Certainly, let’s review the calculations for the potential of an infinitely long charged cylinder, taking into account that in Gaussian units, Gauss’s law is indeed expressed as \Phi_E = 4\pi Q_{\text{enc}}, where Q_{\text{enc}} is the charge enclosed by the Gaussian surface.
The electric flux \Phi_E through a cylindrical Gaussian surface of radius r and length l surrounding the charged cylinder is equal to the electric field E times the surface area of the cylindrical surface:
\Phi_E = E \cdot (2\pi rl).
According to Gauss’s law in Gaussian units, this flux is related to the enclosed charge Q_{\text{enc}} by:
E \cdot (2\pi rl) = 4\pi Q_{\text{enc}}.
For a cylinder with surface charge density \sigma, the total charge per unit length \lambda on the surface is given by:
\lambda = \sigma \cdot (2\pi a),
where a is the radius of the cylinder. The enclosed charge by the Gaussian surface is Q_{\text{enc}} = \lambda l:
Q_{\text{enc}} = \sigma \cdot (2\pi a) \cdot l.
Substituting Q_{\text{enc}} back into Gauss’s law gives us:
E \cdot (2\pi rl) = 4\pi \cdot \sigma \cdot (2\pi a) \cdot l.
Solving for E yields:
E = \frac{2\sigma a}{r}.
Now, let’s calculate the potential \varphi at a distance r from the axis of the cylinder by integrating the electric field from infinity to r:
\varphi(r) - \varphi(\infty) = -\int_{\infty}^{r} E \, dr = -\int_{\infty}^{r} \frac{2\sigma a}{r} \, dr.
Performing the integration:
\varphi(r) - \varphi(\infty) = -2\sigma a \int_{\infty}^{r} \frac{1}{r} \, dr = -2\sigma a \ln \left( \frac{r}{a} \right) + C,
where C is the integration constant. Assuming that the potential at infinity is zero (\varphi(\infty) = 0) and choosing the constant C such that the potential \varphi(a) at the surface of the cylinder is zero, we get:
\varphi(r) = -2\sigma a \ln \left( \frac{r}{a} \right).
This is the correct expression for the potential outside an infinitely long cylinder charged uniformly on its surface in Gaussian units, which should match the value in the book.