## College Physics (7th Edition)

$F_{1}=k\frac{q_{1}q_{2}}{r^{2}}$ Suppose, if we triple each of the charges, and distance is also tripled, then the new force, $F_{2}=k\frac{3q_{1}3q_{2}}{(3r)^{2}}=k\frac{q_{1}q_{2}}{r^{2}}=F_{1}$ Hence (a) is the answer.