The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C.What is the charge on the object? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

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Quizplus · Accepted Answer

The electric field E is given by E = kq/r^2, where E is the electric field, k is Coulomb's constant (8.99 × 10^9 N·m^2/C^2), q is the charge, and r is the distance from the charge. Given E = 180,000 N/C and r = 0.015 m, solving for q gives q = Er^2/k. Substituting the given values, q = (180,000 N/C)(0.015 m)^2 / (8.99 × 10^9 N·m^2/C^2) = -4.5 × 10^-9 C or -4.5 nC, indicating the charge is negative.

The Electric Field 1