Answer
$n = 2.0 \times 10^{10} $
Work Step by Step
To find excess neutron, n when the volume charge density, $\rho$ is uniform, we need to use this equation
$n = \frac{q}{-e} $ and
$dq = \rho A dx$
$\rho=– 4.00 \times 10^{–6} C/m^3$
$A = 4.0cm^2$
$L = 2.00 m$
$n = \frac{dq}{-e} $
$n =\frac{ \rho A}{-e} \int^L_0 dx$
$n =\frac{ |\rho| AL}{-e}$
$n = \frac{(4.00 \times 10^{–6} C/m^3)( 0.0004m^2)(2.0m)}{-1.6 \times 10^{-19} C}$
$n = 2.0 \times 10^{10} $
