A and B are two small spheres separated by a distance of 0,70 m - NSC Physical Sciences - Question 8 - 2017 - Paper 1

To find the net electric field at point P, we must calculate the electric field due to each sphere and then vectorially add them:

1. Electric field due to Sphere A (E_A):

   • Equation:

   E_A = rac{k imes |Q_A|}{r_A^2}

   • Where:
     • $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$ (Coulomb's constant)
     • $Q_A = +1.5 \times 10^{-6} \text{ C}$
     • $r_A = 0.40 \text{ m}$
   • Calculation:

   $E_A = \frac{9 \times 10^9 \times (1.5 \times 10^{-6})}{(0.4)^2} = 8.4375 \times 10^5 \text{ N/C}$

2. Electric field due to Sphere B (E_B):

   • Since Sphere B has a negative charge, the electric field direction will be towards Sphere B:

   $E_B = \frac{k \times |Q_B|}{r_B^2}$

   • Where:
     • $Q_B = -2.0 \times 10^{-6} \text{ C}$
     • $r_B = 0.70 - 0.40 = 0.30 \text{ m}$
   • Calculation:

   $E_B = \frac{9 \times 10^9 \times (2.0 \times 10^{-6})}{(0.3)^2} = 6.6667 \times 10^6 \text{ N/C}$

3. Net Electric Field (E_net):

   $E_{net} = E_A - E_B = 8.4375 \times 10^5 - 6.6667 \times 10^6 = -5.823 \times 10^6 \text{ N/C}$

   Thus, the magnitude of the net electric field at point P is approximately $5.83 \times 10^6 \text{ N/C}$ towards Sphere B.

To find the electrostatic force ($F$) experienced by the charge placed at point P, we can use Coulomb's Law:

$F = Q_{test} \times E_{net}$

Where:

• $Q_{test} = 3.0 \times 10^{-6} \text{ C}$
• $E_{net} = 5.83 \times 10^6 \text{ N/C}$

Calculation:

$F = (3.0 \times 10^{-6}) \times (5.83 \times 10^6) = 0.01749 \text{ N}$

Therefore, the magnitude of the electrostatic force experienced by the charge is approximately $0.0175 \text{ N}$.