Answer:
a
[tex]F = -1.07 *10^{-8} \ N [/tex]
b
[tex]F_r = 1.07 *10^{-8} \ N[/tex]
Explanation:
Generally the force of attraction between this two irons is mathematically represented as
[tex]F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}[/tex]
Here k is known as the proportionality constant with value [tex]k = 2.31 * 10^ {-28} J \cdot m[/tex]
substituting -2 for [tex] Q_{O} [/tex] i.e the charge on oxygen , +1 for [tex] Q_{Li} [/tex] i.e the charge on Lithium and [tex] [0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9} [/tex] for r
So
[tex]F = \frac{ 2.31 * 10^ {-28}* +1 * -2 }{ ( 0.208*10^{-9} )^2 }[/tex]
[tex]F = -1.07 *10^{-8} \ N [/tex]
Generally the force of repulsion will be the magnitude but different direction to the force o attraction
So Force of repulsionn is
[tex]F_r = 1.07 *10^{-8} \ N[/tex]
