two particles each having a mass of 5g and charge of 1.0×10^-7C stay in limiting equilibrium on a horizontal table with a separation of 10cm . The coefficient of friction between them and the table is same . Find the value of this coefficient.

The answer is μ=0.18.
Find the value of this coefficient.
The particles are in equilibrium, meaning the net force on them is zero.
Thus, repulsion due to their charge = friction due to the table.
As the equilibrium is limiting, friction reaches its maximum value.
[tex]\frac{kQ^{2} }{r^{2} }[/tex]=μmg
[tex]k=9*10^{9} Nm^{2} C^{-2}[/tex]The electrostatic constant [tex]Q=10^{-7} C[/tex] is the charge, and r=0.1m is the distance between the charges.
[tex]\frac{(9*10^{2})(10^{-7})^{2} }{(0.1)^{2} }[/tex]=μ*[tex](5*10^{-3})(10)[/tex]
Therefore, The answer is μ=0.18.
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