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ExamsJEE AdvancedPhysics

Two point charges, +100 microC and -4 microC, are placed at points (-2*sqrt(3), 3*sqrt(3), -4) m and (4*sqrt(3), -5*sqrt(3), 6) m respectively in a Cartesian frame. Determine the force vector acting on the -4 microC charge (coordinates in metres, use k = 9*10⁹ N m²/C²).

  1. 9*10⁻⁴ (3*sqrt(3) i - 4*sqrt(3) j + 5k)
  2. 9*10⁻⁴ (-3*sqrt(3) i + 4*sqrt(3) j - 5k)
  3. 2.25*10⁻⁴ (-3*sqrt(3) i + 4*sqrt(3) j - 5k)
  4. 2.25*10⁻⁴ (3*sqrt(3) i - 4*sqrt(3) j + 5k)

Correct answer: 9*10⁻⁴ (-3*sqrt(3) i + 4*sqrt(3) j - 5k)

Solution

The displacement from the +charge to the -charge has magnitude 20 m. Coulomb's law gives a force magnitude of k|q1q2|/r² = 9*10⁻³ N, directed (attraction) from the -4 microC charge toward the +100 microC charge. Writing it in vector form gives 9*10⁻⁴(-3*sqrt(3) i + 4*sqrt(3) j - 5k).

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