Answer
The magnitude of the electric force on the K+ ion due to the Cl- ion is $2.84\times 10^{-12}~N$
Work Step by Step
We can find the electric force on the K+ ion due to the Cl- ion:
$F = \frac{k~q_K~q_{Cl}}{r^2}$
$F = \frac{(9.0\times 10^9~N~m^2/C^2)(1.6\times 10^{-19}~C)(-1.6\times 10^{-19}~C)}{(9.0\times 10^{-9}~m)^2}$
$F = -2.84\times 10^{-12}~N$
The magnitude of the electric force on the K+ ion due to the Cl- ion is $2.84\times 10^{-12}~N$
