As shown in the figure, a particle P carries a charge of -0.01 and...

For the particle to be in equilibrium, the upward electric force must balance the downward gravitational force:

$|q|E = mg$

Since the two identical capacitors are connected in series, the potential difference across each capacitor is $V' = \dfrac{V}{2}$.

The electric field between the plates of the top capacitor is $E = \dfrac{V'}{d} = \dfrac{V}{2d}$.

Substituting this into the equilibrium equation:

$|q|\dfrac{V}{2d} = mg \Rightarrow V = \dfrac{2dmg}{|q|}$

The capacitance of a parallel plate capacitor is $C = \dfrac{\epsilon_0 A}{d}$, which gives $d = \dfrac{\epsilon_0 A}{C}$.

Substituting $d$ into the expression for $V$:

V = \dfrac{2mg}{|q|} \left(\dfrac{\epsilon_0 A}{C} \right)

Substituting the given values ($m = 10^{-5} \text{ kg}$, $g = 9.8 \text{ m/s}^2$, $|q| = 10^{-8} \text{ C}$, $\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$, $A = 10^{-2} \text{ m}^2$, $C = 4 \times 10^{-8} \text{ F}$):

$V =$\dfrac{2 \times 10^{-5} \times 9.8 \times 8.85 \times 10^{-12} \times 10^{-2}}{10^{-8} \times 4 \times 10^{-8}}

$V =$\dfrac{19.6 \times 8.85 \times 10^{-19}}{4 \times 10^{-16}}

$V = 4.9 \times 8.85 \times 10^{-3} \text{ V}$

$V = 43.365 \times 10^{-3} \text{ V} \approx 43 \text{ mV}$

Answer: $43 \text{ mV}$