Answer
The net charge enclosed within the cylinder is $~~2.2\times 10^{-12}~C$
Work Step by Step
In part (a), we found that the flux through the end of the cylinder at $x = 2.0~m$ is $~~0.50~N~m^2/C$
We can find the electric field at $x = 0$:
$E = (x+2)~\hat{i}~N/C$
$E = (0+2)~\hat{i}~N/C$
$E = (2.0~\hat{i})~N/C$
We can find the flux through the end of the cylinder at $x = 0$:
$\Phi = E~\cdot A$
$\Phi = E~\pi~r^2~cos~180^{\circ}$
$\Phi = -(2.0~N/C)~(\pi)~(0.20~m)^2$
$\Phi = -0.25~N~m^2/C$
The net flux through the cylinder is $0.25~N~m^2/C$
We can find the enclosed charge:
$q_{enc} = \epsilon_0~\Phi$
$q_{enc} = (8.854\times 10^{-12}~F/m)(0.25~N~m^2/C)$
$q_{enc} = 2.2\times 10^{-12}~C$
The net charge enclosed within the cylinder is $~~2.2\times 10^{-12}~C$
