A uniform volume charge has constant density \(\rho_{v}=\rho_{0} \mathrm{C} / \mathrm{m}^{3}\) and fills the region \(r

Given the uniform volume charge density \(\rho_v = \rho_0\) inside the sphere of radius a, the total charge enclosed within radius r can be calculated by integrating the charge density over the volume, which is given by: \(Q(r) = \int_{0}^{r} \rho_0 dV = \rho_0 \int_{0}^{r} 4\pi r'^2 dr'\) Now we can calculate the integral: \(Q(r) = \rho_0 \int_{0}^{r} 4\pi r'^2 dr' = \rho_0 \left[ \frac{4\pi r'^3}{3} \right]_0^r = \frac{4\pi \rho_0 r^3}{3}\) #Step 2: Apply Gauss's Law to find the electric field intensity#

According to Gauss's Law, we know that for a spherical symmetric charge distribution like the given one, the electric field intensity multiplied by the area of the Gaussian surface (which is a spherical surface) is equal to the enclosed charge divided by the permittivity ε: \(E(r) \cdot 4\pi r^2 = \frac{Q(r)}{\epsilon} \Rightarrow E(r) = \frac{Q(r)}{4\pi \epsilon r^2}\) From step 1 we found the enclosed charge, so we substitute it into the equation above: \(E(r) = \frac{\frac{4\pi \rho_0 r^3}{3}}{4\pi \epsilon r^2} = \frac{\rho_0 r}{3\epsilon}\) This electric field intensity is valid inside the sphere, i.e., when \(r < a\). Since the conducting spherical shell is at ground potential (zero potential), there is no electric field outside the sphere, i.e., for \(r \geq a\), \(E(r) = 0\). #Step 3: Integrate the electric field to find the potential#

The potential at any point in a radial electric field can be calculated by integrating the electric field intensity over distance. Since there is no electric field outside the sphere, the potential is constant at r ≥ a. As the conducting spherical shell is at ground potential, V(a) = 0, and therefore the potential outside the sphere is also zero. Now, we need to find the potential inside the sphere, for r < a. We need to integrate the electric field we found in step 2: \(V(r) - V(a) = -\int_{a}^{r} E(r') dr' = -\int_{a}^{r} \frac{\rho_0 r'}{3\epsilon} dr'\) \(V(r) = -\int_{a}^{r} \frac{\rho_0 r'}{3\epsilon} dr' + V(a)\) Since the conducting spherical shell is at ground potential (zero potential), V(a) = 0, so the equation becomes: \(V(r) = -\int_{a}^{r} \frac{\rho_0 r'}{3\epsilon} dr'\) Now calculating the integral: \(V(r) = -\int_{a}^{r} \frac{\rho_0 r'}{3\epsilon} dr' = -\frac{\rho_0}{6\epsilon}(r^2 - a^2)\) Therefore, the potential is: $V(r) = \begin{cases} -\frac{\rho_0}{6\epsilon}(r^2 - a^2) & \text{for}\ r < a \\ 0 & \text{for}\ r \geq a \end{cases}$ In conclusion, we have found: 1. The potential everywhere: \((a)\) $V(r) = \begin{cases} -\frac{\rho_0}{6\epsilon}(r^2 - a^2) & \text{for}\ r < a \\ 0 & \text{for}\ r \geq a \end{cases}$ 2. The electric field intensity everywhere: \((b)\) $E(r) = \begin{cases} \frac{\rho_0 r}{3\epsilon} & \text{for}\ r < a \\ 0 & \text{for}\ r \geq a \end{cases}$