Integrated Concepts (a) In Figure 18.59, four equal char? ges? q lie on the corners of a square. A fifth ch? arge? q is on a mass m directly above the center of the square, at a height equal to the length d of one side of the square. Determine the magni?tude of? q in te? rms of? q , m , and d , if the Coulomb force is to equal the weight of m . (b) Is this equilibrium stable or unstable? Discuss.
Solution : Step 1 (a) In the figure 18.59 four equal c arges q lie on the corners of a square, A fift h charge q is ona mass m directly above the center of the square, The height equal to the length d of one side of the square, The ma nitude of q i terms of q , m , and d , Hence the coulomb’s force is equal to weight for m. The charges repel each other and unlike charges attract each other. The force of attraction or repulsion between two charges is given by coulomb’s electrostatic law, We need to calculate the charge q kq q F =ab ra2 b, Here, q aq bs magnitude of two charges, k is constant and r is distance between charges. Step 2 Therefore the four charges of same magnitude q are placed at each corner of the quadrilateral, another charge of magnitude Q is placed above the midpoint at a distance d , here d is equal to the length of side of quadrilateral. Therefore, we need to calculate only one vertical component of force and multiply by four to get the total force. Step 3 From the fig ure distance from any come charge q to the charge Q is calculated from Pythagoras theorem, r = (half of diagonal) + (distance from center of square to charge Q) d 2 Substitute, 2 for half of diagonal and d is for the distance Q r = (d 2 ) + d 2 2 3 = 2 d. Therefore, the magnitude of the force due to charge q coulomb’s law is, kqQ F = r 2 3 Substitute, 2 d for r, in the above equation is 2 qQ F = 3k 2 d