1 State Coulomb's Law. 7.2 Draw a labelled free-body diagram for sphere P. 7.3.1 Calculate the magnitude of the electrostatic force between sphere P and sphere Q.

The charges are $q_P = -30 \times 10^{-6} \text{ C}$ and $q_Q = -20 \times 10^{-6} \text{ C}$.
The distance between them is $r = 80 \text{ cm} = 0.80 \text{ m}$.
Coulomb's constant is $k = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$.
The formula for electrostatic force is $F_e = k \frac{|q_P q_Q|}{r^2}$.

The magnitude of the electrostatic force is $\boxed{\text{}8.44 \text{ N}\text{}}$.

7.3.2

The mass of sphere Q is $m = 90 \text{ g} = 0.090 \text{ kg}$.
The acceleration due to gravity is $g = 9.8 \text{ m/s}^2$.

Sphere Q is in equilibrium, so the net force in both horizontal and vertical directions is zero.
From the free-body diagram for Q (not P, as asked in 7.2, but implied for 7.3.2/7.3.3):

• Horizontal forces: $T \sin \theta = F_e$
• Vertical forces: $T \cos \theta = F_g$ We have $F_e = 8.4375 \text{ N}$ and $F_g = 0.882 \text{ N}$.

Divide the horizontal equilibrium equation by the vertical equilibrium equation:

Using the vertical equilibrium equation:

The tension T in the string is $\boxed{\text{}8.50 \text{ N}\text{}}$.

7.3.3
The angle $\theta$ was calculated in Step 3 of 7.3.2.

The angle $\theta$ is $\boxed{\text{}84.0^\circ\text{}}$.

Question 8

8.1 The electric field at a point is defined as the electrostatic force experienced per unit positive test charge placed at that point.

8.2
The electric field pattern between spheres A (+5 nC) and B (-4 nC) would show:

• Field lines originating from sphere A (positive charge).
• Field lines terminating on sphere B (negative charge).
• Arrows on the field lines pointing away from A and towards B.
• The lines would be denser closer to the charges, especially near A due to its larger magnitude of charge compared to B.
• The lines would not cross each other.

8.3

$q_A = +5 \text{ nC} = +5 \times 10^{-9} \text{ C}$
$q_B = -4 \text{ nC} = -4 \times 10^{-9} \text{ C}$
$k = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$
Distance from X to A is $r_A = r$.
Distance from X to B is $r_B = r + 3r = 4r$.
Net electric field at X is $E_{net} = 4128 \text{ N} \cdot \text{C}^{-1}$ West.

• For $q_A$ (positive), the electric field $E_A$ at X points away from A. Since X is to the left of A, $E_A$ points to the left (West).
• For $q_B$ (negative), the electric field $E_B$ at X points towards B. Since X is to the left of B, $E_B$ points to the right (East).

Since $E_{net}$ is West, $E_A$ (West) must be greater than $E_B$ (East).

The value of r is $\boxed{\text{}0.102 \text{ m}\text{}}$.