A charged sphere M is suspended from a ceiling by a light inextensible, insulated string - NSC Physical Sciences - Question 7 - 2022 - Paper 1

The charge on sphere M is NEGATIVE.

The free-body diagram for sphere N should include:

• The gravitational force (weight) acting downward, labeled as F_g.
• The electrostatic force (F_E) acting upward, labeled as F_E.
• Ensure to indicate the direction of each force clearly.

To find the charge on sphere M, we can use the equation of forces: $F_{E} = m imes g$ Where:

• $m = 2.04 \times 10^3 \text{ kg}$
• $g \approx 9.81 \text{ m/s}^2$

1. Calculate weight:
   $F_g = (2.04 \times 10^3)(9.81) = 20000.4 \text{ N}$
2. Using Coulomb's law, we get: $F_{E} = k \frac{Q_M \cdot Q_N}{r^2}$
   where $r = 0.3 m$ and $k \approx 8.99 \times 10^9 \text{ N m}^2/\text{C}^2$.
3. Substitute values: $F_{E} = k \frac{Q_M \cdot (8.6 \times 10^8)}{(0.3)^2}$
4. Set forces equal:
   $20000.4 = (8.99 \times 10^9) \frac{Q_M \cdot (8.6 \times 10^8)}{(0.3)^2}$
5. Solve for $Q_M$:
   After simplifying and solving, we find: $Q_M = 2.33 \times 10^{-6} \text{ C}$.

7.5.1 Magnitude: The magnitudes of the forces are equal as per Newton's third law.

7.5.2 Direction: The electric force exerted by sphere M on N is upwards, while the force exerted by N on M is downwards.

To calculate the net electric field at point X, determine the electric fields due to spheres M and N:

1. Electric field due to sphere M at point X: $E_M = k \frac{|Q_M|}{(0.3 + 0.1)^2} = k \frac{|Q_M|}{(0.4)^2}$
2. Electric field due to sphere N at point X: $E_N = k \frac{|Q_N|}{(0.1)^2}$
3. Calculate: Substituting the values: $E_M = (8.99 \times 10^9) \frac{2.33 \times 10^{-6}}{(0.4)^2}$ $E_N = (8.99 \times 10^9) \frac{8.6 \times 10^{8}}{(0.1)^2}$
4. Determine the direction of the fields:

• The field due to M is directed towards M (downwards) and that due to N is directed upwards.

5. Net field: $E_{net} = E_N - E_M$
   Substituting the calculated values will give the net electric field at point X.