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Voltage due to point sources Contents: Voltage due to one point charge Whiteboards

Voltage due to point sources Contents: Voltage due to one point charge Whiteboards Voltage due to many point charges Example Whiteboards Cute Voltage problems Example Whiteboards. Voltage due to Point Sources. q 2. q 1. r. Definition: V = Δ E p q

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Voltage due to point sources Contents: Voltage due to one point charge Whiteboards

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  1. Voltage due to point sources • Contents: • Voltage due to one point charge • Whiteboards • Voltage due to many point charges • Example • Whiteboards • Cute Voltage problems • Example • Whiteboards

  2. Voltage due to Point Sources q2 q1 r • Definition: • V = ΔEp • q • ΔEp = W = Fs, but what work to bring q2 from infinity to r? The work is: kq1q2 r It’s simple, just the definite integral of kq1q2 .dr r2 From infinity to r TOC

  3. Voltage due to Point Sources q1 r The voltage at point A is: V = kq r V = voltage at distance r q = charge of q1 r = distance from q1 • Definition: • V = ΔEp • q • ΔEp = W = Fs, A A van de Graaff generator has an 18 cm radius dome, and a charge of 0.83 μC. What is the voltage at the surface of the dome? V = kq/r = (8.99E9)(0.83E-6)/(0.18) = 41453.88889 ≈ 41 kV 10,000 V/in, +q and –q, sheet of rubber TOC

  4. Whiteboards: Voltage due to point charge 1 | 2 | 3 TOC

  5. Lauren Order is 3.45 m from a -150. C charge. What is the voltage at this point? V = kq/r, r = 3.45 m, q = -150x10-6 C V = -3.91x105 V W -3.91x105 V

  6. Alex Tudance measures a voltage of 25,000 volts near a Van de Graaff generator whose dome is 7.8 cm in radius. What is the charge on the dome? V = kq/r, r = .078 m, V = 25,000 V q = 2.17x10-7 C = .22 C W .22 C

  7. Ashley Knott reads a voltage of 10,000. volts at what distance from a 1.00 C charge? V = kq/r, V = 10,000 V, q = 1.00x10-6 C r = .899 m W .899 m

  8. Q2 Q1 Voltages in non linear arrays Find the voltage at point A: A +3.1 C +1.5 C 75 cm 190 cm Voltage is not a vector!!!!!! TOC

  9. Q2 Q1 + 2.6x104 V And The Sum Is… A Find the voltage at point A: +1.5 C +3.1 C 75 cm 190 cm • Voltage at A is scalar sum of V1 and V2: • Voltage due to Q1: • V1 = kq1 = k(1.5x10-6) = 1.27x104 V • r (.752+.752) • Voltage due to Q2: • V2 = kq2 = k(3.1x10-6) = 1.36x104 V • r (.752+1.92) TOC

  10. Whiteboards: Voltage charge arrays 1 | 2 | 3 TOC

  11. Q2 Q1 Find the voltage at point B +1.1 C +2.1 C B 91 cm 138 cm V1 = 10867.03297 V2 = 13680.43478 V1 + V1 = 24547.46775 = 25,000 V W 25,000 V

  12. Q2 Q1 Find the voltage at point C C -4.1 C 38 cm 85 cm +1.1 C V1 = -39587.58847 V2 = +26023.68421 V1 + V1 = -13563.90426 = -14,000 V W -14,000 V

  13. A B Each grid is a meter. If charge A is -14.7 μC, and charge B is +17.2 μC, calculate the voltage at the origin: y V = kq/r + kq/r = k(-14.7E-6)/√(12+42) + k(17.2E-6)/(22+22) = 22617.44323 +2.26E4 V x

  14. Find the voltage in the center of a square 45.0 cm on a side whose corners are occupied by 12.0 C charges. The center is (.452+.452)/2 from all of the charges one charge’s V = 339034.1314 V All four: 1356136.525 V = 1.36x106 V W 1.36x106 V

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