Zero potential between two charges

Potential is a scalar, so V = (1/4πε0)(q₁/r₁ + q₂/r₂). Zero potential needs q₁/r₁ = −q₂/r₂, i.e. the distances satisfy r₁/r₂ = |q₁|/|q₂|.

Given q₁ = +6 μC, q₂ = −24 μC, d = 18 cm. Ratio r₁/r₂ = 6/24 = 1/4. Since the charges are opposite, there are two such points: one between them and one outside, on the side of the smaller charge.

 P2          q1 (+6)     P1                q2 (-24)
  x-----------o-----------x-----------------o
  <-- 6 cm --> <-3.6 cm->  <---- 14.4 cm --->
 (P2 is outside the charges, P1 is between them)

Point between the charges

Let it be x from q₁, so (18 − x) from q₂.

6/x − 24/(18 − x) = 0 → 6(18 − x) = 24x → 108 = 30x → x = 3.6 cm

Point outside, beyond q₁

Let it be x from q₁ (on the side away from q₂); distance from q₂ is (x + 18).

6/x − 24/(x + 18) = 0 → 6(x + 18) = 24x → 108 = 18x → x = 6 cm

Check: at 6 cm, 6/0.06 = 100 and 24/0.24 = 100 (in μC/m), so V = 0.

Answer: V = 0 at 3.6 cm from the 6 μC charge (between the charges) and at 6 cm from the 6 μC charge on the side away from the −24 μC charge (24 cm from it).