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Charge and Polarization

Charge and Polarization. Charge Charge comes in two forms, which Ben Franklin designated as positive (+) and negative(–). Charge is quantized. The smallest possible stable charge, which we designate as e , is the magnitude of the charge on 1 electron or 1 proton.

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Charge and Polarization

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  1. Charge and Polarization

  2. Charge • Charge comes in two forms, which Ben Franklin designated as positive (+) and negative(–). • Charge is quantized. • The smallest possible stable charge, which we designate as e, is the magnitude of the charge on 1 electron or 1 proton. • We say a proton has charge of e, and an electron has a charge of –e. • e is referred to as the “elementary” charge. • e = 1.602 × 10-19 Coulombs. • The coulomb is the SI unit of charge.

  3. Determine the charge on a charged body: N = number of electrons q = charge e = charge of 1 electron, which is -1.60x10-19C Example: 3x1015 electrons are placed on the sphere by induction with a radius of 0.2m. What is the charge on the sphere?

  4. N = q / e • A certain static discharge delivers -0.5 Coulombs of electrical charge. How many electrons are in this discharge? N = (-0.5 C) / (-1.602 x 10-19 C) = 3,121,098,626,716,604,245

  5. 1 mol = 6.02 x1023 molecules 2 mol = 1.204 x1024 molecules • How much positive charge resides in two moles of hydrogen gas (H2)? • How many electrons reside in the hydrogen gas? • How much net charge? Each H atom has 1 proton There is 2 H atoms per molecule # protons = 2 ·1.204 x1024 molecules = 2.408 x 1024 # electrons is equal to the number of protons # electrons = 2.408 x 1024 Net charge is 0 because there is an equal number of protons and electrons

  6. The total charge of a system composed of 1800 particles, all of which are protons or electrons, is 31x10-18 C. • How many protons are in the system? • How many electrons are in the system? q = 31x10-18C e =1.6x10-19C N = 31x 10-18 C / 1.6x10-19 C = 194 N = q / e We must have 194 protons than electrons # protons = 997 # electrons = 803

  7. The atom • The atom has positive charge in the nucleus, located in the protons. The positive charge cannot move from the atom unless there is a nuclear reaction. • The atom has negative charge in the electron cloud on the outside of the atom. Electrons can move from atom to atom without all that much difficulty. • Question • You charge the balloon by rubbing it on hair or on a sweater, and the balloon becomes negative. How can it pick up a neutral tissue?

  8. The electroscope is made from a metal or other conductor, and may be contained within a flask. The vanes are free to move. Pole • This is an electroscope Vanes

  9. 1. Rub the black rod with the fur. Bring the rod toward the pole of the electroscope. What happens to the vanes? 2. Come up with an atomic-level explanation for your observations. - - - - - - - - + - + - + + + + + + Pole - - - - - - - - - - The vanes spread apart - - Vanes Charge is induced

  10. + + + + + Pole + + 1. Rub the glass rod with the silk. Bring the rod toward the pole of the electroscope. What happens to the vanes? 2. Come up with an atomic-level explanation for your observations. + - + - + - - - - - - + + + + + + + + + + Vanes The vanes spread apart Charge is induced

  11. + + What happens when your touch the electroscope with the glass rod? + + + Pole + + + - + - + - - - - - - Electrons flow from the electroscope to the glass rod + + + + + + + + + + Vanes The vanes spread apart Charge is caused through conduction

  12. Charges exert forces on each other. • Like charges (two positives, or two negatives) repel each other, resulting in a repulsive force. • Opposite charges (a positive and a negative) attract each other, resulting in an attractive force.

  13. Coulomb’s Law • Coulomb’s law tells us how the magnitude of the force between two particles varies with their charge and with the distance between them. • k = 8.99 × 109 N m2 / C2 • q1, q2 are charges (C) • r is distance between the charges (m) • F is force (N) • Coulomb’s law applies directly only to spherically symmetric charges.

  14. Sometimes you see Coulomb’s Law written in a slightly different form • ε0 = 8.85 × 10-12 C2/ N m2 permittivity of space constant • q1, q2 are charges (C) • r is distance between the charges (m) • F is force (N) • This version is theoretically derived and less practical than form 1

  15. Spherically Symmetric Forces • Newton’s Law of Gravity • Coulomb’s Law

  16. d = 15cm = 0.15 m • A point charge of positive 12.0 μC experiences an attractive force of 51 mN when it is placed 15 cm from another point charge. What is the other charge? F = 51mN = 0.051 N Charge 1 Charge 2 q1 = 12μC = 12 x 10-6C = -1.06 x10-8 C

  17. q1 = 1.5nC • Calculate the mass of ball B, which is suspended in midair. 1.3 m FBD Fc q2 = -0.5nC Fg Charge is in equilibrium Fc – Fg = 0 Fc – mg = 0 = 4.08 x 10-10 kg

  18. Superposition • Electrical force, like all forces, is a vector quantity. • If a charge is subjected to forces from more than one other charge, vector addition must be performed. • Vector addition to find the resultant vector is sometimes called superposition.

  19. y(m) ΣF = F23 - F13 2m • What is the force on the 4 μC charge? 1 2 FBD 3 F13 F23 1m -3μC 2μC 4μC 0m 0m 1m 2m x(m) = 0.045N to the right

  20. y(m) 1 This vector is on the angle. Must break into components 2m -3μC • What is the force on the 4 μC charge? F13 ΣFx = Fx23 - Fx13 cos(45º) FBD F23 1m ΣFy = Fy13sin(45º) 2 3 2μC 4μC 0m 0m 1m 2m x(m) = +0.03382N to the left

  21. y(m) 1 This vector is on the angle. Must break into components 2m -3μC • What is the force on the 4 μC charge? F13 ΣFx = Fx23 - Fx13 cos(45º) FBD F23 1m ΣFy = Fy13 sin(45º) 2 3 2μC 4μC 0m 0m 1m 2m x(m) = 0.038184N to the right

  22. ΣFy = 0.038184N ΣFx = 0.03382N F = √(Fx + Fy) F = √((0.03382N)2 + (0.038184N)2) = 0.051N θ = Tan-1 (Fy/Fx) θ = Tan-1 (0.03818 N / 0.03382N) = 48.47º

  23. The Electric Field • The presence of + or – charge modifies empty space. This enables the electrical force to act on charged particles without actually touching them. • We say that an “electric field” is created in the space around a charged particle or a configuration of Charges. • If a charged particle is placed in an electric field created by other charges, it will experience a force as a result of the field. • Sometimes we know about the electric field without knowing much about the charge configuration that created it. • We can easily calculate the electric force from the electric field.

  24. Why use fields? • Forces exist only when two or more particles are present. • Fields exist even if no force is present. • The field of one particle only can be calculated.

  25. The arrows in a field are not vectors, they are “lines of force”. • The lines of force indicate the direction of the force on a positive charge placed in the field. • Negative charges experience a force in the opposite direction. + • Field around a charge -

  26. A + + + + + + + + + + + + + + + + • Field between charged plates FB q + FA FA q - FB - - - - - - - - - - - - - - - - - - - - - - - - B

  27. Field Vectors from Field Lines • The electric field at a given point is not the field line itself, but can be determined from the field line. • Vectors of any kind are never curvy • Field Lines and Path of Moving Charge • The electric field lines do notrepresent the patha test charge would travel. • The electric field lines represent the directionof the electric force on a test particle placed in the field.

  28. F F F F F F F

  29. Force from Electric Field • The force on a charged particle placed in an electric field is easily calculated. • F = E q • F: Force (N) • E: Electric Field (N/C) • q: Charge (C)

  30. The electric field in a given region is 4000 N/C pointed toward the north. What is the force exerted on a 400 μg Styrofoam bead bearing 600 excess electrons when placed in the field? - - - - - - E = 4000 N/C q = e-N Fc q = -1.6 x 10 -19 C x 600 = -9.6 x 10 -17 C + + + + + F = E q F = 4000N/C · -9.6 x 10 -17 C = 3.84 x 10-13 N

  31. a = 0m/s2 • A 400 μg Styrofoam bead has 600 excess electrons on its surface. What is the magnitude and direction of the electric field that will suspend the bead in midair? + + + + + ∑F: Fe – mg = ma Fe mg Fe – mg = 0 Fe = qE - - - - - - qE – mg = 0 E = mg / q q = Ne- E = mg / Ne- E = (400 x 10-6 kg · 9.8 m/s2) / (600 ·1.6x10 -19 C) E = 4.083 x 1013 N/C

  32. Superposition

  33. A proton traveling at 440 m/s in the +x direction enters an electric field of magnitude 5400 N/C directed in the +y direction. Find the acceleration. vx constant in the x direction so a = 0 m/s2 - - - - - - - - - - - ∑ F: Fy = ma vy is increasing e+ Fy = ma Fy = qE vx = 440 m/s a = qE / m a = 1.6 x 10 -19 C · 5400 N/C / 6.67x10-27 Kg + + + + + + + + a = 1.295 x 1011 m/s2

  34. For Spherical Electric Fields • The Electric Field surrounding a point charge or a spherical charge can be calculated by: • E = k q / r2 • E: Electric Field (N/C) • k: 8.99 x 109 N m2/C2 • q: Charge (C) • r: distance from center of charge q (m) • Remember that k = 1/4πε0 ∑F = qE • E = k q / r2

  35. Principle of Superposition • When more than one charge contributes to the electric field, the resultant electric field is the vector sum of the electric fields produced by the various charges. • Again, as with force vectors, this is referred to as superposition. • Remember… • Electric field lines are NOT VECTORS, but may be used to derive the direction of electric field vectors at given points. • The resulting vector gives the direction of the electric force on a positive charge placed in the field.

  36. -2 cm 2 cm • A particle bearing -5.0 μC is placed at -2.0 cm, and a particle bearing 5.0 μC is placed at 2.0 cm. What is the field at the origin? E2 ∑E = E1 + E2 E1 E2 -5μC E1 -5μC E = Kq / r2 ∑E = - Kq1 / r21 - Kq2 / r22 ∑E = - K(q1 / r21 + q2 / r22) ∑E = - 9x109 (5x10-6 C / (0.02m)2 + 5x10-6 C / (0.02m)2) E = 2.25 x108 N/C to the left

  37. 2 y q1 r22 = q2 r21 1.0 m 1 5 mC • A particle bearing 10.0 mC is placed at the origin, and a particle bearing 5.0 mC is placed at 1.0 m. Where is the field zero? x q1 (1-x)2 = q2 x2 X 10 mC 1-x q1 (1-2x+x2)= q2 x2 ∑E = 0 10x10-3C - 20x10-3Cx + 10x10-3Cx2 = 5x10-3 x2 E1 + E2 = 0 10x10-3C - 20x10-3Cx +(10x10-3Cx2 - 5x10-3 x2) = 0 Kq1 / r21 – Kq2 / r22 = 0 Kq1 / r21 = Kq2 / r22 r22/ r21 = q2 / q1 10x10-3C - 20x10-3Cx +5x10-3 x2= 0

  38. 5x10-3Cx2 - 20x10-3Cx +10x10-3C = 0

  39. E = 5000N/C +++++++ FBD T θ=50ْ • What is the charge on the bead? Its mass is 32 mg. FE T θ=40ْ FE mg ∑Fx : FE – T cos (40o) = 0 mg ∑Fy : T sin (40o) – mg = 0 qE – T cos (40o) = 0 T = mg / sin (40o) T = qE / cos (40o) qE / cos (40o) = mg / sin (40o) q / = (mg cos (40o)( / (E sin (40o)) q = (0.032Kg ∙ 9.8 m/s2cos (40o)) / (5000 N/C sin (40o)) = 7.47 x10-5 C

  40. Electric Potential Energy • Electrical potential energy is the energy contained in a configuration of charges. • Like all potential energies, when it goes up the configuration is less stable; when it goes down, the configuration is more stable. • The unit is the Joule.

  41. Electrical potential energy increaseswhen charges are brought into less favorableconfigurations Electrical potential energy decreaseswhen charges are brought into more favorableconfigurations. + - + + ΔU > 0 ΔU > 0 + + - + ΔU < 0 ΔU < 0

  42. + + ΔU =___ - + ΔU =___ + - Workmust be done on the charge by an outside force to increasethe electric potential energy.

  43. WORK AND CHARGE • When an outside force is applied to a positivetest charge to move it a distance d, the electric force does negativework. (WE = - FEd) • The electric potential energy has increased and U is positive(U2 > U1) + + + + + + Position 2 + U2 d U1 Position 1 + FE - - - - - - -

  44. WORK AND CHARGE • If a negativecharge moves upward a distance d, the electric force does positivework (WE = +FEd). • The change in the electric potential energy U is negative (U2 < U1) + + + + + + Position 2 - U2 d FE U1 Position 1 - - - - - - - -

  45. Electric Potential • Electric potential is hard to understand, but easy to measure. We commonly call it “voltage”, and its unit is the Volt. • 1 V = 1 J/C • Electric potential is easily related to both the electric potential energy, and to the electric field.

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