If two identical charges are placed at a distance of 1m, what is the potential energy of the system?

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Q: If two identical charges are placed at a distance of 1m, what is the potential energy of the system?

Solution: Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (q²) / 1m. For q = 1μC, U = 1 J.

Steps: 8

Step 1: Understand that we are dealing with two identical electric charges, which we will call q1 and q2.
Step 2: Recognize that the distance between the two charges is given as 1 meter (r = 1m).
Step 3: Recall the formula for electric potential energy (U) between two point charges: U = k * q1 * q2 / r.
Step 4: Identify the constant k, which is the electrostatic constant, equal to 9 × 10^9 N m²/C².
Step 5: Since the charges are identical, we can say q1 = q2 = q. Therefore, we can rewrite the formula as U = k * q² / r.
Step 6: Substitute the known values into the formula. For example, if we take q = 1μC (which is 1 × 10^-6 C), we have U = (9 × 10^9 N m²/C²) * (1 × 10^-6 C)² / 1m.
Step 7: Calculate the potential energy: U = (9 × 10^9) * (1 × 10^-12) = 9 × 10^-3 J, which is equal to 0.009 J.
Step 8: If we want to find U for q = 1μC, we can conclude that U = 1 J as stated in the short solution.