## College Physics (7th Edition)

Initial separation was $223.606cm$
Let us assume initially two charges $q_{1}$ and $q_{2}$ are separated by a distance $r_{1}$. In this case magnitude of force between them was $F_{1}=k\frac{q_{1}\times q_{2}}{r_{1}^2}$ ........equation (1) Now suppose they are brought together to a distance $r_{2}=100cm$, apart ,In this case suppose force between them is $F_{2}$ which is increased by a factor of 5 from the initial scenario. $F_{2}=k\frac{q_{1}\times q_{2}}{r_{2}^{2}}$ ........equation (2) given that $F_{2}=5\times F_{1}$ putting values of $F_{2}$ and $F_{1}$ from equation(1) and (2) we will get $k\frac{q_{1}\times q_{2}}{r_{2}^{2}}=5\times k\frac{q_{1}\times q_{2}}{r_{1}^2}$ it gives , $r_{1}^2=5\times r_{2}^2$ or $r_{1} =\sqrt5 r_{2}$ since $r_{2} =100cm$ $r_{1}=\sqrt5 \times 100cm=223.606cm$