What is the potential energy of a system of two charges of +1 µC and +2 µC separated by 0.1 m?
Practice Questions
1 question
Q1
What is the potential energy of a system of two charges of +1 µC and +2 µC separated by 0.1 m?
180 J
90 J
45 J
36 J
U = k * (q1 * q2) / r = (9 × 10^9) * (1 × 10^-6 * 2 × 10^-6) / 0.1 = 36 J.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the potential energy of a system of two charges of +1 µC and +2 µC separated by 0.1 m?
Solution: U = k * (q1 * q2) / r = (9 × 10^9) * (1 × 10^-6 * 2 × 10^-6) / 0.1 = 36 J.
Steps: 9
Step 1: Identify the values given in the problem. We have two charges: q1 = +1 µC (microcoulomb) and q2 = +2 µC. We also have the distance between them, r = 0.1 m.
Step 2: Convert the charges from microcoulombs to coulombs. 1 µC = 1 × 10^-6 C, so q1 = 1 × 10^-6 C and q2 = 2 × 10^-6 C.
Step 3: Use the formula for potential energy (U) between two point charges: U = k * (q1 * q2) / r, where k is Coulomb's constant, approximately 9 × 10^9 N m²/C².
Step 4: Substitute the values into the formula: U = (9 × 10^9) * (1 × 10^-6 * 2 × 10^-6) / 0.1.
Step 5: Calculate the product of the charges: 1 × 10^-6 * 2 × 10^-6 = 2 × 10^-12.
Step 6: Substitute this value back into the equation: U = (9 × 10^9) * (2 × 10^-12) / 0.1.
