Two charged particles P and Q are separated by a distance of 120 mm - AQA - A-Level Physics - Question 17 - 2021 - Paper 2

To determine the distance from point P to point X, where the electric potential is zero, we can use the formula for electric potential due to point charges.

The electric potential (V) at a point due to a charged particle is given by:

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

where:

• $V$ is the electric potential,
• $k$ is the Coulomb's constant ($8.99 \times 10^9 \text{ N m}^2/\text{C}^2$),
• $q$ is the charge, and
• $r$ is the distance from the charge to the point.

In this case, we have two charges:

• Charge P: $-6 \mu C = -6 \times 10^{-6} C$
• Charge Q: $+4 \mu C = +4 \times 10^{-6} C$

Let the distance from P to X be $d$ mm and thus the distance from X to Q will be $(120 - d)$ mm.

The total potential at point X due to charges P and Q is:

$V_X = k \frac{-6 \times 10^{-6}}{d} + k \frac{4 \times 10^{-6}}{120 - d} = 0$

This leads to:

$\frac{-6}{d} + \frac{4}{120 - d} = 0$

Cross-multiplying to eliminate the fractions gives us:

$-6(120 - d) + 4d = 0$

Expanding this and rearranging results in:

$-720 + 6d + 4d = 0$

Combining like terms:

$10d = 720$

Thus,

$d = \frac{720}{10} = 72 \text{ mm}$

Therefore, the distance from P to X is 72 mm.