Answer:
Electric force = [tex]3.15\times 10^{-11}\textrm{ N}[/tex].
Explanation:
Given:
Charge on one grain, [tex]q_{1}=6\times 10^{-10}\textrm{ C}[/tex]
Charge on second grain, [tex]q_{2}=2.3\times 10^{-15}\textrm{ C}[/tex]
Distance between the two grains is, [tex]r=2\textrm{cm}=0.02\textrm{ m}[/tex]
Electric force acting between two charges [tex]q_{1}\textrm{ and }q_{2}[/tex] separated by a distance [tex]r[/tex] is:
[tex]F_{e}=\frac{kq_{1}q_{2}}{r^2}[/tex]
Where, [tex]k[/tex] is Coulomb's constant equal to [tex]9\times 10^9\textrm{ }Nm^2/C^2[/tex].
Now, plug in the given values and solve for [tex]F{e}[/tex].
[tex]F{e}=\frac{9\times 10^9\times 6.0\times 10^{-10}\times 2.3\times 10^{-15}}{(0.02)^2}\\\\F_{elec}=3.15\times 10^{-11}\textrm{ N}[/tex]
Therefore, the electric force between them is [tex]3.15\times 10^{-11}\textrm{ N}[/tex].
