Consider a system of two equal point charges, each Q=8μC, which are fixed at points (2 m,0) and (-2m, 0). another charge q is held at a point (0, 0.1 m) on the y-axis. Mass of the charge q is 91 mg. At t = 0, q is released from rest and it is observed to oscillate along y-axis in a simple harmonic manner. It is also observed that at t = 0, the force experienced by it is 9 x 10-3 N.Charge q isAmplitude of motion isFrequency (in Hz) of oscillation is

Since Q = + 8 μC, if q is a positive charge, resultant force on it due to Q at A and Q at B will be along positive y-axis and it would move away along y-axis. But the charge q here is observed to oscillate. This is possible only if q is a negative charge so that resultant force on it due to Q at A and Q at B is toward O. Under the action of this force, q moves toward O, crosses O and as it is moving along negative Y direction, resultant force on it will again be toward O. This force retards the motion of q along negative y-axis. It comes to rest at some point and then moves back toward O and so on (figure). Force applied by Q on q has a magnitude.F=14πε0Qqy2+a2Force applied by Q at A on q can be resolved into rectangular components: Fcos⁡θ and Fsin⁡θ. Similarly, force applied by Q at B on q can be resolved into components : Fcos⁡θ and Fsin⁡θ. Fsin⁡θ components of the two forces balance each other so that the net force on q is 2Fcos⁡θ toward O. Therefore, net force on q, Fn=214πε0Qqy2+a2cos⁡θFn=14πε02Qqy2+a2yy2+a21/2=14πε02Qqyy2+a23/2   ........(i)For y << a, net force of on q,Fn=14πε02Qqya3……..(ii)Here, Q=8μC=8×10−6CAt t = 0, q is at y = 0.1 m. Obviously, y < a (-2m). Since the motion is simple harmonic, we can use the approximation y << a so that net force, from Eq. (ii), will be proportional to displacement (y). Initially, i.e., at y = 0 m, force on q is 9 x 10-3 N.Using Eq. (i), we get or  9×10−3=9×10928×10−6q(0.1)(2)3or  q=5×10−6C=5μCThis, in fact, is the magnitude of q. We know that q, as explained earlier, is a negative charge. Hence, q=−5μC.  Since Q = + 8 μC, if q is a positive charge, resultant force on it due to Q at A and Q at B will be along positive y-axis and it would move away along y-axis. But the charge q here is observed to oscillate. This is possible only if q is a negative charge so that resultant force on it due to Q at A and Q at B is toward O. Under the action of this force, q moves toward O, crosses O and as it is moving along negative Y direction, resultant force on it will again be toward O. This force retards the motion of q along negative y-axis. It comes to rest at some point and then moves back toward O and so on (figure). Force applied by Q on q has a magnitude.F=14πε0Qqy2+a2Force applied by Q at A on q can be resolved into rectangular components: Fcos⁡θ and Fsin⁡θ. Similarly, force applied by Q at B on q can be resolved into components : Fcos⁡θ and Fsin⁡θ. Fsin⁡θ components of the two forces balance each other so that the net force on q is 2Fcos⁡θ toward O. Therefore, net force on q, Fn=214πε0Qqy2+a2cos⁡θFn=14πε02Qqy2+a2yy2+a21/2=14πε02Qqyy2+a23/2   ........(i)For y << a, net force of on q,Fn=14πε02Qqya3……..(ii)Here, Q=8μC=8×10−6CAt t = 0, q is at y = 0.1 m. Obviously, y < a (-2m). Since the motion is simple harmonic, we can use the approximation y << a so that net force, from Eq. (ii), will be proportional to displacement (y). Initially, i.e., at y = 0 m, force on q is 9 x 10-3 N.Using Eq. (i), we get or  9×10−3=9×10928×10−6q(0.1)(2)3or  q=5×10−6C=5μCThis, in fact, is the magnitude of q. We know that q, as explained earlier, is a negative charge. At t = 0, q is released at a point 0.1 m from O on y-axis. As it oscillates, its other extreme position will be 0.1 m from O on the negative y-axis, assuming undamped simple harmonic motion. Hence, amplitude of oscillations is 0.1 m or 10 cm. Since Q = + 8 μC, if q is a positive charge, resultant force on it due to Q at A and Q at B will be along positive y-axis and it would move away along y-axis. But the charge q here is observed to oscillate. This is possible only if q is a negative charge so that resultant force on it due to Q at A and Q at B is toward O. Under the action of this force, q moves toward O, crosses O and as it is moving along negative Y direction, resultant force on it will again be toward O. This force retards the motion of q along negative y-axis. It comes to rest at some point and then moves back toward O and so on (figure). Force applied by Q on q has a magnitude.F=14πε0Qqy2+a2Force applied by Q at A on q can be resolved into rectangular components: Fcos⁡θ and Fsin⁡θ. Similarly, force applied by Q at B on q can be resolved into components : Fcos⁡θ and Fsin⁡θ. Fsin⁡θ components of the two forces balance each other so that the net force on q is 2Fcos⁡θ toward O. Therefore, net force on q, Fn=214πε0Qqy2+a2cos⁡θFn=14πε02Qqy2+a2yy2+a21/2=14πε02Qqyy2+a23/2   ........(i)For y << a, net force of on q,Fn=14πε02Qqya3……..(ii)Here, Q=8μC=8×10−6CAt t = 0, q is at y = 0.1 m. Obviously, y < a (-2m). Since the motion is simple harmonic, we can use the approximation y << a so that net force, from Eq. (ii), will be proportional to displacement (y). Initially, i.e., at y = 0 m, force on q is 9 x 10-3 N.Using Eq. (i), we get or  9×10−3=9×10928×10−6q(0.1)(2)3or  q=5×10−6C=5μCThis, in fact, is the magnitude of q. We know that q, as explained earlier, is a negative charge. Hence, q=−5μC. At t = 0, q is released at a point 0.1 m from O on y-axis. As it oscillates, its other extreme position will be 0.1 m from O on the negative y-axis, assuming undamped simple harmonic motion. Hence, amplitude of oscillations is 0.1 m or 10 cm. From figure, we getFn=14πε02Qqya3=kywhere k=14πε02Qqa3Thus, Fn∝y. We also know that Fn always acts toward O (mean position). Time period of resulting SHM will be T=2πm/k of frequency is 12πk/mf=12π14πε02Qqma3=1(2×3.14)9×109(2)8×10−65×10−691×10−6(2)3=5 Hz                                                             m=91mg=91×10-6kg