Three point charges are located on the x-axis at the following positions: Q1 = +2.00 μC is at x = 1.00 m, Q2 = +3.00 μC is at x = 0.00, and Q3 = -5.00 μC is at x = -1.00 m. What is the magnitude and direction of the electric force on Q2? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)

Question

Three point charges are located on the x-axis at the following positions: Q1 = +2.00 μC is at x = 1.00 m, Q2 = +3.00 μC is at x = 0.00, and Q3 = -5.00 μC is at x = -1.00 m. What is the magnitude and direction of the electric force on Q2?  (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)

BlackTom AI · Accepted Answer

We’re asked to find the electric force on Q2 due to the other two charges, Q1 and Q3, along the x-axis.
First, note positions and signs: Q1 = +2.00 μC at x = +1.00 m, Q2 = +3.00 μC at x = 0.00 m, Q3 = −5.00 μC at x = −1.00 m. The force between point charges follows F = k q1 q2 / r^2 along the line joining them, with direction away from like charges and toward opposite charges.
Calculate the force on Q2 due to Q1:
- Distance r between Q1 and Q2 is 1.00 m.
- q1 q2 = (2.00 μC)(3.00 μC) = 6.00 μC^2; in SI: 6.00 × 10^−12 C^2.
- Magnitude F_Q1_on_Q2 = k (6.00 × 10^−12) / (1.00)^2 = (8.99 × 10^9)(6.00 × 10^−12) ≈ 0.0539 N.
- Since Q1 and Q2 have like charges, the force on Q2 due to Q1 is repulsive, pushing Q2 to the left (toward negative x).
Compute the force on Q2 due to Q3:
- Distance r between Q3 and Q2 is also 1.00 m.
- q2 q3 = (3.00 μC)(−5.00 μC) = −15.00 μC^2; magnitude 15.00 × 10^−12 C^2.
- Magnitude F_Q3_on_Q2 = k (15.0 × 10^−12) / (1.00)^2 = (8.99 × 10^9)(15.0 × 10^−12) ≈ 0.1349 N.
- Since Q3 is negative and Q2 is positive, they attract, so the force on Q2 due to Q3 is toward Q3, i.e., to the left (negative x).
Now sum the contributions along the x-axis:
- Both forces on Q2 point to the left (negative x direction).
- F_total = 0.0539 N + 0.1349 N ≈ 0.1888 N to the left.
Compare with the answer choices:
- Option 1: 0.158 N, to the right — incorrect magnitude and opposite direction.
- Option 2: 0.189 N, to the left — matches the calculated total magnitude (approximately 0.189 N) and direction (left).
- Option 3: 0.0540 N, to the right — incorrect magnitude and opposite direction.
- Option 4: 0.189 N, to the right — correct magnitude but incorrect direction.
- Option 5: 0.158 N, to the left — incorrect magnitude.
Thus, the correct physical result corresponds to the scenario where the net force points left with a magnitude near 0.189 N, arising from the additive leftward forces due to both Q1 and Q3 on Q2.

Three point charges are located on the x-axis at the following positions: Q1 = +2.00 μC is at x = 1.00 m, Q2 = +3.00 μC is at x = 0.00, and Q3 = -5.00 μC is at x = -1.00 m. What is the magnitude and direction of the electric force on Q2? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)单项选择题

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