What is the magnitude of the electric field 2.8 cm from a tiny object that carries an excess charge of -16 nC? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²) A)180,000 N/C B)5100 N/C C)1.8 × 10¹⁴ N/C D)-180,000 N/C E)-5100 N/C

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Quizplus · Accepted Answer

The magnitude of the electric field (E) created by a point charge (q) at a distance (r) is given by the formula E = k|q|/r^2, where k is Coulomb's constant (9.0 × 10^9 N·m^2/C^2). Substituting the given values (q = -16 nC = -16 × 10^-9 C, r = 2.8 cm = 0.028 m) into the formula, we find E = (9.0 × 10^9 N·m^2/C^2)(16 × 10^-9 C)/(0.028 m)^2, which calculates to approximately 180,000 N/C. The negative sign of the charge affects the direction of the electric field, not its magnitude, hence the correct answer is the positive magnitude.

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