What is the potential energy of a system of two charges +3μC and -5μC separated

₹0.0

Question: What is the potential energy of a system of two charges +3μC and -5μC separated by 0.3m?

Options:

Correct Answer: -0.45 J

Solution:

Potential energy U = k * q1 * q2 / r = (9 × 10^9) * (3 × 10^-6) * (-5 × 10^-6) / 0.3 = -0.45 J.

What is the potential energy of a system of two charges +3μC and -5μC separated by 0.3m?

What is the potential energy of a system of two charges +3μC and -5μC separated by 0.3m?

Step 1: Identify the values of the charges. We have charge q1 = +3μC and charge q2 = -5μC.
Step 2: Convert the microcoulombs (μC) to coulombs (C). +3μC = 3 × 10^-6 C and -5μC = -5 × 10^-6 C.
Step 3: Identify the distance between the charges. The distance r = 0.3m.
Step 4: Use the formula for potential energy U = k * q1 * q2 / r, where k = 9 × 10^9 N m²/C² (Coulomb's constant).
Step 5: Substitute the values into the formula: U = (9 × 10^9) * (3 × 10^-6) * (-5 × 10^-6) / 0.3.
Step 6: Calculate the numerator: (9 × 10^9) * (3 × 10^-6) * (-5 × 10^-6) = -135 × 10^-3 = -0.135.
Step 7: Divide the result by the distance (0.3): U = -0.135 / 0.3 = -0.45 J.
Step 8: Conclude that the potential energy of the system is -0.45 J.

No concepts available.