Two charged spheres, R and S, are both stationary on a smooth, insulated surface inclined at an angle of 25° to the horizontal - NSC Physical Sciences - Question 7 - 2021 - Paper 1

Using Coulomb's law, we have: $F = k \frac{|Q_1 Q_2|}{r^2}$ Where:

• $F = 1.2 \times 10^3 N$ (the force)
• $k = 8.99 \times 10^9 Nm^2/C^2$ (Coulomb's constant)
• $Q_1 = 5 \times 10^{-9} C$
• $Q_2 = -6 \times 10^{-9} C$

Rearranging the formula to solve for $r$, we have: $r = \sqrt{\frac{k \cdot |Q_1 Q_2|}{F}}$

Substituting the values, we find: $r = \sqrt{\frac{(8.99 \times 10^9)(5 \times 10^{-9})(6 \times 10^{-9})}{1.2 \times 10^3}} \approx 0.015 m (or 0.02 m)$

The free-body diagram for sphere S includes the following forces:

• The weight ($W$) acting downwards, $W = mg$
• The normal force ($N$) acting perpendicular to the surface
• The tension ($T$) in the string acting upwards along the string
• The electrostatic force ($F_E$) acting horizontally due to sphere R, directed away from sphere R.

The diagram can be represented as follows:

      N
      | 
      |      T
      |      |
      |      |
     ---    F_E
      |          
      W

Where the lengths are not to scale.

Using Newton's second law, we set up the equation: $F_{net} = ma$ Since the net force is equal to the tension minus the weight component acting along the incline: $T - F_E - W_i = 0$ Where $W_i$ is the component of the weight acting parallel to the incline, given by: $W_i = mg \sin(25°)$

Substituting the forces: $T = F_E + mg \sin(25°)$ Where:

• $F_E = 1.2 \times 10^3 N$
• $m = 0.01 kg$
• $g = 9.8 m/s^2$

Hence: $T \approx 1.2 \times 10^3 N + (0.01)(9.8)\sin(25°) \approx 1203.87 N$

To find the net electric field at point P, we use the formula: $E_{net} = k \frac{Q}{r^2}$ For sphere R: $E_R = k \frac{5 \times 10^{-9}}{(0.03)^2}$ And for sphere S: $E_S = k \frac{-6 \times 10^{-9}}{(0.03)^2}$ Calculating these gives: $E_{net} = 8.99 \times 10^9 \left(\frac{(5 \times 10^{-9}) + (-6 \times 10^{-9})}{(0.03)^2}\right)$ Calculating the electric field results in: $E_{net} \approx -3.78 \times 10^7 N/C$ This indicates the direction away from the negative charge.