A small glass bead has been charged to 1.9 nC.What is the strength of the electric field 2.0 cm from the center of the bead? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

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Quizplus · Accepted Answer

The strength of the electric field $E$ created by a point charge $q$ at a distance $r$ is given by $E = \frac{kq}{r^2}$, where $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$. Substituting $q = 1.9 \, \text{nC} = 1.9 \times 10^{-9} \, \text{C}$ and $r = 2.0 \, \text{cm} = 0.02 \, \text{m}$ into the formula gives $E = \frac{8.99 \times 10^9 \times 1.9 \times 10^{-9}}{(0.02)^2} = 4.3 \times 10^{14} \, \text{N/C}$.

A Small Glass Bead Has Been Charged to 1