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Electric Field Contents: Basic Concept Calculating electric Field example Whiteboards PowerPoint Presentation
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Electric Field Contents: Basic Concept Calculating electric Field example Whiteboards

Electric Field Contents: Basic Concept Calculating electric Field example Whiteboards

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Electric Field Contents: Basic Concept Calculating electric Field example Whiteboards

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  1. Electric Field • Contents: • Basic Concept • Calculating electric Field example • Whiteboards • Point Charges • Examples • Whiteboards • Electric Fields in Charge Arrays

  2. Gravitational Field Electrical Field • Force on mass Force on charge • g = Field strength E = Field strength • g = 9.81 N/kg (on earth) E is in N/C • F = mg F = qE (E = F/q) • Direction: Uh… down Direction: F on + Charge • Force: Force: • F = Gm1m2 F = kq1q2 • r2 r2 • Field: Field: • g = F = Gm E = F = kq • m r2q r2 • G = 6.67x10-11 Nm2kg-2 k = 8.99x109 Nm2C-2 TOC

  3. Electric Field Direction: Force This Way +q E Force This Way -q Example 1 - A +125 C charge experiences a force to the right of .0175 N. What is the Electric field, and its direction? E = F/q = .0175 N/125x10-6 C = 140 N/C to the right TOC

  4. Which way is the electric field? (wwpcd?) + + + + + - - - - -

  5. Which way is the electric field? - - - - - + + + + +

  6. Which way is the electric field? +

  7. Which way is the electric field? + -

  8. Electric Field Example 2 - An electron travels through a region where there is a downward electric field of 325 N/C. What force in what direction acts on the electron, and what is its acceleration? F = Eq = (325 N/C down)(-1.602x10-19C) = -5.21x10-17N down F = ma, a = F/m = (5.21x10-17 N)/(9.11x10-31kg) = 5.72x1013 m/s/s TOC

  9. Whiteboards: Uniform field 1 | 2 | 3 | 4 TOC

  10. Ishunta Dunnit notices that a charge of -125 C experiences a force of .15 N to the right. What is the electric field and its direction? E = F/q = (.15 N)/(-125x10-6 N) = -1200 N/C right or 1200 N/C left W 1200 N/C left

  11. Doan Botherme places a +12 mC charge into an upward 160 N/C electric field. What force in what direction does it experience? E = F/q, q = 12x10-3 C, E = 160 N/C up F = 1.92 = 1.9 N up W 1.9 N up

  12. Alfred O. Dadark is on a planet where a mass of 0.12 kg experiences a downward force of 7.80 N. What is the gravitational field on the surface of this planet? g = F/m, m = 1.12 kg, E = 7.80 N down g = (7.80 N down)/(0.12 kg) = 65 N/kg down W 65 N/kg down

  13. Telly Vishun places an unknown charge into a known upward electric field of 612 N/C, and the charge experiences a downward force of .851 N. What is the charge? E = F/q, E = 612 N/C up, F = .851 N down q = -.0013905 C = -1.39 mC W -1.39 mC

  14. Sal F. Hone levitates a .00125 kg ball with an upward electric field of 590 N/C. What is the charge on the ball? (Hint gravity = electrical force) Eq = mg E = F/q, F = Eq, F = mg, m = .00125 kg, g = 9.80 N/kg, E = 590 N/C q = 2.07627E-05 = +20.8 C W +20.8 C

  15. Electric fields around point charges Away from positive Toward negative Direction of E field is: TOC

  16. Fields around points of mass or charge Example: What is the electric field 2.0 m to the right of a -21 μC charge? E = kq/r2 = (8.99E9)(-21E-6)/(22) = -47,197.5 N/C (away) or 47,000 N/C toward a, b, c are points, Q is a charge TOC

  17. Whiteboards: Field relative to points 1 | 2 | 3 TOC

  18. Vera Similitude measures the electric field 13.5 m to the right of a -1.45 C charge. What electric field in what direction? • E for a point charge: • E = kq • r2 • k = 8.99x109 Nm2C-2, q = -1.45x10-6 C, r = 13.5 m • E = -71.5 N/C away from the charge, or to the left (toward the negative charge) W 71.5 N/C to the left

  19. Vesta Buhl measures an electric field of 2,120 N/C, 67 cm from a charge of unknown value. The electric field is away from the charge. What is the charge? • E for a point charge: • E = kq • r2 • k = 8.99x109 Nm2C-2, E = 2,120 N/C, r = .67 m • q = 1.06x10-7 C = +.11 C. It is a positive charge as the E-field is away from it W +.11 C

  20. Amelia Rate measures a gravitational field of 3.4 N/kg. What distance is she from the center of the earth? (Me = 5.98 x 1024 kg. ) • g for a point mass: • g = Gm • r2 • G = 6.67x10-11 Nm2kg-2, g = 3.4 N/kg, m = 5.98x1024 kg • r = 10831137.03 m = 10.8 x 106 m (re = 6.38 x 106 m) W 1.1 x 107 m

  21. Tara Bull measures an electric field of 10. N/C what distance from an electron? • E for a point charge: • E = kq • r2 • k = 8.99x109 Nm2C-2, E = 10 N/C, q = 1.602x10-19 C • r = 1.20008x10-5 m = 12 m W 12 m

  22. A C +12.5 C -187 C x 72.1 cm 11.2 cm What is the electric field at the x? (Magnitude and direction) • EA = kqA = 8.958E6 N/C to the left (away from +) • r2 • EC = kqC = 2.423E6 N/C to the right (toward -) • r2 • = - 8.958E6 N/C + 2.423E6 N/C = -6.54E6 N/C (Left) W 6.54E6 N/C to the left

  23. A B Each grid is a meter. If charge A is -14.7 μC, and charge B is +17.2 μC, calculate the electric field at the origin: y x

  24. mag angle x y A 7773.7 165.96 -7541.6 1885.4 B 19328.5 45 13667.3 13667.3 6125.7 15552.7 Mag 16,715 N/C angle 68.5º (trig angle)

  25. Q2 Q1 Electric Fields in Non Linear Arrays Find the electric Field at point A: A +3.1 C +1.5 C 75 cm 190 cm • Calculate the individual E-Fields • Figure out direction (angle) • Add E-Fields (as vectors) TOC

  26. Q2 Q1 Electric Fields in Non Linear Arrays E1 E2 A • Calculate the individual E-Fields • Figure out direction (angle) • Add E-Fields (as vectors) +3.1 C +1.5 C 75 cm 190 cm • E1 = kq1 = k(1.5C) = 11,987 N/C • r2 (.752 + .752) • E2 = kq2 = k(3.1C) = 6,679.2 N/C • r2 (1.92 + .752) TOC

  27. Q2 Q1 Electric Fields in Non Linear Arrays E1= 11,987 N/C E2= 6,679.2 N/C A 1 2 +3.1 C +1.5 C 75 cm 190 cm • Figuring out direction: • 1 = 45o • 2 = Tan-1(.75/1.9) = 21.54o • T = 180-21.54 = 158.46o Really hoping I don’t need to show any more… TOC