Answer
$Q = 0.056~nC$
Work Step by Step
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.
$\lambda = \frac{Q}{L}$
$Q = \lambda~L$
$Q = (5.6\times 10^{-9}~C/m)(0.010~m)$
$Q = 0.056\times 10^{-9}~C$
$Q = 0.056~nC$
