A charge of 2 X 10-7 C is acted upon by a force of 0.1 N. Determine the distance to the other charge of 4.5 X 10-7 C with both the charges in vacuum.

Question

BlackTom AI · Accepted Answer

Coulomb’s law gives the magnitude of the electrostatic force: F = k q1 q2 / r^2, where k = 8.99 × 10^9 N·m^2/C^2 in vacuum. Here q1 = 2.0 × 10^-7 C, q2 = 4.5 × 10^-7 C, and F = 0.1 N. Solve for r:

1) Compute the product k q1 q2:
   q1 q2 = (2.0 × 10^-7)(4.5 × 10^-7) = 9.0 × 10^-14 C^2.
   k q1 q2 = (8.99 × 10^9)(9.0 × 10^-14) = 8.99 × 9.0 × 10^(9-14)
            = 80.91 × 10^-5 = 8.091 × 10^-4 (units: N·m^2).

2) Rearrange Coulomb’s law to find r:
   r^2 = k q1 q2 / F = (8.091 × 10^-4) / 0.1 = 8.091 × 10^-3.
   r = sqrt(8.091 × 10^-3) ≈ 0.089964 m ≈ 0.09 m.

Why this is correct: plugging r = 0.09 m back into Coulomb’s law yields F ≈ (8.091×10^-4)/(0.09^2) = (8.091×10^-4)/0.0081 ≈ 0.0999 N ≈ 0.1 N, matching the given force.

Why the other choices are incorrect:
- a) 0.05 m: r^2 = 0.0025 ⇒ F = 8.091×10^-4 / 0.0025 ≈ 0.324 N (much larger than 0.1 N).
- c) 0.07 m: r^2 = 0.0049 ⇒ F ≈ 8.091×10^-4 / 0.0049 ≈ 0.165 N (too large).
- d) 0.03 m: r^2 = 0.0009 ⇒ F ≈ 8.991×10^-4 / 0.0009 ≈ 0.899 N (far too large).

Conclusion: The distance is approximately 0.09 m, so the correct answer is b. 0.09.

A charge of 2 X 10-7 C is acted upon by a force of 0.1 N. Determine the distance to the other charge of 4.5 X 10-7 C with both the charges in vacuum.单项选择题

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