Answer
$E = 19.2~N/C$
Work Step by Step
We can find the charge density from $r = 0$ to $r = 0.040~m$:
$\sigma = \int_{0}^{0.040~m}~Ar^2~dr$
$\sigma = \frac{Ar^3}{3}\Big\vert_{0}^{0.040~m}$
$\sigma = \frac{(2.5\times 10^{-6}~C/m^5)(0.040~m)^3}{3} - \frac{A(0)^3}{3}$
$\sigma = 5.33\times 10^{-11}~C/m^2$
We can find the linear charge density $\lambda$:
$\lambda = (5.33\times 10^{-11}~C/m^2)(1.0~m) = 5.33\times 10^{-11}~C/m$
We can use Equation (23-12) to find the electric field:
$E = \frac{\lambda}{2\pi~\epsilon_0~r}$
$E = \frac{5.33\times 10^{-11}~C/m}{(2\pi)~(8.854\times 10^{-12}~F/m)~(0.050~m)}$
$E = 19.2~N/C$
