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

To find the distance from P to 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 charge Q is given by:

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

where:

• $k$ is the electrostatic constant,
• $Q$ is the charge, and
• $r$ is the distance from the charge to the point.

In this case, we have:

• Charge P = -6 µC
• Charge Q = +4 µC
• Distance between P and Q = 120 mm.

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

We set the sum of the potentials due to both charges equal to zero:

$V_P + V_Q = 0$

Substituting the values:

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

This simplifies to:

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

Cross-multiplying gives:

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

Expanding this:

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

Combining terms leads to:

$10d = 720$

Solving for $d$:

$d = 72 \text{ mm}$

Thus, the correct answer is D 72 mm.