Two charges P and Q are 100 mm apart - AQA - A-Level Physics - Question 18 - 2020 - Paper 2

To find the distance from point P to point X where the electric potential is 0 V, we can use the principle of superposition of electric potentials due to point charges. The electric potential V at a point due to a point charge is given by the formula:

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

where:

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

Let the distance from P to X be $d$. Therefore, the distance from Q to X will be $(100 - d)$ mm.

The total potential at point X due to both charges can be expressed as:

$V_X = k \frac{2 \mu C}{d} + k \frac{-3 \mu C}{100 - d} = 0$

Removing the constant $k$ from both sides:

$\frac{2 \mu C}{d} - \frac{3 \mu C}{100 - d} = 0$

Rearranging gives us:

$\frac{2}{d} = \frac{3}{100 - d}$

Cross-multiplying leads to:

$2(100 - d) = 3d$

Expanding this:

$200 - 2d = 3d$

Rearranging yields:

$200 = 5d$

Thus,

$d = \frac{200}{5} = 40 \text{ mm}$

So, the distance from P to X is 40 mm, which corresponds to option B.