The charge on sphere A is +5.0 Nc while that on sphere B is – 4.0 Nc. a) Draw the sphere and show the electric field pattern around them. b) Determine the magnitude of the electric force between the two spheres. c) What is the value of the angle θ if sphere A has a mass of 4.5 × 10 kg?

Sphere A is positively charged ($+5.0 \text{ Nc}$), and sphere B is negatively charged ($-4.0 \text{ Nc}$). Electric field lines originate from positive charges and terminate on negative charges.

• The electric field lines start from sphere A and end on sphere B, curving between them.
• The lines are perpendicular to the surface of both spheres.
• The density of the field lines is highest in the region between the spheres, indicating a stronger electric field there.
• Since the magnitude of charge on A ($+5.0 \text{ Nc}$) is greater than on B ($-4.0 \text{ Nc}$), some field lines will originate from A and extend outwards into space, not terminating on B.

The magnitude of the electric force between two point charges is given by Coulomb's Law:

$$F = k \frac{|q_A q_B|}{r^2}$$

where $k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2$ is Coulomb's constant, $q_A$ and $q_B$ are the magnitudes of the charges, and $r$ is the distance between them.

Given:
$q_A = +5.0 \text{ Nc} = +5.0 \times 10^{-9} \text{ C}$
$q_B = -4.0 \text{ Nc} = -4.0 \times 10^{-9} \text{ C}$
$r = 3.5 \text{ cm} = 0.035 \text{ m}$

Substitute the values into Coulomb's Law:
\begin{align*} F &= (8.99 \times 10^9 N m^2/C^2) \frac{|(5.0 \times 10^{-9} C)(-4.0 \times 10^{-9} C)|}{(0.035 m)^2} \ &= (8.99 \times 10^9) \frac{20.0 \times 10^{-18}}{(0.035)^2} N \ &= (8.99 \times 10^9) \frac{2.0 \times 10^{-17}}{0.001225} N \ &= \frac{1.798 \times 10^{-7}}{0.001225} N \ &\approx 1.4677 \times 10^{-4} N \end{align*}
The magnitude of the electric force is $\boxed{\mathbf{1.47 \times 10^{-4} \text{ N}}}$.

Sphere A is in equilibrium under the influence of three forces:

1. Gravitational force (weight) $W = mg$, acting vertically downwards.
2. Electric force $F_e$, acting horizontally towards sphere B (attraction).
3. Tension $T$ in the string, acting along the string at an angle $\theta$ to the vertical.

For equilibrium, the net force in both horizontal and vertical directions must be zero.
Given:
Mass of sphere A, $m = 4.5 \times 10 \text{ kg} = 45 \text{ kg}$
Acceleration due to gravity, $g = 9.81 \text{ m/s}^2$
Electric force, $F_e = 1.4677 \times 10^{-4} \text{ N}$ (from Part b)

First, calculate the weight of sphere A:

$$W = mg = (45 \text{ kg})(9.81 \text{ m/s}^2) = 441.45 \text{ N}$$

Resolve the tension $T$ into its components:

• Vertical component: $T \cos \theta$
• Horizontal component: $T \sin \theta$

Equilibrium equations:

• Vertical forces: $T \cos \theta = W = 441.45 \text{ N}$
• Horizontal forces: $T \sin \theta = F_e = 1.4677 \times 10^{-4} \text{ N}$

Divide the horizontal equation by the vertical equation:

$$\frac{T \sin \theta}{T \cos \theta} = \frac{F_e}{W}$$ $$\tan \theta = \frac{F_e}{W}$$

Substitute the values:

$$\tan \theta = \frac{1.4677 \times 10^{-4} \text{ N}}{441.45 \text{ N}}$$ $$\tan \theta \approx 3.3248 \times 10^{-7}$$

Now, calculate $\theta$:

$$\theta = \arctan(3.3248 \times 10^{-7})$$ $$\theta \approx 1.905 \times 10^{-5} \text{ degrees}$$

The value of the angle $\theta$ is $\boxed{\mathbf{1.91 \times 10^{-5} \text{ degrees}}}$.