A particle, P, with a charge of + 5 × 10^−6 C, is located 1.0 m along a straight line from particle V, with a charge of +7 × 10^−6 C - NSC Physical Sciences - Question 7 - 2018 - Paper 1

Coulomb's law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

To find the distance x, we can use Coulomb's law. Let F_PQ be the force exerted by P on Q and F_VQ be the force exerted by V on Q.

Given the charges:

• Charge of P: $q_1 = 5 × 10^{-6} C$
• Charge of V: $q_2 = 7 × 10^{-6} C$
• Distance PQ: 1 m
• Distance QP: $x$ m
• Distance QV: $(1 - x)$ m.

By Coulomb's law, we write:

$F_{PQ} = k \frac{q_1 \cdot Q}{x^2}$
$F_{VQ} = k \frac{q_2 \cdot Q}{(1-x)^2}$

Setting the forces equal because the net force is zero:
$k \frac{q_1 \cdot Q}{x^2} = k \frac{q_2 \cdot Q}{(1-x)^2}$

Cancelling out the common terms and simplifying leads to:

$\frac{5 \times 10^{-6}}{x^2} = \frac{7 \times 10^{-6}}{(1 - x)^2}$

Cross-multiplying and simplifying further gives: $45 × 10^3 (1 - x)^2 = 63 × 10^3 x^2$

Solving this quadratic equation will yield the value of x. After calculations, we find:

$x ≈ 0.458 m$

This means that particle Q is approximately 0.458 m away from P.