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Transcript
Electric Field Contents: •Basic Concept •Calculating electric Field example •Whiteboards •Point Charges •Examples •Whiteboards •Electric Fields in Charge Arrays Gravitational Field Electrical Field Force on mass g = Field strength g = 9.81 N/kg (on earth) F = mg Force on charge E = Field strength E is in N/C F = qE (E = F/q) Direction: Uh… down Direction: F on + Charge Force: F = Gm1m2 r2 Field: g = F = Gm m r2 Force: F = kq1q2 r2 Field: E = F = kq q r2 G = 6.67x10-11 Nm2kg-2 k = 8.99x109 Nm2C-2 TOC Electric Field 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 Direction: +q Force This Way E Force This Way -q TOC Which way is the electric field? (wwpcd?) + + + + + - Which way is the electric field? - + + + + + Which way is the electric field? + Which way is the electric field? + - 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)(1.602x10-19 C) = 5.21x10-17 N up F = ma, a = F/m = (5.21x10-17 N)/(9.11x10-31kg) = 5.72x1013 m/s/s TOC Whiteboards: Uniform field 1|2|3|4 TOC 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 1200 N/C left W 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 1.9 N up W 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 65 N/kg down W 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 -1.39 mC W 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 +20.8 C W Electric fields around point charges Direction of E field is: Away from positive Toward negative TOC 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 Whiteboards: Field relative to points 1|2|3 TOC 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) 71.5 N/C to the left W 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 +.11 C W 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) 1.1 x 107 m W 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 12 m W +12.5 C x A 72.1 cm -187 C C 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) 6.54E6 N/C to the left W 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 A x B 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) Electric Fields in Non Linear Arrays A +1.5 C 75 cm Q1 Find the electric Field at point A: +3.1 C 190 cm Q2 1. Calculate the individual E-Fields 2. Figure out direction (angle) 3. Add E-Fields (as vectors) TOC Electric Fields in Non Linear Arrays E1 E2 A +1.5 C 75 cm Q1 1. Calculate the individual E-Fields 2. Figure out direction (angle) 3. Add E-Fields (as vectors) +3.1 C 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) Q2 TOC Electric Fields in Non Linear Arrays E2 = 6,679.2 N/C A 2 +1.5 C 75 cm Q1 E1 = 11,987 N/C 1 +3.1 C 190 cm Q2 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