The $\vec { E }$ field at a point 2.8 cm from a small object points toward the object and is equal to 180,000 N/C.What is the object's charge q? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

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The electric field $E$ is given by $E = \frac{kq}{r^2}$, where $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, $q$ is the charge, and $r$ is the distance from the charge. Given $E = 180,000 \, \text{N/C}$ and $r = 2.8 \, \text{cm} = 0.028 \, \text{m}$, we can solve for $q$: $q = \frac{E \cdot r^2}{k}$. 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 and pointing towards the object.

The E⃗\vec { E }E Field at a Point 2

The $\vec { E }$ Field at a Point 2