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

Answer

`mu = 0.18`

Explanation

Here `q_(1) = q_(2) = 10^(-7) C, r = 10 cm = 10^(-1) m`
`m = 5g = 5xx10^(-3) kg, mu = ?`
For limiting equilibrium,
`F = mu R = mu mg`
`(q_(1) q_(2))/(4pi in_(0) r^(2)) = mu mg`
`mu = (q_(1) q_(2))/(4pi in_(0) r^(2) mg) = (10^(-7) xx10^(-7)xx9xx10^(9))/((10^(-1))^(2) xx5xx10^(-3)xx9.8)`
= 0.18