lect05 - • Physics 227: Lecture 5 Gauss’s Law...

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Unformatted text preview: • Physics 227: Lecture 5 Gauss’s Law Leftovers, Work / Potential Energy in Electrostatics Lecture 4 review: • Gauss’s Law: for a closed surface, Φ = ∫ E.dA = ∫ E⊥dA = ∫ E cosφ dA = qenclosed/ε0. • • • The flux does not depend on the size or shape of the Gaussian surface, or the position of the charges, it depends only on the total enclosed charge. Examples: point charge, inside uniform sphere, inside spherical shell, infinitely long line charge, infinitely long cylinder of charge, inside the cylinder of charge, infinite plane of charge, two planes of charge, ... Conductors: electrons move around so that the field inside a conductor vanishes. Thus requires all net charge is on the surface. Monday, September 19, 2011 Some Geometry r dθ ˆ n dφ dθ r dφ r • The area corresponding to an angle range dθ x dφ is r2 dθ dφ, if the surface is perpendicular to r vector. • If the surface is tilted by an angle α - ˆ os α = r.n - its actual cˆ area in the range of angles dθ x dφ increases by a factor 1/cosα. • With charge density λ, the total charge on the surface is: dq = rdθ rdφ λ / cosα. Monday, September 19, 2011 • Field Inside Spherical Shell For an spherical or infinite cylindrical shell of charge, the field inside is 0. • Here we show this occurs because the fields from opposite parts of the sphere or the shell cancel. • What is the field at the diamond a distance z above the center of the circle a 2d slice through a cylinder or sphere. θ z α α r0 r0 • To get the field, we need to integrate o ver angles: ``E = ∫kdq/r2’’. The geometry is shown for a particular angle θ. (Angles are measured from (0,z), not the origin.) • The field from a small area at the upper right, a distance r away from (0,z), w ill be kdq/r2 = k (rdθ rdφ λ / cosα) /r2 = k (dθ dφ λ / cosα). This is independent of r! And independent of α for opposite points from (0,z)! • The field from the opposite direction is exactly the same magnitude, so the two cancel! This only works for the circular geometry - for, e.g., a cube the factors of cosα would generally be different. Monday, September 19, 2011 Field at Surface of a Charged Conductor • What is the field at the surface of a charged conductor? • Draw a small Gaussian box across the surface. The box should be small enough that the surface is flat and the charge density is constant, to a good approximation, and should be || / ⊥ to the local surface. • Gauss’s Law: Φ = ∫E.dA = 4.0.A + 0.A + E.A = EA • Φ = q/ε0 = σA/ε0 • ➮ E = σ/ε0 • This is a factor of 2 larger than for a plane of charge, because all the field is to one side. Monday, September 19, 2011 E Field in a Idealized Parallel Plate Capacitor • (Superposition) Adding the fields from two infinite planes of charge yields the idealized result. • Or we can use Gaussian cylinders + properties of conductors + symmetry: E = E0ŷ, but E = 0 at the end of the Gauusian cylinder inside the conductor, and E⊥ = 0 for sides of the Gaussian cylinders. • • • For S2 and S3, qenclosed = 0 ➮ Φ = 0 ➮ Eoutside.A = 0 ➮ E = 0. For S1, qenclosed = σA ➮ Φ = σA/ε0 ➮ Einside.A = σA/ε0 ➮ Einside = σ/ε0 ŷ. For S4, qenclosed = -σA ➮ Φ = -σA/ε0 ➮ Einside.A = -σA/ε0 ➮ Einside = σ/ε0 ŷ. (Note: here the normal is in the -ŷ direction, so the negative signs cancel.) • Do not get confused - we have ``3’’ ways of getting the same result. Do not a dd them! Even though there is ±σ, you can get the field using one of them! The other is already included in your work in using Eoutside, in conductor = 0. Monday, September 19, 2011 iClicker: What is the Surface Charge on a Conductor Consider a conducting sphere, w ith a cavity inside. 5 μC We insert a 5 μC charge into the cavity. What is the charge on the inner and outer surfaces of the conductor? A. 0 μC, 0 μC. B. -5 μC, -5 μC. C. -5 μC, 5 μC. D. 5 μC, -5 μC. E. 5 μC, 5 μC. Monday, September 19, 2011 iClicker: What is the Surface Charge on a Conductor Consider a conducting sphere, w ith a cavity inside. 5 μC We insert a 5 μC charge into the cavity. What is the charge on the inner and outer surfaces of the conductor? We need -5 μC on the inside surface to stop the field lines from the +5 μC charge in the cavity. With a neutral conductor, that requires we have +5 μC on the outer surface. If the conductor had total charge Q μC, we would have Q+5 μC on the outer surface. A. 0 μC, 0 μC. B. -5 μC, -5 μC. C. -5 μC, 5 μC. D. 5 μC, -5 μC. E. 5 μC, 5 μC. Monday, September 19, 2011 Conservative Forces • Gravity and Electric Forces are conservative forces. • • • You can relate the force to a potential / potential energy. Energy is converted from potential to kinetic. No energy is lost. Changes in the potential / potential energy, depend only on the initial and final points of a trajectory, not on the intermediate path taken. Monday, September 19, 2011 Force / Work / Potential Energy / Field / Potential • Let’s review these concepts from gravity first - for a mass m in a constant gravitational field, like near the surface of the earth ￿ ￿ Fg = −mg z ˆ Force of gravity integrate differentiate ￿z F · d￿ = −mg ∆z W= ∆U = − ￿ Work done by gravity ￿z F · d￿ = mg ∆z Potential energy from gravity often used as U = mgz divide by m ￿ ￿ = Fg /m = −g z g ˆ Gravitational field Monday, September 19, 2011 integrate differentiate ∆V = − ￿ divide by m ￿ · d￿ = g ∆z gz Just as the field = the force / m, the potential = the potential energy / m Force / Work / Potential Energy / Field / Potential • Now we move to a charge q in a constant electric field, as with an infinite plane of charge. σ ￿ ￿ z ˆ Fc = q E = −q 2￿0 Electric Force integrate differentiate W= ￿ ￿ Fc · d￿ = −qE ∆z z Work done by electric force ∆U = − ￿ ￿ Fc · d￿ = qE ∆z z Potential energy from electric force often used as U = qEz divide by q σ ￿ ￿ z ˆ E = F /q = − 2￿0 Electric field Monday, September 19, 2011 integrate differentiate ∆V = − ￿ divide by q ￿z E · d￿ = E ∆z Just as the field = the force / q, the potential = the potential energy / q Force / Work / Potential Energy / Field / Potential • Now we move to a charge q in a constant electric field, as with an infinite plane of charge. σ ￿ ￿ z ˆ Fc = q E = −q Electric Force in 2￿0 units of Newtons integrate differentiate W= ￿ ￿ Fc · d￿ = −qE ∆z z ￿Work in units of Joules ￿ ∆U = − Fc · d￿ = qE ∆z z Energy in units of Joules (or “electron Volts”) U = qEz divide by q σ ￿ ￿ z ˆ E = F /q = − 2￿0 Electric field units: Volts/m or Newtons/Coulomb Monday, September 19, 2011 integrate differentiate ∆V = − ￿ divide by q ￿z E · d￿ = E ∆z Potential V in units of Volts = J/C iClicker: Work and Potential energy 5 μC v E Note: in all those formulas you have to watch your limits of integration / signs. Let’s see about your intuition! The field points left. The 5 μC charge moves right. Which of the following is correct? A. The field does positive work on the charge and potential energy increases. B. The field does positive work on the charge and potential energy decreases. C. The field does negative work on the charge and potential energy increases. D. The field does negative work on the charge and potential energy decreases. E. The field does no work on the charge and potential energy increases. Monday, September 19, 2011 iClicker: Work and Potential energy 5 μC v E E points “downhill”, so the potential energy for q>0 increases. Going “uphill” slows the charge down, so the work done by the field is negative. Note: in all those formulas you have to watch your limits of integration / signs. Let’s see about your intuition! The field points left. The 5 μC charge moves right. Which of the following is correct? A. The field does positive work on the charge and potential energy increases. B. The field does positive work on the charge and potential energy decreases. C. The field does negative work on the charge and potential energy increases. D. The field does negative work on the charge and potential energy decreases. E. The field does no work on the charge and potential energy increases. Monday, September 19, 2011 Force / Work / Potential Energy / Field / Potential for Two Point Charges • Let’s put a charge Q at the origin, generating an E field, and move a charge q around. W= ￿ Fon q kQq ￿ = qE = 2 r ˆ r Electric Force integrate differentiate ￿ F = −∇U divide by q integrate ￿ Ef rom Q ￿ b a ∆U = − ￿c · d￿ = ( kQq − kQq ) F l ra rb Work done by electric force ￿ b a ￿c · d￿ = ( kQq − kQq ) F l rb ra Set Potential energy = 0 at ∞ ￿ a ￿c · −d￿ = kQq U =− F r ra ∞ ￿ divide by q b kQ kQ ￿l E · d￿ = ( − ) rb ra ￿ ∆V = − Fon q kQ = = 2 r differentiate ˆ a q r Just as the field = the force / q, the Electric field Monday, September 19, 2011 ￿ E = −∇V potential = the potential energy / q More than Two Point Charges • To get the total potential energy of a charge q0 interacting w ith several other charges, add up the several potential energies n n ￿ kq0 qi ￿ qi U0 = = kq0 r0 i r0 i i=1,... i=1,... • To get the total potential energy of the entire system of charges, add up the potential energies of all the pairs of charges U0 = ￿ kqi qj i<j rij We need i<j, or we will count each pair twice, or even get ∞’s from ri2/0! Monday, September 19, 2011 Potential Energy of a 3-charge System -q A charge -q is brought from ∞ so that it and two fixed charges, of ±q, form an equilateral triangle. What is the potential energy of the third charge? What is the potential energy of the system? -q A. U3 = 2kq2/d, Usys=6kq2/d. q -q B. U3 = 0, Usys=kq2/d. C. U3 = 2kq2/d, Usys=3kq2/d. D. U3 = 0, Usys=-kq2/d. E. U3 = 0, Usys=-2kq2/d. Monday, September 19, 2011 Potential Energy of a 3-charge System -q A charge -q is brought from ∞ so that it and two fixed charges, of ±q, form an equilateral triangle. What is the potential energy of the third charge? What is the potential energy of the system? -q A. U3 = 2kq2/d, Usys=6kq2/d. -q q Of the 3 pairs, 2 are +w ith U<0, and 1 is -- with U>0, so Utotal = -kq2/d. The 3rd charge interacts with + and -, so U3 = 0. Monday, September 19, 2011 B. U3 = 0, Usys=kq2/d. C. U3 = 2kq2/d, Usys=3kq2/d. D. U3 = 0, Usys=-kq2/d. E. U3 = 0, Usys=-2kq2/d. Potential Energy of a 3-charge System -q Do these two systems have the same, or opposite, system potential energies? One is +q, -q, -q, the other is +q, +q, -q. -q -q -q q In both cases, we have 2 +- pairs with U<0, and 1 same sign pair U>0, so Utotal = -kq2/d is the same. Monday, September 19, 2011 -q q q Thank you, and See you next Thursday Monday, September 19, 2011 ...
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