Two small charged spheres, A and B, are placed on insulated stands, 0,2 m apart, as shown in the diagram below. They carry charges of -4 x 10⁻⁶ C and +3 x 10⁻⁶ C respectively.

The charge on sphere A is $Q_A = -4 \times 10^{-6}$ C. The charge of a single electron is $e = -1.6 \times 10^{-19}$ C.
The number of electrons in excess is given by:

The number of electrons in excess on sphere A is $\boxed{\text{}2.5 \times 10^{13}\text{}}$.

The charges are $Q_A = -4 \times 10^{-6}$ C and $Q_B = +3 \times 10^{-6}$ C. The distance between them is $r = 0.2$ m. Coulomb's constant is $k = 9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$.
The magnitude of the electrostatic force is given by Coulomb's Law:

The magnitude of the electrostatic force is $\boxed{\text{}2.7\text{ N}\text{}}$.

An electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects. It is defined as the electrostatic force per unit positive charge.

Point M is 0.1 m to the right of sphere B.
The distance from A to M is $r_{AM} = 0.2\text{ m} + 0.1\text{ m} = 0.3$ m.
The distance from B to M is $r_{BM} = 0.1$ m.
The electric field due to sphere A at M ($E_A$) is:

Since $Q_A$ is negative, $E_A$ points towards A (to the left).

The electric field due to sphere B at M ($E_B$) is:

Since $Q_B$ is positive, $E_B$ points away from B (to the right).

The net electric field at M is the vector sum of $E_A$ and $E_B$. Since they are in opposite directions, we subtract their magnitudes. Let's take right as positive:

The magnitude of the net electric field at point M is $\boxed{\text{}2.3 \times 10^6\text{ N/C}\text{}}$.

Sphere A has a negative charge ($Q_A = -4 \times 10^{-6}$ C). Sphere B has a positive charge ($Q_B = +3 \times 10^{-6}$ C).
The force exerted by B on A ($F_{AB}$) is attractive, so it points to the right (positive x-direction).
The net electrostatic force experienced by sphere A is 7.69 N in the direction shown, which has both a positive x-component and a positive y-component.
The force exerted by D on A ($F_{AD}$) acts along the y-axis because D is located vertically above A.
Since the net force on A has an upward (positive y) component, the force $F_{AD}$ must be directed upwards.
For a negative charge A to experience an upward force from D, D must repel A. This means D must also be a negative charge.
Therefore, the charge on sphere D is $\boxed{\text{}NEGATIVE\text{}}$.

The x-component of the net force on A is solely due to sphere B, as D is on the y-axis relative to A.
From Step 2, the magnitude of the force $F_{AB}$ is 2.7 N. This is the x-component of the net force on A.

The magnitude of the net force on A is $F_{net} = 7.69$ N.
We can find the y-component of the net force using the Pythagorean theorem:

This y-component of the net force is the force exerted by D on A, $F_{AD}$.

The distance between A and D is $r_{AD} = 0.15$ m.
Using Coulomb's Law for $F_{AD}$:

We can rearrange to solve for $|Q_D|$:

The magnitude of the charge on sphere D is $\boxed{\text{}4.5 \times 10^{-6}\text{ C}\text{}}$.