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

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

Solution: U = k * (q1 * q2) / r = (9 × 10^9) * (3 × 10^-6 * 4 × 10^-6) / 0.2 = 0.54 J.

Steps: 9

Step 1: Identify the values given in the problem. We have two charges: q1 = +3μC and q2 = +4μC. The distance between them is r = 0.2m.
Step 2: Convert the microcoulombs (μC) to coulombs (C). 3μC = 3 × 10^-6 C and 4μC = 4 × 10^-6 C.
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) * (3 × 10^-6 * 4 × 10^-6) / 0.2.
Step 5: Calculate the product of the charges: 3 × 10^-6 * 4 × 10^-6 = 12 × 10^-12 C².
Step 6: Now substitute this back into the equation: U = (9 × 10^9) * (12 × 10^-12) / 0.2.
Step 7: Calculate the numerator: (9 × 10^9) * (12 × 10^-12) = 108 × 10^-3 = 0.108.
Step 8: Divide by the distance: U = 0.108 / 0.2 = 0.54 J.
Step 9: Therefore, the potential energy of the system is 0.54 Joules.