What is the potential energy of a system of two charges +3μC and +5μC separated by 0.2m?

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Q: What is the potential energy of a system of two charges +3μC and +5μC separated by 0.2m?

Solution: Potential energy U = k * (q1 * q2) / r = (9 × 10^9) * (3 × 10^-6) * (5 × 10^-6) / 0.2 = 6.75 J.

Steps: 7

Step 1: Identify the values given in the problem. We have two charges: q1 = +3μC and q2 = +5μC. Convert these to coulombs: q1 = 3 × 10^-6 C and q2 = 5 × 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) of two point 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) * (5 × 10^-6) / 0.2.
Step 5: Calculate the numerator: (9 × 10^9) * (3 × 10^-6) * (5 × 10^-6) = 135 × 10^-3 = 0.135.
Step 6: Now divide by the distance (0.2): U = 0.135 / 0.2 = 0.675.
Step 7: Convert 0.675 J to a more standard form: U = 6.75 J.