## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$Q = 0.056~nC$
We can use the equation for the electric field strength at a distance $r$ from a charged wire to find the linear charge density: $E = \frac{\lambda}{2\pi ~\epsilon_0~r}$ $\lambda = E~2\pi ~\epsilon_0~r$ $\lambda = (2000~N/C)~(2\pi)(8.85\times 10^{-12}~C^2/N~m^2)(0.050~m)$ $\lambda = 5.6\times 10^{-9}~C/m$ We can find the charge on a 1.0 cm segment of wire. $\lambda = \frac{Q}{L}$ $Q = \lambda~L$ $Q = (5.6\times 10^{-9}~C/m)(0.010~m)$ $Q = 0.056\times 10^{-9}~C$ $Q = 0.056~nC$