7.1 Two small, identical spheres, P and T, are placed a distance of 0.1 m apart, as shown in the diagram below - NSC Physical Sciences - Question 7 - 2022 - Paper 1

The electric field lines should be drawn starting from charge P and ending at charge T. The lines should show that the field points away from the positive charge (P) and towards the negative charge (T), indicating the direction of the force on a positive test charge placed in the field.

Charge Q_s is POSITIVE. This is inferred from the net electrostatic force experienced by sphere S, which has a magnitude of 10 N directed toward sphere T, indicating an attraction to the negatively charged sphere T.

Using Coulomb’s law: $F_{net} = k \frac{Q_1 Q_2}{r^2}$ Given that the net force is 10 N and the distance between the charges is 0.15 m, we first find the magnitude of charge Q_s:

1. Calculate using the known values: $10 = \frac{(9 \times 10^9)(3 \times 10^{-6})(Q_s)}{(0.15)^2}$
2. Solving for Q_s: $Q_s = 4.887 \times 10^{-6} C$
3. To find the number of electrons: $n = \frac{Q_s}{e} = \frac{4.887 \times 10^{-6}}{1.6 \times 10^{-19}} = 3.05 \times 10^{13} electrons$

The electric field E is directly proportional to ( \frac{1}{r^2} ), meaning as the distance r increases, the electric field strength decreases.

Given that the gradient (slope) of the graph represents ( \frac{E_A}{0.04^2} ):

1. Set up the equation: $680 = \frac{E_A}{0.04^2}$
2. Rearranging gives: $E_A = 680 \times 0.04^2 = 1.088 \, N/C$

The charge on sphere B is SMALLER THAN the charge on sphere A. This conclusion is based on the electric field produced by sphere B being less than that produced by sphere A at the same distance, indicating a smaller charge magnitude.