What is the electric potential energy of a system of two charges, +1 µC and -1 µ

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Question: What is the electric potential energy of a system of two charges, +1 µC and -1 µC, separated by 0.1 m?

Options:

Correct Answer: -0.09 J

Solution:

Potential energy U = k * q1 * q2 / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) * (-1 x 10^-6 C) / 0.1 m = -0.09 J.

What is the electric potential energy of a system of two charges, +1 µC and -1 µC, separated by 0.1 m?

What is the electric potential energy of a system of two charges, +1 µC and -1 µC, separated by 0.1 m?

Step 1: Identify the values given in the problem. We have two charges: q1 = +1 µC (microcoulomb) and q2 = -1 µC. Convert these to coulombs: q1 = 1 x 10^-6 C and q2 = -1 x 10^-6 C.
Step 2: Identify the distance between the charges, which is given as r = 0.1 m.
Step 3: Use the formula for electric potential energy (U) of two point charges: U = k * q1 * q2 / r, where k is the Coulomb's constant, approximately 8.99 x 10^9 N m²/C².
Step 4: Substitute the values into the formula: U = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) * (-1 x 10^-6 C) / 0.1 m.
Step 5: Calculate the numerator: (8.99 x 10^9) * (1 x 10^-6) * (-1 x 10^-6) = -8.99 x 10^-3.
Step 6: Divide the result by the distance (0.1 m): U = -8.99 x 10^-3 / 0.1 = -0.0899 J.
Step 7: Round the result to two decimal places: U ≈ -0.09 J.

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