Find the energy stored in a sphere of charge \(q\) and radius \(a\) with uniform charge density, and show that infinite energy is required to compress the sphere to a point. Find also the stored energy in the case where the charge is uniformly spread over the surface of the sphere.

First, let's calculate electric potential inside (\(r < a\)) and outside (\(r > a\)) the uniformly charged sphere. For a sphere with a charge density \(\rho = \frac{3q}{4\pi a^3}\) and radius \(a\), - Inside: \(V_{in}(r) = \frac{1}{4\pi\epsilon_0}\frac{q}{2a^3}(3a^2 - r^2)\) - Outside: \(V_{out}(r) = \frac{1}{4\pi\epsilon_0}\frac{q}{r}\)

Now, we can use the expression for the potential energy: \(U = \frac{1}{2} \int_{V} \rho(\vec{r}) V(\vec{r}) d^3\vec{r}\) We separate the integral into two parts (inside and outside the sphere) and consider that \(\rho_{out} = 0\): \(U = \frac{1}{2} \int_{V_{in}} \rho(\vec{r}) V_{in}(\vec{r}) d^3\vec{r} + 0\) Now, substitute the values for \(\rho(\vec{r})\) and \(V_{in}(\vec{r})\): \(U = \frac{1}{2} \int_{V_{in}} \frac{3q}{4\pi a^3} \cdot \frac{1}{4\pi\epsilon_0}\frac{q}{2a^3}(3a^2 - r^2) d^3\vec{r}\) Now integrate: \(U = \frac{3q^2}{40\pi\epsilon_0 a^5} \int_{0}^{a} (3a^2 - r^2) r^2\, 4\pi r^2 dr\) \(U = \frac{3q^2}{10\epsilon_0 a^5} \int_{0}^{a} (3a^2 - r^2) r^4 dr\) After integrating over the limits, we get: \(U = \frac{3q^2}{5\epsilon_0 a}\) This expression shows the energy stored in a uniformly charged sphere.

To analyze the energy required to compress the sphere to a point, we can take the limit as the sphere's radius approaches zero: \(\lim_{a \to 0} \frac{3q^2}{5\epsilon_0 a}\) This limit is infinite, which means that infinite energy is required to compress the sphere to a point.

Now let's consider the case when the charge is uniformly spread over the surface of the sphere. The surface charge density is given by: \(\sigma = \frac{q}{4\pi a^2}\) The expression for potential energy in this case is: \(U = \frac{1}{2} \int_{S} \sigma(\vec{r}) V(\vec{r}) d^2\vec{r}\) While the sphere has a uniform surface charge distribution, the electric potential is constant across the sphere: \(V(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{a}\) Now substitute the values for \(\sigma(\vec{r})\) and \(V(\vec{r})\) and integrate over the sphere's surface: \(U = \frac{1}{2} \int_{S} \frac{q}{4\pi a^2} \cdot \frac{1}{4\pi\epsilon_0}\frac{q}{a} d^2\vec{r}\) \(U = \frac{q^2}{32\pi^2\epsilon_0 a^3} \int_{S} d^2\vec{r}\) After integrating over the surface, we get: \(U = \frac{q^2}{8\pi\epsilon_0 a}\) This expression shows the energy stored in a sphere with uniform charge distribution on its surface.