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Unformatted text preview: Lecture slides are posted each Thursday for that week. Thursday, September 9, 2010 What is an electric flux through spheres of radii r1 and r2= 2r1 surrounding a charge of q=-3 e? A : Φ1 = Φ2 ￿= 0 B : Φ1 = Φ2 = 0 C : Φ1 = Φ2 /2 r2 -3e r1 D : Φ1 = 2Φ2 E: Φ1 = Φ2 /4 2 Thursday, September 9, 2010 What is a electric flux through spheres of radii r1 and r2= 2r1 surrounding a charge of q=-3 e? A : Φ1 = Φ2 ￿= 0 B : Φ1 = Φ2 = 0 C : Φ1 = Φ2 /2 D : Φ1 = 2Φ2 E: Φ1 = Φ2 /4 r2 -3e r1 equal: A 3 Thursday, September 9, 2010 iClicker What is the flux through the spherical surface encompassing two charges q1= 3q, q2=-2q A q 5 ￿0 B C q − q ￿0 ￿0 D C 3q -2q q 3 ￿0 0 4 Thursday, September 9, 2010 iClicker What is the flux through the spherical surface encompassing two charges q1= 3q, q2=-2q A q 5 ￿0 B C q − q ￿0 ￿0 D C 3q q 3 ￿0 -2q qtot = 3q − 2q = q 0 5 Thursday, September 9, 2010 iClicker The total electric flux through two closed surfaces is A: equal B: not equal 6 Thursday, September 9, 2010 iClicker The total flux of electric field through the closed surface is A: > 0 B: =0 C: <0 +3q +3q +3q +3q +3q -q 7 Thursday, September 9, 2010 Gauss’s Law for Various Charge We have applied Gauss’s Law to a point charge and showed that we get Coulomb’s Law Now let’s look at more complicated distributions of charge and calculate the resulting electric field We will use a “charge density” to describe the distribution of charge This charge density will be different depending on the geometry August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 64 Cylindrical Symmetry (1) Let’s calculate the electric field from a conducting wire with charge per unit length λ using Gauss’s Law ￿ q E · dA = ￿0 We start by assuming a Gaussian surface in the form of a right cylinder with radius r and length L placed around the wire such that the wire is along the axis of the cylinder August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 65 Cylindrical Symmetry (2) From symmetry we can see that the electric field will extend radially from the wire E How? • If we rotate the wire along its axis, the electric field must look the same • Cylindrical symmetry • If we imagine a very long wire, the electric field cannot be different anywhere along the length of the wire • Translational symmetry Thus our assumption of a right cylinder as a Gaussian surface is perfectly suited for the calculation of the electric field using Gauss’ Law August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 66 Cylindrical Symmetry (3) The electric flux through the ends of the cylinder is zero because the electric field is always parallel to the ends The electric field is always perpendicular to the wall of the cylinder so … and now solve for the electric field August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 67 Planar Symmetry (1) Assume that we have a thin, infinite non-conducting sheet of positive charge E σ The charge density in this case is the charge per unit area, σ From symmetry, we can see that the electric field will be perpendicular to the surface of the sheet August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 68 Planar Symmetry (2) To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height 2r, chosen to cut through the plane perpendicularly. Because the electric field is perpendicular to the plane everywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the ends of the cylinder. Using Gauss’ Law we get … so the electric field from an infinite non-conducting sheet with charge density σ is August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 69 Spherical Symmetry (1) Now let’s calculate the electric field from charge distributed as a spherical shell. Assume that we have a spherical shell of charge q with radius rS (gray). + + + + We will assume two different spherical Gaussian surfaces • r > rS (blue) i.e., outside + + + + • r < rS (red) i.e., inside August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 72 Spherical Symmetry (2) Let’s start with the Gaussian surface outside the sphere of charge, r > rS (blue) We know from symmetry arguments that the electric field will be radial outside the charged sphere • If we rotate the sphere, the electric field cannot change • Spherical symmetry Thus we can apply Gauss’ Law and get . … so the electric field is August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 73 Spherical Symmetry (3) Let’s take the Gaussian surface inside the sphere of charge, r < rS (red) We know that the enclosed charge is zero so By symmetry, we find that the electric field is zero everywhere inside spherical shell of charge Thus we obtain two results • The electric field outside a spherical shell of charge is the same as that of a point charge • The electric field inside a spherical shell of charge is zero August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 74 Spherical Symmetry: Uniform Distribution Next, let’s calculate the electric field from charge distributed uniformly throughout a sphere. Assume that we have a solid sphere of charge Q with radius R with constant charge density per unit volume ρ We will assume two different spherical Gaussian surfaces • r2 > R (outside) • r1 < R (inside) R Charge density inside ρ (C/m^3) August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 75 Spherical Symmetry: Uniform Consider a Gaussian surface with radius r2 > R Again by spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface. Gauss’ Law gives us total charge area R Solving for E we find same as a point charge! August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 78 Spherical Symmetry: Uniform Let’s start with a Gaussian surface with r1 < R From spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface. Gauss’s Law gives us volume . area R Solving for E we find August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 76 Spherical Symmetry: Uniform R In terms of the total charge Q … August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 77 E-field for charged sphere with constant density of charges E/E(R) 1.0 0.8 0.6 0.4 0.2 0.5 1.0 1.5 2.0 2.5 3.0 r/R 21 Thursday, September 9, 2010 Gauss theorem in differential form ￿ q E · dA = ￿0 q q q= ￿ ρdV dV = r2 dΩdr dA = r2 dn 1 div E = ρ 4π￿0 12 E · dA = r E · dn = ρr dΩdr ￿0 1 Divergence of electric field equals E · dn = dΩdr ￿0 local charge density 2 div E = ∂x Ex + ∂y Ey + ∂z Ez 22 Thursday, September 9, 2010 Electric Field from a Ring of Charge We can’t solve it by Gauss’s Law Use integration … Total charge Q is distributed homogeneously over a ring of radius R. Find E-field on the axis at distance z. August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 79 dφ Take section Its charge is dφ dφ dq = Q 2π It produces E-field Edq dq dφ Q =k 2 =k 2 R +z 2π R2 + z 2 From symmetry: only Ez is important Ez,dq dφ Q z √ = Edq cos θ = k 2 + z2 2π R R2 + z 2 24 Thursday, September 9, 2010 dφ Q z √ Ez,dq = Edq cos θ = k 2π R 2 + z 2 R 2 + z 2 Qz Integrate over dφ E = k 2 + z 2 ) 3 /2 (R E 0.3 ∝ 1/z 0.2 2 0.1 z/R 0.5 1.0 1.5 2.0 2.5 3.0 25 Thursday, September 9, 2010 Shielding An interesting application of Gauss’ Law: The electric field inside a charged conductor is zero • movie from Wikipedia. August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 60 Shielding An interesting application of Gauss’ Law: The electric field inside a charged conductor is zero • movie from Wikipedia. August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 60 Pointed Surfaces E field is always perpendicular to the surface of a charged conductor Now consider a sharp point. The field lines are much closer together, i.e., the field is much stronger and looks much more like the field of a point charge. August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 22 80 Find electric field between two plates of area A opposite charges q, distance d apart. q A:E= Ad￿0 q B:E= 2A￿0 q C:E= A￿0 qd D:E= A￿0 q E:E= dA￿0 Thursday, September 9, 2010 iClicker Find electric field between two plates of area A opposite charges q q A:E= Ad￿0 q B:E= 2A￿0 q C:E= A￿0 qd D:E= A￿0 q E:E= dA￿0 Thursday, September 9, 2010 From single plane: Total E-field q E= A￿ 0 Electric Potential Energy Electrostatic force is conservative force • For a conservative force, the work is path-independent • Work over closed path is 0 B A E E • circulation of electrostatic field is zero ￿ E · dl = 0 August 12, 2010 Thursday, September 9, 2010 NB: there is inductive E-field, for which work over closed path is not 0 (circulation not 0) 2 Electric Potential Energy Electrostatic force is conservative force • For a conservative force, the work is path-independent • Work over closed path is 0 B dl A E E • circulation of electrostatic field is zero ￿ E · dl = 0 August 12, 2010 Thursday, September 9, 2010 NB: there is inductive E-field, for which work over closed path is not 0 (circulation not 0) 2 Is this a possible configuration of static E-field? Ex(y) Thursday, September 9, 2010 Is this a possible configuration of static E-field? Ex(y) E.dl>0, large E.dl=0 E.dl< 0, small Thursday, September 9, 2010 NO, since E.dl=0￿ E · dl = 0 Is this a possible configuration of static E-field? Ex(y) E.dl>0, large E.dl=0 E.dl< 0, small NO, since E.dl=0￿ E · dl = 0 This is a possible configuration of non-electrostatic fields Thursday, September 9, 2010 Line of Electrostatic field cannot be closed on themselves. E ￿ NB: there is inductive E-field, for which work over closed path is not 0 (circulation not 0, field lines do close on themselves) Thursday, September 9, 2010 E · dl = 0 Line of Electrostatic field cannot be closed on themselves. E ￿ NB: there is inductive E-field, for which work over closed path is not 0 (circulation not 0, field lines do close on themselves) Thursday, September 9, 2010 E · dl = 0 Electric potential b aq E Thursday, September 9, 2010 Charge q moves from point a at x=0 to point b at x=L in constant Efield E. How does it’s energy change? Electric potential y L x b aq E Thursday, September 9, 2010 Charge q moves from point a at x=0 to point b at x=L in constant Efield E. How does it’s energy change? Dificult way: y L ￿ m￿ 2 2 E= vx + vy 2 vy = const = vy,0 x b a E Thursday, September 9, 2010 vx = vx,0 + at = vx,0 + F qE t = vx,0 + t m m x = vx,0 t + at2 /2 = L ￿ 2 −mvx,0 + m2 vx,0 + 2qELm t= qE ￿ 2qEL 2 vx = + vx,0 m ￿ m￿ 2 2 E= v + vx,0 + qEL 2 y,0 ∆E = E − E0 = qEL ∆EK = qEL E-field did work on a particle When moved along the field, energy of a positively charged particle increased, energy of a negatively charged particle decreased Thursday, September 9, 2010 Electric potential Electric potential = ability of E-field to do potential energy work. U = mgh Recall gravity: Φ V g = gh kinetic energy Gravity (in constant grav. field): • potential V= g h • potential energy U = m V Thursday, September 9, 2010 Electric potential potential energy ΦE V Electric potential kinetic energy y L a E Thursday, September 9, 2010 x Energy conservation: change in kinetic energy of a particle = - change in potential energy of a particle change in kinetic energy of a particle = work done by E-field Thursday, September 9, 2010 Electric potential Φ ∆EK = qEL ∆ E k = − ∆E p ∆Ep = q ∆Φ V ∆Φ = −E ∆x V Electric potential = change in energy of unit charge, when moved distance -∆x along electric field Only difference is important -> Electric potential difference = change in energy of unit charge, when moved distance - ∆x along electric field Thursday, September 9, 2010 Potential is useful when we do not care about direction of velocity, only energy v v0 a E Thursday, September 9, 2010 b Electric Potential Energy (2) Like gravitational or mechanical potential energy, we must define a reference point from which to define the electric potential energy Typically, the electric potential energy is defined to be zero when all charges are infinitely far apart We can then write a simpler definition of the electric potential taking the initial potential energy to be zero, The negative sign on the work: • If E does positive work then U < 0 • If E does negative work then U > 0 August 12, 2010 Thursday, September 9, 2010 3 Constant Electric Field Let’s look at the electric potential energy when we move a charge q by a distance d in a constant electric field The definition of work is W =F·d For a constant electric field the force is … so the work done by the electric field on the charge is W = eE · d = qEd cos θ August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 23 4 Constant Electric Field Let’s look at the electric potential energy when we move a charge q by a distance d in a constant electric field The definition of work is W =F·d For a constant electric field the force is … so the work done by the electric field on the charge is W = eE · d = qEd cos θ August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 23 4 Constant Electric Field - Special Cases If the displacement is in the same direction as the electric field • A positive charge loses potential energy when it moves in the direction of the electric field. If the displacement is in the direction opposite to the electric field • A positive charge gains potential energy when it moves in the direction opposite to the electric field. August 12, 2010 Thursday, September 9, 2010 University Physics, Chapter 23 5 Definition of the Electric Potential The electric potential energy of a charged particle in an electric field depends not only on the electric field but on the charge of the particle We want to define a quantity to probe the electric field that is independent of the charge of the probe We define the electric potential as “potential energy per unit charge of a test particle” Unlike the electric field, which is a vector, the electric potential is a scalar • The electric potential has a value everywhere in space but has no direction • Units: Volt, symbol V, 1V = 1J/C August 12, 2010 Thursday, September 9, 2010 6 Electric Potential V The electric potential, V, is defined as the electric potential energy, U, per unit charge, V=U/q Actually, U = q V , potential energy is charge times potential The electric potential is a characteristic of the electric field, regardless of whether a charged object has been placed in that field (because U ∝ q) The electric potential is a scalar The electric potential is defined everywhere in space as a value, but has no direction The electric potential is defined up to an arbitrary constant August 12, 2010 Thursday, September 9, 2010 7 Electric potential ~ “mountain” in energy space. Positive charges what to roll along the steepest decent. ∆Φ = − E ∆x → V 1.0 y 0 ￿2 E = −∇Φ V 2 E ￿0.5 0.0 ￿2 Thursday, September 9, 2010 0 x 2 Negative charges what to roll along the steepest decent of minus potential ∆Φ = − E ∆x → V E = −∇Φ V 2 0 x 2￿ 0.0 5.0￿ E 2 y 0 2￿ 0.1 Thursday, September 9, 2010 Math Reminder - Partial Derivatives Given a function V(x,y,z), the partial derivatives are they act on x, y, and z independently Example: V(x,y,z) = 2xy2 + z3 Meaning: partial derivatives give the slope along the respective direction August 12, 2010 Thursday, September 9, 2010 33 Calculating the Field from the Potential (2) We can calculate any component of the electric field by taking the partial derivative of the potential along the direction of that component We can write the components of the electric field in terms of partial derivatives of the potential as In terms of graphical representations of the electric potential, we can get an approximate value for the electric field by measuring the gradient of the potential perpendicular to an equipotential line August 12, 2010 Thursday, September 9, 2010 34 iClicker An electric potential is in volts. Which of the following expressions describes the associated electric field, in units of volts per meter? a) b) c) d) e) Thursday, September 9, 2010 reminder: E = −∇V iClicker An electric potential is in volts. Which of the following expressions describes the associated electric field, in units of volts per meter? reminder: a) b) c) d) e) E = −∇V Ex = −∂x V = 10x Ey = − ∂ y V = 1 Ez = − ∂ z V = 1 Thursday, September 9, 2010 The Volt The commonly encountered unit joules/coulomb is called the volt, abbreviated V, after the Italian physicist Alessandro Volta (1745-1827) With this definition of the volt, we can express the units of the electric field as For the remainder of our studies, we will use the unit V/m for the electric field August 12, 2010 Thursday, September 9, 2010 10 Example: Energy Gain of a Proton (1) A proton is placed between two parallel conducting plates in a vacuum as shown. + - Distance between the plates is d, E-field inside is E. The proton is released from rest close to the positive plate. Question: What is the kinetic energy of the proton when it reaches the negative plate? August 12, 2010 Thursday, September 9, 2010 11 Example: Energy Gain of a Proton (1) A proton is placed between two parallel conducting plates in a vacuum as shown. + - Distance between the plates is d, E-field inside is E. The proton is released from rest close to the positive plate. Question: What is the kinetic energy of the proton when it reaches the negative plate? August 12, 2010 Thursday, September 9, 2010 11 Two ways to solve the problem E d ∆Φ = EL Vd Dynamical v = at, x = at2 /2, E = mv 2 /2 qE a= m qE t2 x= m2 qE T 2 L= d m2 ￿ 2d Lm T= qE ￿ d qE 2Lm v= m qE E = mv 2 /2 = qEL d Thursday, September 9, 2010 Energetic E = q ∆Φ V Two ways to solve the problem + E - d ∆Φ = EL Vd Dynamical v = at, x = at2 /2, E = mv 2 /2 qE a= m qE t2 x= m2 qE T 2 L= d m2 ￿ 2d Lm T= qE ￿ d qE 2Lm v= m qE E = mv 2 /2 = qEL d Thursday, September 9, 2010 Energetic E = q ∆Φ V Example: Energy Gain of a Proton (2) initial final Conservation of energy ΔK = − ΔU = + 450 eV Because the acceleration of a charged particle across a potential difference is often used in nuclear and high energy physics, the energy unit electron-volt (eV) is common An eV is the energy gained by a charge e that accelerates across an electric potential of 1 volt August 12, 2010 Thursday, September 9, 2010 12 ...
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