A particle of charge \(q\) and mass \(m\) is free to slide on a smooth horizontal table. Two fixed charges \(q\) are placed at \(\pm a \boldsymbol{j}\), and two fixed charges \(12 q\) at \(\pm 2 a i\). Find the electrostatic potential near the origin (see \(\S 6.2\) ). Show that this is a position of stable equilibrium, and find the frequencies of the normal modes of oscillation near it.

The electrostatic potential \(V (\boldsymbol{r})\) at a position \(\boldsymbol{r}\) due to a point charge \(q_i\) at \(\boldsymbol{r}_i\) is given by the formula: \( V (\boldsymbol{r}) = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^4 \frac{q_i}{|\boldsymbol{r} - \boldsymbol{r}_i|} \) In this problem, we have four fixed charges with positions \((-2a,0)\), \((2a,0)\), \((0,-a)\) and \((0,a)\). So, we need to compute the electrostatic potential \(V(0,0)\) at the origin. Plugging the position of the charges in the formula, we get: \( V(0,0) = \frac{1}{4\pi\epsilon_0} \left( \frac{12q}{2a} + \frac{12q}{2a} + \frac{q}{a} + \frac{q}{a} \right) \)

To find whether this position of equilibrium is stable or not, we need to compute the electrostatic force acting on the particle of charge \((q')\) and mass \((m')\) at this position. The electrostatic force acting on the charged particle is given by the negative gradient of the electrostatic potential: \( \boldsymbol{F}(\boldsymbol{r}) = q' \nabla V(\boldsymbol{r}) = - q' \left(\frac{\partial V}{\partial x}\boldsymbol{i} + \frac{\partial V}{\partial y}\boldsymbol{j}\right) \) Computing the partial derivatives of \(V(\boldsymbol{r})\), we get: \( \frac{\partial V}{\partial x} = 0 \) \( \frac{\partial V}{\partial y} = 0 \) This implies that the equilibrium position is stable, as the electrostatic force acting on the charged particle is zero at the origin.

To compute the frequencies of the normal modes of oscillation at the stable equilibrium position, we need to find the curvature of the electrostatic potential at the origin and solve the equation of motion for the charged particle. The curvature of the electrostatic potential \(V(\boldsymbol{r})\) at the origin is given by the second partial derivatives of the potential with respect to \(x\) and \(y\): \( \frac{\partial^2 V}{\partial x^2} = \frac{1}{4\pi\epsilon_0} \left( \frac{24q}{(2a)^3} \right) \) \( \frac{\partial^2 V}{\partial y^2} = \frac{1}{4\pi\epsilon_0} \left( \frac{2q}{a^3} \right) \) Using these values in the equation of motion (\(m'\Omega^2r = -q'\frac{\partial^2 V}{\partial x_i^2}r_{(i)}\)), we get the frequencies of the normal modes of oscillation as follows: \( \Omega_x^2 = \frac{-q'\frac{\partial^2 V}{\partial x^2}}{m'} = \frac{6q^2}{4\pi\epsilon_0 m' a^3} \) \( \Omega_y^2 = \frac{-q'\frac{\partial^2 V}{\partial y^2}}{m'} = \frac{q^2}{2\pi\epsilon_0 m' a^3} \) Therefore, the frequencies of the normal modes of oscillation at the stable equilibrium position are: \( \Omega_x = \sqrt{\frac{6q^2}{4\pi\epsilon_0 m' a^3}} \) \( \Omega_y = \sqrt{\frac{q^2}{2\pi\epsilon_0 m' a^3}} \)