Find out the points on the line joining two charges `+qand -3q`(kept at as distance of `1.0 m`) where electric potential is zero.

Explanation

To find the points on the line joining two charges \( +q \) and \( -3q \) (kept at a distance of \( 1.0 \, m \)) where the electric potential is zero, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two charges: \( +q \) at point A and \( -3q \) at point B, separated by a distance of \( 1.0 \, m \).
- We need to find points along the line joining these two charges where the electric potential \( V \) is zero.

2. **Electric Potential Formula**:
- The electric potential \( V \) due to a point charge is given by:
\[ V = \frac{k \cdot Q}{r} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge.

3. **Setting Up the Equation**:
- Let’s denote the distance from charge \( +q \) to the point where the potential is zero as \( r \). The distance from charge \( -3q \) to the same point will then be \( 1 - r \).
- The total electric potential at that point is:
\[ V = V_{+q} + V_{-3q} = \frac{k \cdot q}{r} - \frac{k \cdot 3q}{1 - r} \]
- Setting this equal to zero gives us:
\[ \frac{k \cdot q}{r} - \frac{k \cdot 3q}{1 - r} = 0 \]

4. **Simplifying the Equation**:
- We can cancel \( k \cdot q \) (assuming \( q \neq 0 \)):
\[ \frac{1}{r} - \frac{3}{1 - r} = 0 \]
- Rearranging gives:
\[ \frac{1}{r} = \frac{3}{1 - r} \]

5. **Cross Multiplying**:
- Cross-multiplying gives:
\[ 1 - r = 3r \]
- Rearranging leads to:
\[ 1 = 4r \quad \Rightarrow \quad r = \frac{1}{4} = 0.25 \, m \]

6. **Finding the Second Point**:
- Now, we check for a point outside the segment \( AB \). Let’s denote the distance from \( -3q \) to the point as \( r \) (to the left of \( +q \)). The distance from \( +q \) will then be \( 1 + r \).
- The potential equation becomes:
\[ \frac{k \cdot q}{1 + r} - \frac{k \cdot 3q}{r} = 0 \]
- Canceling \( k \cdot q \) gives:
\[ \frac{1}{1 + r} = \frac{3}{r} \]

7. **Cross Multiplying Again**:
- Cross-multiplying gives:
\[ r = 3(1 + r) \]
- Rearranging leads to:
\[ r = 3 + 3r \quad \Rightarrow \quad -2r = 3 \quad \Rightarrow \quad r = -\frac{3}{2} = -1.5 \, m \]
- This means the second point is \( 0.5 \, m \) to the left of \( -3q \).

### Final Answer:
- The points where the electric potential is zero are:
- \( 0.25 \, m \) from \( +q \) (inside the segment)
- \( 0.5 \, m \) to the left of \( -3q \) (outside the segment)