According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by

F = \frac{1}{4 π ϵ_{0} K} \times \frac{q_{1} q_{2}}{d^{2}}

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F = \frac{1}{4 π ϵ_{0} K} \times \frac{q_{1} q_{2}}{(r^{+} + r^{-})^{2}}

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F = 4 π ϵ_{0} K \times \frac{q_{1} q_{2}}{d}

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F = 1 4 π ϵ_{0} K \times \frac{q_{1} q_{2}}{(r^{+} + r^{-})}

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Solution

The correct options are
A $F = \frac{1}{4 π ϵ_{0} K} \times \frac{q_{1} q_{2}}{d^{2}}$
B $F = \frac{1}{4 π ϵ_{0} K} \times \frac{q_{1} q_{2}}{(r^{+} + r^{-})^{2}}$
According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by
$F = \frac{1}{4 π ϵ_{0} K} \times \frac{q_{1} q_{2}}{d^{2}}$ or $F = \frac{1}{4 π ϵ_{0} K} \times \frac{q_{1} q_{2}}{(r^{+} + r^{-})^{2}}$

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According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by

The correct options are A F=14πϵ0K×q1q2d2 B F=14πϵ0K×q1q2(r++r−)2According to Coulomb's law, the force of attraction (F) between two oppositely charged ions separated by a distance d in air is given by F=14πϵ0K×q1q2d2 or F=14πϵ0K×q1q2(r++r−)2