The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C.What is the object's charge q? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²) A)-16 nC B)+16 nC C)-17 nC D)+17 nC

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The electric field E is given by $E = \frac{kq}{r^2}$, where $E = 180,000 \, \text{N/C}$, $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, and $r = 0.028 \, \text{m}$. Solving for $q$, we find $q = \frac{E \cdot r^2}{k}$. Substituting the given values, $q = \frac{180,000 \cdot (0.028)^2}{8.99 \times 10^9}$, which calculates to approximately $-16 \times 10^{-9}$ C or $-16 \, \text{nC}$, indicating the charge is negative because the electric field points toward the object.

The Electric Field at a Point 2