Plot a graph showing the variation of coulomb force (F) versus (1/r2} where r is the distance between the two charges of each pair of charges : (1 J.1C, 2 J.1C) and (2 J.tC, - 3 J.1C). Interpret the graphs obtained.

F1= K (C)(2C)/r2 = 2KC2 /r2 [repulsive force]Compare this with y = mx + cF1 is on y-axis,1/r2 is on x axis2KC2 is the slope. The graph passes through origin----F2 = K(C)(-3C)/ r2 = -3KC2 /r2 [attractive]here only the slope is different (-3KC2 )Let us plot the magnitude of force versus 1/r2 such that both attractive and repulsive forces are plotted in first quadrant with positive slopes. In that case, we have two straight lines that pass through origin with slopes 2KC2 and 3KC2 , the part KC2is common in the slopes, so just draw them as 2 and 3.The line corresponding to F2 is steeper. That means attractive force is greater than repulsive force since magnitude of the slope is more for attraction.