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

The electric potential V due to a point charge is given by the formula:

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

where:

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

For point X, which is at a distance of 'd' from charge P (2 µC), the distance from charge Q (-3 µC) is (100 mm - d).

Thus, the potential at point X (V_X) can be expressed as:

$V_X = k \frac{2 \times 10^{-6}}{d} + k \frac{-3 \times 10^{-6}}{100 - d}$

Setting V_X = 0:

$0 = k \frac{2 \times 10^{-6}}{d} - k \frac{3 \times 10^{-6}}{100 - d}$

Cancelling out k, we get:

$2 \times 10^{-6} (100 - d) = 3 \times 10^{-6} d$