---
title: "Zero potential between two charges"
subject: "Engineering Physics"
chapter: 7
key: "semester_1/engineering_physics:zero-potential-two-point-charges"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/zero-potential-two-point-charges
site: BCT Engineering
---

# Zero potential between two charges

Engineering Physics · Chapter 7: Electrostatics

- Asked in: 2074 Asoj, 5 marks
- Video: [Two charges 3 x 10⁻⁸ C and -2 x 10⁻⁸ C are located 15 cm apart. At what point on the line joining th](https://www.youtube.com/watch?v=ciiYpr-ROic) — Onyou Class
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/zero-potential-two-point-charges
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7

## Question

Two point charges 6μC and −24μC are 18 cm apart in air. Locate the positions of zero potential on the line joining the charges.

## Short answer

Zero potential occurs where 6/x = 24/(18−x) between charges giving x = 3.6 cm from the 6 μC charge, and outside beyond the 6 μC charge where 6/x = 24/(x+18) giving x = 6 cm from it (24 cm from the −24 μC charge).

## Answer

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.

```text
 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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For q1 and q2 of opposite sign, what condition makes V = 0?
   A. r1 = r2
   B. r1 r2 = q1 q2
   C. r1/r2 = |q2|/|q1|
   D. r1/r2 = |q1|/|q2|
   Answer: D. q1/r1 = -q2/r2 requires distances proportional to the charge magnitudes. (Key point: Condition)

2. How many zero-potential points lie on the line, and where?
   A. One, between the charges
   B. Two, both between the charges
   C. Two: one between the charges and one outside, on the side of the smaller charge
   D. Two, outside beyond each charge
   Answer: C. The smaller charge must be the nearer one, so the outside point is beyond the 6 μC charge. (Key point: Number of points)

3. How far from the 6 μC charge is the point between them (d = 18 cm)?
   A. 9 cm
   B. 3.6 cm
   C. 6 cm
   D. 14.4 cm
   Answer: B. 6/x = 24/(18 - x) gives 30x = 108, so x = 3.6 cm. (Key point: Point between)

4. How far from the 6 μC charge is the outside point, and on which side?
   A. 6 cm, on the side away from the -24 μC charge
   B. 6 cm, on the side toward the -24 μC charge
   C. 24 cm, beyond the -24 μC charge
   D. 3.6 cm, on the side away from the -24 μC charge
   Answer: A. 6/x = 24/(x + 18) gives 18x = 108, x = 6 cm. (Key point: Outside point)

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