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Solving Voltage and Charge Movement Problems in Electric Fields

Explore the intricacies of electric potentials and work done in moving point charges within an electric field. We tackle problems involving a 6 µC charge being brought from infinity to various positions relative to two +1.5 µC charges, calculate initial and final voltages, and determine the work required for such movements. Key formulas like ( V = k cdot q / r ) and energy change calculations are employed. We also analyze situations with both positive and negative charges to find voltage at specific points and the energy needed for charged particle movements.

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Solving Voltage and Charge Movement Problems in Electric Fields

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  1. Cute problems with voltage - Movement q q What work to bring a 6 C charge from infinity to halfway between the other two charges? +1.5 C +1.5 C 24.0 cm • Find initial voltage • Find final voltage • V = final - initial • ΔEp = W = ΔVq TOC

  2. Cute problems with voltage q q +1.5 C +1.5 C 12.0 cm 12.0 cm • Voltage at midpoint due to one of charges: • V = kq = k(1.5x10-6) = 112,400 V • r .12 • Total potential is twice that: = 224,800 V • V = w/q • The charge is then released. What is its velocity when it is far away if it has a mass of 12 grams? (.012 kg) • Vq = 1/2mv2 TOC

  3. q q q q q q +1.5 C +1.5 C +6 C 16.0 cm 8.0 cm What work to bring a 6 C charge from halfway between the other two charges to 8.0 cm from the left charge? +6 C +1.5 C +1.5 C 24.0 cm • Find initial voltage • Find final voltage • V = final - initial • ΔEp = W = ΔVq TOC

  4. Try this one q q q What work to bring a 13.0 C charge from halfway between the other two charges to 6.0 cm from the positive and 18 cm from the negative? +3.20 C +13.0 C -4.10 C 12.0 cm 12.0 cm Initial V -67425 V Final V 274700. V Change in V 342100. V Work 4.448 J TOC +4.4 J

  5. Each grid is a meter. If charge A is -14.7 μC, and charge B is +17.2 μC, calculate the voltage at points g and h, and the amount of work needed to move a +3.40 μC charge from g to h. A B y Vg = kq/r + kq/r = k(-14.7E-6)/√(12+22) + k(17.2E-6)/√(12+52) = -28775.6 V Vh = kq/r + kq/r = k(-14.7E-6)/√(72+32) + k(17.2E-6)/√(52+02) = 13573.1 V ΔV = Vh – Vg = 42348.6 V W = ΔEp = ΔVq = (42348.6 V)(+3.40E-6 C) = 0.14399 J ≈ 0.144 J g x h 0.144 J

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