e 5e 3e 0.7 23 7e -7 e 9e -9 e 0.8 24 7e -9 e