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

Explore the concept of voltage due to point charges, whiteboards with examples, cute voltage problems, voltage at specific points, and practical calculations. Learn to calculate voltage in non-linear arrays and charge arrays for a better understanding.

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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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