Two point charges of magnitudes +4 nC and -6 nC are placed at points A and C respectively - NSC Physical Sciences - Question 10 - 2016 - Paper 1

To calculate the electric field at point R due to the charges, we use the formula:

$E = k \frac{Q}{r^2}$

Where:

• $k$ = $9 \times 10^9 \text{ N\cdot m}^2/\text{C}^2$
• $Q$ is the charge in Coulombs
• $r$ is the distance from the charge to point R.

1. For charge at A (+4 nC, 20 mm = 0.02 m):
   $E_A = k \frac{Q_A}{r_A^2} = 9 \times 10^9 \frac{4 \times 10^{-9}}{(0.02)^2} = 90,000 \text{ N/C}$ (to the right)

2. For charge at C (-6 nC, 25 mm = 0.025 m):
   $E_C = k \frac{|Q_C|}{r_C^2} = 9 \times 10^9 \frac{6 \times 10^{-9}}{(0.025)^2} = 86,400 \text{ N/C}$ (to the left)

Now, the net electric field at R: $E_{NET} = E_A + E_C = 90,000 + 86,400 = 176,400 \text{ N/C}$ (to the right)

The new distance between the charges becomes:

• $25 \text{ mm} - 15 \text{ mm} = 10 \text{ mm} = 0.01 \text{ m}$.

The force can be calculated using Coulomb's Law:

$F = k \frac{|Q_A Q_C|}{r^2}$

Substituting the values: $F = 9 \times 10^9 \frac{(4 \times 10^{-9})(-6 \times 10^{-9})}{(0.01)^2}$

This results in: $F = 7.2 \times 10^{-2} \text{ N (attraction)}$