Two point charges `Q and -Q //4` placed along the x - axis are separated by a distance `r`. Take `-Q//4` as origin and it is placed at the right of Q. Then the potential is zero.

Options

A. at `x = r//3` only
B. at `x = -r//5` only
C. both at `x = r//3` and at `x = -r//5`
D. there exist two points on the axis where the electric field is zero.

Explanation

To solve the problem of finding the points where the electric potential is zero due to two point charges \( Q \) and \( -\frac{Q}{4} \) placed along the x-axis, we can follow these steps:

### Step 1: Understand the Setup
- We have two charges: \( Q \) is placed at position \( x = r \) and \( -\frac{Q}{4} \) is placed at the origin \( x = 0 \).
- We need to find the points along the x-axis where the electric potential \( V \) is zero.

### Step 2: Write the Expression for Electric Potential
The electric potential \( V \) at a point \( x \) due to a point charge is given by:
\[ V = k \frac{Q}{d} \]
where \( k \) is Coulomb's constant and \( d \) is the distance from the charge to the point where we are calculating the potential.

For our two charges, the total potential \( V \) at a point \( x \) is:
\[ V(x) = k \frac{Q}{r - x} + k \left(-\frac{Q}{4}\right) \frac{1}{x} \]
This simplifies to:
\[ V(x) = k \left( \frac{Q}{r - x} - \frac{Q}{4x} \right) \]

### Step 3: Set the Total Potential to Zero
To find the points where the potential is zero, we set the equation to zero:
\[ \frac{Q}{r - x} - \frac{Q}{4x} = 0 \]

### Step 4: Solve for \( x \)
We can simplify the equation:
\[ \frac{Q}{r - x} = \frac{Q}{4x} \]
Cancelling \( Q \) from both sides (assuming \( Q \neq 0 \)):
\[ \frac{1}{r - x} = \frac{1}{4x} \]
Cross-multiplying gives:
\[ 4x = r - x \]
Rearranging this equation:
\[ 4x + x = r \implies 5x = r \implies x = \frac{r}{5} \]

### Step 5: Check the Validity of the Solution
Since \( -\frac{Q}{4} \) is at the origin and \( Q \) is at \( r \), we need to check if \( x = \frac{r}{5} \) lies in the range between the two charges. Since \( \frac{r}{5} \) is positive and less than \( r \), it is valid.

### Step 6: Consider Other Possible Points
Next, we should also check if there are potential points on the left side of the origin (i.e., \( x < 0 \)):
\[ V(x) = k \left( \frac{Q}{r - x} - \frac{Q}{4x} \right) \]
Setting this to zero:
\[ \frac{Q}{r - x} = \frac{Q}{4x} \]
This leads to the same equation as before, but we need to check if it can yield a valid negative solution.

Cross-multiplying:
\[ 4x = r - x \implies 5x = r \implies x = -\frac{r}{5} \]
This point is also valid since it lies to the left of the origin.

### Conclusion
The two points where the electric potential is zero are:
1. \( x = \frac{r}{5} \) (between the charges)
2. \( x = -\frac{r}{5} \) (to the left of the origin)