# Solution ## (1) **Electric Field Inside and Outside the Cylinder Using Gauss's Law** Given: - Cylinder radius: \( R \) - Volume charge density: \( \rho = A s^2 \), where \( s \) is the radial distance from the axis ### **Inside the Cylinder (\( s < R \)):** **Step 1: Find Enclosed Charge** Consider a cylindrical Gaussian surface of radius \( s \) and length \( L \): \[ q_{\text{enc}} = \int \rho \, dV \] \[ dV = 2\pi s' L ds' \] \[ q_{\text{enc}} = \int_0^s A {s'}^2 (2\pi s' L) ds' = 2\pi A L \int_0^s {s'}^3 ds' \] \[ = 2\pi A L \left[ \frac{{s'}^4}{4} \right]_0^s = 2\pi A L \frac{s^4}{4} \] **Step 2: Apply Gauss’s Law** \[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0} \] \[ E(2\pi s L) = \frac{2\pi A L s^4}{4\epsilon_0} \] \[ E = \frac{A s^3}{4 \epsilon_0} \] **Final Expression (Inside):** \[ \boxed{E_{\text{in}}(s) = \frac{A s^3}{4 \epsilon_0}, \quad s < R} \] --- ### **Outside the Cylinder (\( s > R \)):** **Step 1: Find Total Charge Enclosed by Cylinder** \[ q_{\text{enc}} = \int_0^R \rho \cdot 2\pi s' L ds' = 2\pi A L \int_0^R {s'}^3 ds' = 2\pi A L \frac{R^4}{4} \] \[ = \frac{\pi A L R^4}{2} \] **Step 2: Apply Gauss’s Law** \[ E(2\pi s L) = \frac{q_{\text{enc}}}{\epsilon_0} = \frac{\pi A L R^4}{2\epsilon_0} \] \[ E = \frac{A R^4}{4 \epsilon_0 s} \] **Final Expression (Outside):** \[ \boxed{E_{\text{out}}(s) = \frac{A R^4}{4 \epsilon_0 s}, \quad s > R} \] --- ## (2) **Potential Outside the Cylinder** The electric potential difference is given by: \[ V(s) - V(\infty) = -\int_{\infty}^{s} \vec{E} \cdot d\vec{l} \] Taking \( V(\infty) = 0 \), and using \( E_{\text{out}}(s) \) from part 1: \[ V(s) = -\int_{\infty}^{s} \frac{A R^4}{4 \epsilon_0 s'} ds' \] \[ = -\frac{A R^4}{4 \epsilon_0} \int_{\infty}^{s} \frac{1}{s'} ds' \] \[ = -\frac{A R^4}{4 \epsilon_0} [\ln s']_{\infty}^{s} \] \[ = -\frac{A R^4}{4 \epsilon_0} (\ln s - \ln \infty) \] Since \( \ln \infty \to \infty \), but we set \( V(\infty) = 0 \), so the potential is: \[ V(s) = \frac{A R^4}{4 \epsilon_0} \ln \left( \frac{\infty}{s} \right) \to -\frac{A R^4}{4 \epsilon_0} \ln s \] **Final Expression:** \[ \boxed{V(s) = -\frac{A R^4}{4\epsilon_0}\ln(s) + C \qquad (s > R)} \] where \( C \) is set by the reference point for the potential (typically \( V(\infty) = 0 \)).