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 \( 1.0 \, \text{m} \)) where the electric potential is zero, we can follow these steps:

### Step 1: Understand the setup
We have two charges:
- Charge \( +q \) located at point A (let's say at \( x = 0 \))
- Charge \( -3q \) located at point B (at \( x = 1 \, \text{m} \))

We need to find the points along the line joining these two charges where the electric potential \( V \) is zero.

### Step 2: Write the expression for electric potential
The electric potential \( V \) at a point \( P \) (at a distance \( x \) from charge \( +q \) and \( 1 - x \) from charge \( -3q \)) due to both charges is given by:
\[ V = V_{+q} + V_{-3q} = \frac{kq}{x} - \frac{k(3q)}{1 - x} \]
where \( k \) is the electrostatic constant.

### Step 3: Set the total potential to zero
To find the points where the total electric potential is zero, we set the equation to zero:
\[ \frac{kq}{x} - \frac{k(3q)}{1 - x} = 0 \]

### Step 4: Simplify the equation
We can cancel \( kq \) from both sides (assuming \( q \neq 0 \)):
\[ \frac{1}{x} = \frac{3}{1 - x} \]

### Step 5: Cross-multiply to solve for \( x \)
Cross-multiplying gives us:
\[ 1 - x = 3x \]
Rearranging this, we get:
\[ 1 = 4x \implies x = \frac{1}{4} \, \text{m} \]

### Step 6: Identify the first point
This means that one point where the potential is zero is at \( x = 0.25 \, \text{m} \) from charge \( +q \).

### Step 7: Find the second point
Now, we need to check if there is a point outside the two charges where the potential is also zero. Let's consider a point \( P' \) to the left of charge \( +q \) (at \( x = -d \)):
- The distance from \( +q \) is \( d \)
- The distance from \( -3q \) is \( d + 1 \)

The potential at this point \( P' \) is:
\[ V = \frac{kq}{d} - \frac{k(3q)}{d + 1} \]
Setting this equal to zero:
\[ \frac{1}{d} = \frac{3}{d + 1} \]
Cross-multiplying gives:
\[ d + 1 = 3d \implies 1 = 2d \implies d = \frac{1}{2} \, \text{m} \]

### Step 8: Identify the second point
Thus, the second point where the potential is zero is at \( x = -0.5 \, \text{m} \) from charge \( +q \).

### Final Answer
The points where the electric potential is zero are:
- \( 0.25 \, \text{m} \) to the right of charge \( +q \)
- \( 0.5 \, \text{m} \) to the left of charge \( +q \)