Two equal magnitude negative charges \((-q\) and \(-q)\) are fixed at coordinates \((-d, 0)\) and \((d, 0) .\) A positive charge of the same magnitude, \(q\), and with mass \(m\) is placed at coordinate \((0,0),\) midway between the two negative charges. If the positive charge is moved a distance \(\delta \ll d\) in the positive \(y\) -direction and then released, the resulting motion will be that of a harmonic oscillator-the positive charge will oscillate between coordinates \((0, \delta)\) and \((0,-\delta)\). Find the net force acting on the positive charge when it moves to \((0, \delta)\) and use the binomial expansion \((1+x)^{n} \approx 1+n x,\) for \(x \ll 1,\) to find an expression for the frequency of the resulting oscillation.

When the positive charge is at the coordinate \((0, \delta)\), it will experience a force from each of the fixed negative charges. Using Coulomb's law, we can find the force acting on the positive charge due to each of the negative charges: \(F_{1}=\frac{k(-q)q}{[(0+d)^2+\delta^2]}=\frac{-kq^2}{(d^2+\delta^2)}\) \(F_{2}=\frac{k(-q)q}{[(0-d)^2+\delta^2]}=\frac{-kq^2}{(d^2+\delta^2)}\) Here, \(F_{1}\) and \(F_{2}\) are the forces acting on the positive charge due to the negative charges at coordinates \((-d, 0)\) and \((d, 0)\) respectively, and \(k\) is the electrostatic constant. Notice that the magnitudes of the forces are the same, but they have different directions.

The net force acting on the positive charge will be the vector sum of the forces \(F_{1}\) and \(F_{2}\). Let's choose x and y axes such that the negative charges are on the x-axis and the positive charge moves along the y-axis. The horizontal components of the forces cancel each other out since the magnitudes of the forces are the same and their directions are symmetrical about the y-axis. Therefore, only the vertical components of the forces need to be considered. The angle \(\theta\) between the vertical axis and the forces can be calculated as follows: \(\theta = \tan^{-1}(\frac{\delta}{d})\) The vertical component of each force can now be calculated as follows: \(F_{1y} = F_{1} \cos(\theta) = \frac{-kq^2}{(d^2+\delta^2)}\cos(\tan^{-1}(\frac{\delta}{d}))\) As \(\frac{\delta}{d} \ll 1\), we can approximate \(\cos(\tan^{-1}(\frac{\delta}{d}))\) with the binomial expansion: \(\cos(\tan^{-1}(\frac{\delta}{d})) \approx 1 - \frac{1}{2}(\frac{\delta}{d})^2\) Now using this approximate value, we can find \(F_{1y}\): \(F_{1y} \approx -kq^2 (1-\frac{1}{2}(\frac{\delta}{d})^2) \frac{1}{d^2+\delta^2}\) Since the positive charge is symmetrically located between the negative charges, the vertical component of force \(F_{2}\) acting on it will be the same as \(F_{1y}\). The net vertical force, \(F_{net}\), will be the sum of the vertical components of \(F_{1}\) and \(F_{2}\): \(F_{net} = 2 F_{1y} \approx -\frac{2kq^2}{d^2+\delta^2}(1-\frac{1}{2}(\frac{\delta}{d})^2)\)

The resulting motion of the positive charge is that of a harmonic oscillator. In this case, the net force acting on the charge is proportional to its displacement from the equilibrium position: \(F_{net} = -k_{eff}\delta\) where \(k_{eff}\) is the effective spring constant for the motion. Now using the expression for the net force found in step 2, we can find \(k_{eff}\): \(k_{eff} \approx \frac{2kq^2}{d^2+\delta^2}(1-\frac{1}{2}(\frac{\delta}{d})^2)\)

In a harmonic oscillator, the frequency of oscillation can be related to the effective spring constant and the mass of the oscillating object: \(\omega = \sqrt{\frac{k_{eff}}{m}}\) We can substitute the expression for \(k_{eff}\) from step 3 into this equation to find the frequency of the resulting oscillation \(\omega\): \(\omega \approx \sqrt{\frac{2kq^2}{m(d^2+\delta^2)}(1-\frac{1}{2}(\frac{\delta}{d})^2)}\) This is the expression for the frequency of the resulting oscillation.