What is the potential energy of a system of two charges +3 μC and +4 μC separated by 0.2 m? (2023)

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Q: What is the potential energy of a system of two charges +3 μC and +4 μC separated by 0.2 m? (2023)

Solution: Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / 0.2 m = 54 J.

Steps: 7

Step 1: Identify the values given in the problem. We have two charges: q1 = +3 μC (microcoulombs) and q2 = +4 μC. The distance between them is r = 0.2 m.
Step 2: Convert the charges from microcoulombs to coulombs. 3 μC = 3 × 10^-6 C and 4 μC = 4 × 10^-6 C.
Step 3: Use the formula for potential energy (U) of two charges: U = k * q1 * q2 / r, where k is the Coulomb's constant, k = 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.
Step 7: Convert 5.4 × 10^5 to joules: U = 54 J.