Q.Two charges $3 \times 10^{-8}\ \text{C}$ and $-2 \times 10^{-8}\ \text{C}$ are located $15\ \text{cm}$ apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Concept: Electric Potential — a scalar that adds algebraically. Put $q_1=+3\times10^{-8}\,\text{C}$ at $x=0$ and $q_2=-2\times10^{-8}\,\text{C}$ at $x=0.15\,\text{m}$. For opposite charges the total potential vanishes at two points on the line — one between them and one beyond the smaller charge.

Setting $\dfrac{kq_1}{x}+\dfrac{kq_2}{|0.15-x|}=0$ gives $\dfrac{3}{x}=\dfrac{2}{|0.15-x|}$.

• Between the charges ($0<x<0.15$): $3(0.15-x)=2x\Rightarrow x=0.09\,\text{m}=9\,\text{cm}$. …

Concept first

Electric potential is a scalar: the total at a point is the algebraic sum of $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ from each charge, sign included. Zero total potential means the positive contribution of $+3\times10^{-8}\,\text{C}$ exactly cancels the negative contribution of $-2\times10^{-8}\,\text{C}$. For opposite charges this can occur at two places on the line.

Step 1 — Coordinates

Place $q_1=+3\times10^{-8}\,\text{C}$ at $x=0$ and $q_2=-2\times10^{-8}\,\text{C}$ at $x=0.15\,\text{m}$. Let $P$ be at distance $x$ from $q_1$; its distance from $q_2$ is $|0.15-x|$.

Step 2 — Zero-potential condition

Step 3 — Point between the charges ($0<x<0.15$)

Here $|0.15-x|=0.15-x$:

This point is $9\,\text{cm}$ from the positive charge and $6\,\text{cm}$ from the negative charge.

Step 4 — Point beyond the negative charge ($x>0.15$)

Here $|0.15-x|=x-0.15$: …

Method: Finding Where the Potential Is Zero on the Line Joining Two Charges

This method applies whenever two point charges of opposite sign are given and you must find the point(s) on the line joining them where the net potential is zero.

Steps

Step 1: Set up a coordinate axis along the line joining the charges

Place one charge at the origin and the other at $x = d$ (the given separation). Let $x$ be the distance of the test point from the first charge.

Step 2: Write the zero-potential condition

Potential is a scalar, so the contributions simply add:

Since the charges have opposite sign, this ratio can be positive, so a real solution exists.

Step 3: Look for a solution between the charges

Between the charges, $r_1 + r_2 = d$. Substitute $r_2 = d - r_1$ and solve the resulting linear equation for $r_1$. This region always has exactly one solution when the charges have opposite sign, because the potential smoothly changes sign as you cross from one charge's side to the other.

Step 4: Look for a solution beyond the smaller-magnitude charge …