7.1 In an experiment to verify the relationship between the electrostatic force, $F_e$, and distance, $r$, between two identical, positively charged spheres, the graph below was obtained - NSC Physical Sciences - Question 7 - 2016 - Paper 1

The dependent variable of the experiment is the electrostatic force, $F_e$.

The graph indicates that the electrostatic force $F_e$ is inversely proportional to the square of the distance, meaning that as the distance increases, the force decreases.

From the graph, the slope is calculated as rac{\Delta F_e}{\Delta (\frac{1}{r^2})} = 4.82 \times 10^3 \text{ N m}^{2}/\text{C}^{2}. Using Coulomb's law, $F_e = k \frac{Q^2}{r^2}$, where $k = 8.9875 \times 10^9 \text{ N m}^{2}/\text{C}^{2}$, we can deduce:

$Q^2 = ext{slope} \times r^2$ Since $r$ is represented in meters, we can pick a value for $r$ and calculate $Q$, yielding $Q = 7.32 \times 10^{-7} \text{ C}$.

The diagram should depict the electric field lines radiating outward from sphere $A$, indicating its positive charge.

The electric field due to sphere $A$ at point $P$ is given by: $E_A = \frac{k |Q_A|}{r^2}$ Substituting the values: $E_A = \frac{8.99 \times 10^9 \times 0.75 \times 10^{-6}}{(0.09)^2} = 8.33 \times 10^6 \text{ N/C}$
For sphere $B$: $E_B = \frac{k |Q_B|}{(d + d')^2}$
Thus: $E_B = \frac{8.99 \times 10^9 \times 0.8 \times 10^{-6}}{0.12^2} = 6.25 \times 10^6 \text{ N/C}$
The direction of $E_B$ is towards sphere $B$, so: $E_{net} = E_A - E_B = 8.33 \times 10^6 - 6.25 \times 10^6 = 2.08 \times 10^6 \text{ N/C}$