Three charges are placed on a straight line: Q1=+3 microcoulomb Q2=-4 microcoulomb Q3=+6 microcoulomb The distance between Q1 and Q2 is 15 cm, while the distance between Q2 and Q3 is 25 cm. calculate the resultant electrostatic force acting on Q2, stating both its magnitude and direction.

Since $Q_1$ is positive and $Q_2$ is negative, they will attract each other. The force $F_{12}$ on $Q_2$ will be directed towards $Q_1$ (to the left).
Using Coulomb's Law:

$$F_{12} = k \frac{|Q_1 Q_2|}{r_{12}^2}$$ $$F_{12} = (8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2) \frac{|(3 \times 10^{-6} \text{ C})(-4 \times 10^{-6} \text{ C})|}{(0.15 \text{ m})^2}$$ $$F_{12} = (8.99 \times 10^9) \frac{12 \times 10^{-12}}{0.0225}$$ $$F_{12} = 4.7946 \text{ N}$$

The force $F_{12}$ is $4.7946 \text{ N}$ to the left.

Since $Q_3$ is positive and $Q_2$ is negative, they will attract each other. The force $F_{32}$ on $Q_2$ will be directed towards $Q_3$ (to the right).
Using Coulomb's Law:

$$F_{32} = k \frac{|Q_3 Q_2|}{r_{23}^2}$$ $$F_{32} = (8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2) \frac{|(6 \times 10^{-6} \text{ C})(-4 \times 10^{-6} \text{ C})|}{(0.25 \text{ m})^2}$$ $$F_{32} = (8.99 \times 10^9) \frac{24 \times 10^{-12}}{0.0625}$$ $$F_{32} = 3.45216 \text{ N}$$

The force $F_{32}$ is $3.45216 \text{ N}$ to the right.

The forces $F_{12}$ and $F_{32}$ act in opposite directions on $Q_2$. Let's define the direction to the right as positive.

$$F_{\text{net}} = F_{32} - F_{12}$$ $$F_{\text{net}} = 3.45216 \text{ N} - 4.7946 \text{ N}$$ $$F_{\text{net}} = -1.34244 \text{ N}$$

The negative sign indicates that the resultant force is directed to the left.

The magnitude of the resultant force is $1.34 \text{ N}$ (rounded to two decimal places).
The direction of the resultant force is towards $Q_1$ (to the left).

The resultant electrostatic force acting on $Q_2$ is $\boxed{\text{1.34 N to the left}}$.