Find the quadrupole moment of a distribution of charge on the surface of a sphere of radius \(a\) with surface charge density \(\sigma=\sigma_{0}\left(\frac{3}{2} \cos ^{2} \theta-\frac{1}{2}\right)\). Find the total energy stored in this distribution.

To find the charge distribution on the sphere's surface, we multiply the surface charge density formula by the surface area element \(dS\) which is given by \(a^2 \sin \theta d\theta d\phi\). This gives: $$d Q = \sigma dS = \sigma_0 \left(\frac{3}{2}\cos^2\theta - \frac{1}{2}\right) a^2 \sin \theta d\theta d\phi$$

The components of the quadrupole moment tensor are given by the following integral: $$ Q_{lm} = \frac{1}{2}\int \rho(\vec{r}) (3x_l x_m - r^2 \delta_{lm}) d^3 r$$ In spherical coordinates, the tensor components are given by: $$ Q_{\theta\theta}=\frac{1}{2}\int dQ(3a^2 \cos^2\theta - a^2)$$ $$ Q_{\phi\phi}=\frac{1}{2}\int dQ a^2(3 \cos^2\theta \sin^2\theta - \sin^2\theta)$$ $$ Q_{\theta\phi}=\frac{1}{2}\int dQ(3a^2 \cos\theta \sin\theta \cos\phi \sin\theta \sin\phi)$$

Now, let's compute each component of the quadrupole moment tensor. First, we calculate \(Q_{\theta \theta}\): $$ Q_{\theta\theta}=\frac{1}{2}\int_{0}^{\pi}\int_{0}^{2\pi} dQ (3a^2 \cos^2\theta - a^2) =\frac{3}{2} a^2 \sigma_0 \int_{0}^{\pi}\int_{0}^{2\pi} (\cos^2\theta\sin\theta-\frac{1}{3}\sin\theta)d\theta d\phi = 0$$ Next, let's compute \(Q_{\phi\phi}\): $$ Q_{\phi\phi}=\frac{1}{2}\int_{0}^{\pi}\int_{0}^{2\pi} dQ a^2(3 \cos^2\theta \sin^2\theta - \sin^2\theta) = a^2 \sigma_0\int_{0}^{\pi}\int_{0}^{2\pi} (\frac{3}{2}\cos^2\theta\sin^3\theta-\frac{1}{2}\sin^3\theta)d\theta d\phi = 0$$ Lastly, we compute \(Q_{\theta\phi}\): $$ Q_{\theta\phi}=\frac{1}{2}\int_{0}^{\pi}\int_{0}^{2\pi} dQ(3a^2 \cos\theta \sin\theta \cos\phi \sin\theta \sin\phi) = 0$$ Since all components of the quadrupole moment tensor are zero, the quadrupole moment is also zero.

Now that we have found the quadrupole moment tensor, we can find the total energy stored in the distribution of charge. To find the total stored energy, we can use the following formula: $$ W = \frac{1}{2}\frac{1}{4\pi\epsilon_0}\int d^3 r \rho(\vec{r}) V(\vec{r})$$ However, since the quadrupole moment tensor is zero, the potential \(V(\vec{r})\) due to the quadrupole moment is also zero. Therefore, there is no energy stored in the distribution of charge due to the quadrupole moment.$$ W = 0$$.