A particle A with a charge of 2.0 imes 10^ -6 C is held fixed on a...

The maximum force of static friction acting on the second particle is given by

$f = \mu m g$

Substituting the given values:

$f = 0.2 \times 0.08 \times 9.8 = 0.1568 \text{ N}$

The electrostatic force between the two charges is

$F = \dfrac{1}{4 \pi \epsilon_0} \dfrac{|q_1 q_2|}{r^2}$

Substituting the known values:

$F = \dfrac{9 \times 10^9 \times 2.0 \times 10^{-6} \times |q_2|}{(0.1)^2}$

$F = 1.8 \times 10^6 |q_2|$

For the second particle to remain in equilibrium, the electrostatic force must be less than or equal to the maximum static friction:

$F \le f$

$1.8 \times 10^6 |q_2| \le 0.1568$

$|q_2| \le \dfrac{0.1568}{1.8 \times 10^6}$

$|q_2| \le 8.71 \times 10^{-8} \text{ C}$

Thus, the charge of the second particle must lie between $\pm 8.71 \times 10^{-8} \text{ C}$.