What is the potential energy of a system of two charges of +1μC and -1μC separated by 0.1m?
Practice Questions
1 question
Q1
What is the potential energy of a system of two charges of +1μC and -1μC separated by 0.1m?
-0.09 J
0.09 J
-0.18 J
0.18 J
Potential energy U = k * q1 * q2 / r = (9 × 10^9) * (1 × 10^-6) * (-1 × 10^-6) / 0.1 = -0.09 J.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the potential energy of a system of two charges of +1μC and -1μC separated by 0.1m?
Solution: Potential energy U = k * q1 * q2 / r = (9 × 10^9) * (1 × 10^-6) * (-1 × 10^-6) / 0.1 = -0.09 J.
Steps: 6
Step 1: Identify the values given in the problem. We have two charges: q1 = +1μC (which is 1 × 10^-6 C) and q2 = -1μC (which is -1 × 10^-6 C). The distance between them, r, is 0.1m.
Step 2: Use the formula for potential energy (U) between two charges: U = k * q1 * q2 / r, where k is the Coulomb's constant (approximately 9 × 10^9 N m²/C²).
Step 3: Substitute the values into the formula: U = (9 × 10^9) * (1 × 10^-6) * (-1 × 10^-6) / 0.1.
