Calculate the electric field at a point 2.0 m away from a 4.0 nC charge, and the forces between multiple charges.

The point charge is $q = 4.0 \text{ nC} = 4.0 \times 10^{-9} \text{ C}$.
The distance from the charge to the field point is $r = 2.0 \text{ m}$.
Coulomb's constant is $k = 8.987 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$.

The magnitude of the electric field $E$ at a distance $r$ from a point charge $q$ is given by:

The magnitude of the electric field is $\boxed{\text{8.99 N/C}}$.

.4:

$q_1 = +4.00 \text{ nC} = +4.00 \times 10^{-9} \text{ C}$ at $x_1 = 0 \text{ m}$.
$q_2 = -1.20 \text{ nC} = -1.20 \times 10^{-9} \text{ C}$ at $x_2 = 6.00 \text{ cm} = 0.06 \text{ m}$.
$q_3 = -4.00 \text{ nC} = -4.00 \times 10^{-9} \text{ C}$ at $x_3 = 4.00 \text{ cm} = 0.04 \text{ m}$.
Coulomb's constant $k = 8.987 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$.

The distance between $q_1$ and $q_3$ is $r_{13} = |x_3 - x_1| = |0.04 \text{ m} - 0 \text{ m}| = 0.04 \text{ m}$.
Since $q_1$ is positive and $q_3$ is negative, they attract. $q_3$ is to the right of $q_1$, so $F_{13}$ is in the positive x-direction.

So, $\vec{F}_{13} = +8.987 \times 10^{-5} \text{ N}$ (in the positive x-direction).

The distance between $q_2$ and $q_3$ is $r_{23} = |x_3 - x_2| = |0.04 \text{ m} - 0.06 \text{ m}| = 0.02 \text{ m}$.
Since $q_2$ is negative and $q_3$ is negative, they repel. $q_3$ is to the left of $q_2$, so $F_{23}$ is in the negative x-direction.

So, $\vec{F}_{23} = -1.07844 \times 10^{-4} \text{ N}$ (in the negative x-direction).

The total force is the vector sum of $F_{13}$ and $F_{23}$.

The total electric force on $q_3$ is \boxed{\text{-1.80 \times 10^{-5} N}}. The negative sign indicates the force is in the negative x-direction.