91Ó°ÊÓ

To find the electric field inside and outside the cylinder, we will use Gauss's Law, which states that the total electric flux through a closed Gaussian surface is equal to the charge enclosed by it divided by the vacuum permittivity constant ε₀. First, consider a Gaussian surface of a cylinder with radius r that encloses the given cylinder (i.e., \(0 < r < R\)), so that we can find the electric field inside the cylinder. Since there is no charge enclosed within the Gaussian surface that is inside the given cylinder, the electric field inside the given cylinder should be zero. Therefore, \(E_{inside} = 0\) Now, let's consider a Gaussian surface of a cylinder with radius r that encloses the given cylinder (i.e., \(r>R\)), so that we can find the electric field outside the cylinder. For this Gaussian surface, the charge enclosed would be the charge uniformly distributed on the surface of the cylinder, which can be calculated as: \(Q_{enc} = λL\) where L is the length of the cylinder. Now, the electric flux through this Gaussian surface can be given by: \(Φ_E = E_{outside} *A\) where A is the surface area of the Gaussian cylindrical surface, and E is the resultant electric field at a distance r from the center of the cylinder. A = 2πrL, so \(E_{outside} * 2πrL = \frac{λL}{ε₀}\) \(E_{outside} = \frac{λ}{2πε₀r}\) So, our electric field values are: \(E_{inside} = 0\) \(E_{outside} = \frac{λ}{2πε₀r}\)

Now let's find the electric potential inside and outside the cylinder. First, let's calculate the electric potential outside the cylinder. Let V(r) be the electric potential at a distance r from the axis of the cylinder. The electric field is the negative gradient of the electric potential: \(E = -\frac{dV(r)}{dr}\) Now, let's integrate: \(-\int_{V(R)}^{V(r)} dV = \int_{R}^{r} \frac{λ}{2πε₀r'} dr'\) \(V(R) - V(r) = -\frac{λ}{2πε₀} \ln(\frac{r}{R})\) \(V(r) = V(R) + \frac{λ}{2πε₀} \ln(\frac{r}{R})\) For the electric potential inside the cylinder, since the electric field is zero, the electric potential is constant and result: \(V_{inside}(r) = V(R)\) So, our electric potential values are: \(V_{inside}(r) = V(R)\) \(V_{outside}(r) = V(R) + \frac{λ}{2πε₀}\ln(\frac{r}{R})\)

To plot the electric field and electric potential as a function of distance r from the center of the cylinder, we will express their values as follows: For Electric Field: \(E(r) = \begin{cases} 0 & 0 \leq r R \end{cases} \) For Electric Potential: \(V(r) = \begin{cases} V(R) & 0 \leq r R \end{cases} \) Plot these functions for different values of \(r\) to visualize the electric field and electric potential as a function of distance from the center of the rod. Keep in mind that these are only the analytical expressions, and the plotting process would require computation in software or a graphic calculator.