## Chemistry (4th Edition)

$E_{el}= k\frac{Q_{1}Q_{2}}{d}$ (a) When both charges are doubled, $E_{el}=k\frac{2Q_{1}\times2Q_{2}}{d}=4\times k\frac{Q_{1}Q_{2}}{d}$ The magnitude of electrostatic attraction is 4 times greater. (b) When we double one of the charges, $E_{el}=k\frac{2Q_{1}Q_{2}}{d}=2\times k\frac{Q_{1}Q_{2}}{d}$ The magnitude of electrostatic attraction doubles. (c) When distance between the charges are doubled, $E_{el}=k\frac{Q_{1}Q_{2}}{2d}=\frac{1}{2}\times k\frac{Q_{1}Q_{2}}{d}$ (d) When distance is reduced to half, $E_{el}=k\frac{Q_{1}Q_{2}}{\frac{1}{2}d}=2\times k\frac{Q_{1}Q_{2}}{d}$ The magnitude of electrostatic attraction doubles. (e) When both charges are doubled and the distance between them are also doubled, $E_{el}=k\frac{2Q_{1}\times2Q_{2}}{2d}=2\times k\frac{Q_{1}Q_{2}}{d}$ Again, the magnitude of electrostatic attraction doubles. Thus the correct options are (b), (d) and (e).