A small glass bead has been charged to $1.9 \mathrm { nC }$. What is the strength of the electric field $2.0 \mathrm {~cm}$ from the center of the bead? $\left( k = 1 / 4 \pi \varepsilon _ { 0 } = 8.99 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right)$

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The electric field $ E $ due to a point charge is given by $ E = \frac{k \cdot |q|}{r^2} $. Substituting the given values: $ k = 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} $, $ q = 1.9 \times 10^{-9} \, \mathrm{C} $, and $ r = 0.02 \, \mathrm{m} $, we find $ E = \frac{8.99 \times 10^9 \times 1.9 \times 10^{-9}}{(0.02)^2} = 4.3 \times 10^4 \, \mathrm{N/C} $.

A Small Glass Bead Has Been Charged To 1.9nC1.9 \mathrm { nC }1.9nC

A Small Glass Bead Has Been Charged To $1.9 \mathrm { nC }$