12. A point charge is placed at point X, as shown - Edexcel - A-Level Physics - Question 12 - 2023 - Paper 4

Using Coulomb's Law, the magnitude of the force can be calculated using the formula:

$F = k \frac{|q_1 q_2|}{r^2}$

Where:

• $k = 8.85 \times 10^{-12} \text{C}^2/\text{N m}^2$
• $q_1 = -4.5 \text{nC} = -4.5 \times 10^{-9} \text{C}$
• $q_2 = +7.0 \text{nC} = 7.0 \times 10^{-9} \text{C}$
• $r = 4.0 \text{cm} = 0.040 \text{m}$

Substituting the values, we get:

$F = 8.85 \times 10^{-12} \frac{|(-4.5 \times 10^{-9})(7.0 \times 10^{-9})|}{(0.040)^2}$

Calculating gives:

$F = 1.77 \times 10^{-4} \text{N}$

The work done can be calculated by determining the change in electric potential energy while moving the charge:

$W = q(V_f - V_i)$

Where:

• $q = -4.5 \text{nC} = -4.5 \times 10^{-9} \text{C}$
• The potential difference can be calculated using:

$V = k \frac{q}{r}$

Substituting $V_f$ at 9.0 cm and $V_i$ at 4.0 cm gives:

1. Calculate $V_i$ at 4.0 cm:

$V_i = 8.85 \times 10^{-12} \frac{7.0 \times 10^{-9}}{0.040}$

2. Calculate $V_f$ at 9.0 cm:

$V_f = 8.85 \times 10^{-12} \frac{7.0 \times 10^{-9}}{0.090}$

Using these, the work done is:

W = 3.9 \times 10^{-5} \text{J}$$