What is the electric potential energy of a system of two charges of +2 μC and -2 μC separated by 0.5 m?
Practice Questions
1 question
Q1
What is the electric potential energy of a system of two charges of +2 μC and -2 μC separated by 0.5 m?
-72 J
72 J
0 J
-36 J
U = k * (q1 * q2) / r = (9 × 10^9 N m²/C²) * (2 × 10^-6 C * -2 × 10^-6 C) / 0.5 m = -72 J.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the electric potential energy of a system of two charges of +2 μC and -2 μC separated by 0.5 m?
Solution: U = k * (q1 * q2) / r = (9 × 10^9 N m²/C²) * (2 × 10^-6 C * -2 × 10^-6 C) / 0.5 m = -72 J.
Steps: 9
Step 1: Identify the values given in the problem. We have two charges: q1 = +2 μC and q2 = -2 μC. Convert these to Coulombs: q1 = 2 × 10^-6 C and q2 = -2 × 10^-6 C.
Step 2: Identify the distance between the charges, which is given as r = 0.5 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 9 × 10^9 N m²/C²).
Step 4: Substitute the values into the formula: U = (9 × 10^9 N m²/C²) * (2 × 10^-6 C * -2 × 10^-6 C) / 0.5 m.
Step 5: Calculate the product of the charges: (2 × 10^-6 C) * (-2 × 10^-6 C) = -4 × 10^-12 C².
Step 6: Substitute this value back into the equation: U = (9 × 10^9 N m²/C²) * (-4 × 10^-12 C²) / 0.5 m.
Step 7: Calculate the numerator: (9 × 10^9) * (-4 × 10^-12) = -36 × 10^-3 N m².
Step 8: Now divide by 0.5 m: U = -36 × 10^-3 N m² / 0.5 = -72 J.
Step 9: Conclude that the electric potential energy of the system is U = -72 J.
