What is the potential energy of two charges +3 μC and +4 μC separated by 0.2 m?

Practice Questions

Q1
What is the potential energy of two charges +3 μC and +4 μC separated by 0.2 m? (2023)
  1. -54 J
  2. 54 J
  3. 0.54 J
  4. 5.4 J

Questions & Step-by-Step Solutions

What is the potential energy of two charges +3 μC and +4 μC separated by 0.2 m? (2023)
  • Step 1: Identify the values given in the problem. We have two charges: q1 = +3 μC and q2 = +4 μC. Convert these to coulombs: q1 = 3 × 10^-6 C and q2 = 4 × 10^-6 C.
  • Step 2: Identify the distance between the charges, which is given as r = 0.2 m.
  • Step 3: Use the formula for potential energy (U) between two charges: U = k * q1 * q2 / r, where k is the Coulomb's constant, approximately 9 × 10^9 N m²/C².
  • Step 4: Substitute the values into the formula: U = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / (0.2 m).
  • Step 5: Calculate the numerator: (9 × 10^9) * (3 × 10^-6) * (4 × 10^-6) = 108 × 10^3 = 1.08 × 10^5.
  • Step 6: Divide the result by the distance (0.2 m): U = (1.08 × 10^5) / (0.2) = 5.4 × 10^5 = 54 J.
  • Step 7: Conclude that the potential energy of the two charges is 54 Joules.
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