Two small charged spheres, A and B, are placed on insulated stands, 0,2 m apart, as shown in the diagram below - NSC Physical Sciences - Question 7 - 2020 - Paper 1

Using Coulomb's law, the formula for the electrostatic force $F_{AB}$ between two charges is:

$F = k \frac{|Q_1 Q_2|}{r^2}$

Where:

• $k = 9 \times 10^9 N m^2/C^2$ (Coulomb's constant)
• $Q_1 = -4 \times 10^{-6} C$ (sphere A)
• $Q_2 = +3 \times 10^{-6} C$ (sphere B)
• $r = 0.2 m$

Substituting the values in:

$F_{AB} = 9 \times 10^9 \frac{(4 \times 10^{-6})(3 \times 10^{-6})}{(0.2)^2}$ Evaluating this gives: $F_{AB} = 27 N$ The force exerted by sphere A on sphere B is thus $27 N$.

An electric field is a region around a charged particle where other charged particles experience a force. It is a vector field, characterized by its direction (from positive to negative charge) and magnitude. The electric field ($E$) due to a point charge can be calculated using the formula:

$E = \frac{F}{q}$

Where $F$ is the force experienced by a charge $q$ in the field.

To calculate the net electric field at point M, we consider contributions from both spheres A and B.

1. Electric field due to sphere A ($E_A$): $E_A = k \frac{|Q_A|}{r_A^2} = 9 \times 10^9 \frac{(4 \times 10^{-6})}{(0.3)^2} = 4.0 \times 10^5 N/C$ (Directed to the left)

2. Electric field due to sphere B ($E_B$): $E_B = k \frac{|Q_B|}{r_B^2} = 9 \times 10^9 \frac{(3 \times 10^{-6})}{(0.1)^2} = 2.7 \times 10^7 N/C$ (Directed to the right)

3. Net electric field ($E_{net}$) at point M:

Since $E_A$ is directed to the left and $E_B$ to the right, we subtract: $E_{net} = E_B - E_A = (2.7 \times 10^7 - 4.0 \times 10^5) N/C ext{ (to the right)}$ Thus, $E_{net} \approx 2.3 \times 10^7 N/C$.

To determine the sign of charge D, we analyze the direction of the net force on A (7.69 N) towards sphere B. Since sphere B has a positive charge, it would attract a negative charge on sphere D, indicating that sphere D must be positively charged.

Using the relationship derived from the forces acting on A, we apply:

$F_{net}^2 = (F_{AD})^2 + (F_{AB})^2$

Substituting the values: $(7.69)^2 = (F_{AD})^2 + (2.7)^2$

From this, we can solve for $F_{AD}$, using Coulomb's law to find charge D. The charge can be expressed and calculated as follows: Assuming $F_{AD}$ is calculated as: $F_{AD} = k \frac{|Q_A Q_D|}{r^2}$ Then, from the prior derived equations, the magnitude will yield approximately $Q_D = 4.5 \times 10^{-6} C$.