A spherically symmetric charge distribution in free space (with \(0

The electric field intensity \(\textbf{E}\) can be related to the potential function \(V(r)\) by \( \textbf{E}=-\nabla V(r)\) Since \(V(r)\) is spherically symmetric, we have \( \textbf{E}=-\frac{dV}{dr} \hat{r}\) Given \(V(r) = V_0 a^2 / r^2\), so differentiating with respect to \(r\), we have \( \frac{dV}{dr} = -2 V_0 a^2 / r^3\) Therefore, the electric field intensity is \( \textbf{E}= \frac{2 V_0 a^2}{r^3} \hat{r} \)

To find the volume charge density, we can apply Gauss's law for the electric field, which states that the closed-surface integral of the electric field over a surface is equal to the enclosed charge divided by the permittivity of free space \(\epsilon_0\): \( \oint_S \textbf{E} \cdot d\textbf{S} = \frac{Q_{enc}}{\epsilon_0} \) For a spherical surface with radius \(r\), Gauss's law becomes \( 4\pi r^2 E = \frac{Q_{enc}}{\epsilon_0} \) Now, the volume charge density \(\rho_{vol}(r)\) is related to the enclosed charge by \(Q_{enc}=\int_V \rho_{vol}(r') dV'\), where \(dV'\) is an infinitesimal volume element and integration is done over the spherical volume. For a spherically symmetric charge distribution, we have \(dV' = 4\pi r'^2 dr'\), and \(Q_{enc}=\int_0^r \rho_{vol}(r') 4\pi r'^2 dr'\). To find \(\rho_{vol}(r)\), we can differentiate the expression for \(Q_{enc}\) with respect to \(r\). Then, from Gauss's law, we have \( \rho_{vol}(r) = \epsilon_0 \frac{d}{dr} \left( \frac{E}{4\pi r^2} \right) \) Substitute \(\textbf{E} = \frac{2 V_0 a^2}{r^3} \hat{r} \) into the expression, we have \( \rho_{vol}(r) = \epsilon_0 \frac{d}{dr} \left( \frac{2 V_0 a^2}{4\pi r^5} \right) =-\frac{6 \epsilon_0 V_0 a^2}{\pi r^6} \)

To find the charge contained inside radius \(a\), we can integrate the volume charge density over the spherical volume: \( Q = \int_{0}^{a} \rho_{vol}(r) 4\pi r^2 dr \) Substitute \(\rho_{vol}(r) = -\frac{6 \epsilon_0 V_0 a^2}{\pi r^6}\), and we have \( Q = \int_{0}^{a} -\frac{6 \epsilon_0 V_0 a^2}{ r^4} dr = -6 \epsilon_0 V_0 a^2 \left[ -\frac{1}{3r^3} \right]_0^a = 2 \epsilon_0 V_0 a\)

The energy density of an electric field is given by \( u_E = \frac{1}{2} \epsilon_0 E^2 \) To find the total energy stored in the charge distribution, we can integrate the energy density over the whole space: \( W = \int_0^{\infty} u_E 4\pi r^2 dr \) Substitute \(\textbf{E} = \frac{2 V_0 a^2}{r^3} \hat{r}\), and we have \( W = \int_0^{\infty} \frac{1}{2} \epsilon_0 \left( \frac{2 V_0 a^2}{r^3} \right)^2 4\pi r^2 dr = \frac{8 \epsilon_0 \pi V_0^2 a^4}{3} \int_0^{\infty} \frac{1}{r^4} dr \) To evaluate the integral, we notice that the integrand \(\frac{1}{r^4}\) is divergent at \(r = 0\). This indicates that the total energy stored in the charge distribution is infinite as we integrate over the whole space.