Cylinder E-Field

The cylinder can be parameterised using cylindrical coordinates. That means, a point on the cylinder has the coordinates:

$\vec{r}_c = \cos{\theta} \hat {i} + \sin{\theta} \hat {j} + z \hat {k}$

The point at which the electric field is to be evaluated is:

$\vec{r}_p = 0.5 \hat {i} + 0 \hat {j} + 0 \hat {k}$

The vector joining these two points and directed towards the point os interest $P$ is:

$\vec{r}=\vec{r}_p-\vec{r}_c$

Consider a surface area element of area $dS = d\theta \ dz$ at the point $\vec{r}_c$ on the cylinder. The charge on this elementary surface is also $dQ = \sigma d\theta \ dz$ , where $\sigma = 1$ . The field at the point $P$ due to this elementary surface is:

$d\vec{E} = \frac{1}{4\pi \epsilon_o}\frac{dQ}{\lvert \vec{r} \rvert^3}\vec{r}$

Substituting ll expression above gives us elementary electric fields along the X,Y and Z directions. Recognising the symmetry of the geometry, the Y and Z components are ignored as they will eventually cancel out. This leaves us with the X component which is:

$dE_x = \frac{1}{4\pi \epsilon_o}\frac{\left(0.5 - \cos{\theta}\right)\sigma d\theta \ dz}{\left(\left(0.5 - \cos{\theta}\right)^2 + \sin^2{\theta} + z^2\right)^{3/2}}$

Therefore:

$\epsilon_o E_x = \frac{1}{4\pi} \int_{-1}^{1} \int_{0}^{2 \pi} \frac{\left(0.5 - \cos{\theta}\right)\sigma d\theta \ dz}{\left(\left(0.5 - \cos{\theta}\right)^2 + \sin^2{\theta} + z^2\right)^{3/2}} = \approx -0.08988$

From here, the magnitude of the above value is that answer.

By symmetry, it suffices to consider the $x$ -component. Now, by Coulomb's Law, $\epsilon_0 E_T=\frac{1}{4\pi}\int_0^{2\pi}\int_{-1}^1\frac{\cos \theta -0.5}{\left((\cos \theta -0.5)^2+\sin^2 \theta+s^2\right)^{3/2}}ds\ d\theta\approx \boxed{0.09}$