Problems

The force \(F\) of repulsion between two particles with positive charges \(Q\) and \(Q'\) is given by \(F=kQQ'/r^{2},\) where \(k\) is a positive constant and \(r\) is the distance between the particles. Two small beads \(P_{1}\) and \(P_{2}\) are fixed to a straight horizontal smooth wire, a distance \(d\) apart. A third bead \(P_{3}\) of mass \(m\) is free to move along the wire between \(P_{1}\) and \(P_{3}.\) The beads carry positive electrical charges \(Q_{1},Q_{2}\) and \(Q_{3}.\) If \(P_{3}\) is in equilibrium at a distance \(a\) from \(P_{1},\) show that \[ a=\frac{d\sqrt{Q_{1}}}{\sqrt{Q_{1}}+\sqrt{Q_{2}}}. \] Suppose that \(P_{3}\) is displaced slightly from its equilibrium position and released from rest. Show that it performs approximate simple harmonic motion with period \[ \frac{\pi d}{(\sqrt{Q_{1}}+\sqrt{Q_{2}})^{2}}\sqrt{\frac{2md\sqrt{Q_{1}Q_{2}}}{kQ_{3}}.} \] {[}You may use the fact that \(\dfrac{1}{(a+y)^{2}}\approx\dfrac{1}{a^{2}}-\dfrac{2y}{a^{3}}\) for small \(y.\){]}