91Ó°ÊÓ

In spherical coordinates, Poisson's equation is given by: $$\nabla^2 V = -\frac{\rho(r)}{\epsilon_0}$$ Where \(\nabla^2\) is the Laplacian operator in spherical coordinates and is given by: $$\nabla^2 V = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial V}{\partial r}\right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left(\sin \theta \frac{\partial V}{\partial \theta}\right) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 V}{\partial \phi^2}$$

In our case, \(\theta\) and \(\phi\) do not appear in the expression for \(V(r)\), so the derivatives with respect to these variables will be zero: $$\nabla^2 V = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial V}{\partial r}\right)$$

Region (r <= a): $$V_1(r) = \left[\rho_{0} /\left(6 \epsilon_{0}\right)\right]\left[3 a^{2}-r^{2}\right]$$ Let's calculate the first and second derivatives of \(V_1(r)\) with respect to \(r\): $$\frac{\partial V_1}{\partial r} = -\frac{2r\rho_0}{6\epsilon_0}$$ $$\frac{\partial^2 V_1}{\partial r^2} = -\frac{2\rho_0}{6\epsilon_0}$$ Now let's plug these expressions into the expression for the Laplacian: $$\nabla^2 V_1 = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial V_1}{\partial r}\right)= \frac{1}{r^2} \frac{\partial}{\partial r}\left(-r^2 \frac{2\rho_0}{6\epsilon_0}\right) = -2\frac{\rho_0}{3\epsilon_0}$$ Region (r >= a): $$V_2(r) = \left(a^{3} \rho_{0}\right) /\left(3 \epsilon_{0} r\right)$$ Let's calculate the first and second derivatives of \(V_2(r)\) with respect to \(r\): $$\frac{\partial V_2}{\partial r} = -\frac{a^3 \rho_0}{3 \epsilon_0 r^2}$$ $$\frac{\partial^2 V_2}{\partial r^2} = \frac{2 a^3 \rho_0}{3 \epsilon_0 r^3}$$ Now let's plug these expressions into the expression for the Laplacian: $$\nabla^2 V_2 = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial V_2}{\partial r}\right) = \frac{1}{r^2} \frac{\partial}{\partial r}\left(- a^3\frac{\rho_0}{3\epsilon_0}\right) = -2\frac{\rho_0}{3\epsilon_0}\frac{a^3}{r^3}$$

Region (r <= a): From Poisson's equation, we have: $$-\frac{\rho_1(r)}{\epsilon_0} = -2\frac{\rho_0}{3\epsilon_0}$$ or $$\rho_1(r) = \frac{2}{3}\rho_0$$ Region (r >= a): From Poisson's equation, we have: $$-\frac{\rho_2(r)}{\epsilon_0} = -2\frac{\rho_0}{3\epsilon_0}\frac{a^3}{r^3}$$ or $$\rho_2(r) = \frac{2}{3}\rho_0\frac{a^3}{r^3}$$ So, we have the expressions for the volume charge density for both regions: $$\rho(r) =\left\\{\begin{array}{c}{\frac{2}{3}\rho_0 \quad (r \leq a)} \\\ \frac{2}{3}\rho_0\frac{a^3}{r^3} \quad (r \geq a)\end{array}\right.$$

To calculate the total charge present, we need to integrate the volume charge density over the volume: $$Q = \int \rho(r) dV = \int_{0}^{a} \rho_1(r) dV + \int_{a}^{\infty} \rho_2(r) dV$$ Since the charge density is spherically symmetric, we can calculate the integral in spherical coordinates: $$Q = \int_{0}^{a} \rho_1(r) 4\pi r^2 dr + \int_{a}^{\infty} \rho_2(r) 4\pi r^2 dr$$ Substitute the expressions for \(\rho_1(r)\) and \(\rho_2(r)\), we have: $$Q = 4\pi\left(\int_{0}^{a} \frac{2}{3}\rho_0 r^2 dr + \int_{a}^{\infty} \frac{2}{3}\rho_0\frac{a^3}{r} dr\right)$$ Now we can evaluate the integrals: $$Q = 4\pi\frac{2}{3}\rho_0\left(\left[\frac{1}{3} r^3\right]_{0}^{a} + \left[-a^3 \ln(r)\right]_{a}^{\infty}\right)$$ The second integral converges for the given function. Therefore, we get: $$Q = 4\pi\frac{2}{3}\rho_0\left(\frac{1}{3} a^3 - 0 + a^3\right) = 4\pi\frac{2}{3}\rho_0 a^3\left(\frac{1}{3} + 1\right)$$ So, the total charge present is: $$Q = \frac{8}{3}\pi\rho_0 a^3$$