Electric potential is a scalar quantity. Due to a point charge charge q at distance r, the potential is given by `V=(q)/(4pi in_(0)r)`. A point charge q is placed at `(3a, 0)` and another charge `-2q` is placed at `(-3a, 0)`.
If we plot a graph of potential (V) on x-axis it will be like:

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Answer

To solve the problem of finding the electric potential due to two point charges, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- We have a point charge \( q \) located at \( (3a, 0) \).
- We have another point charge \( -2q \) located at \( (-3a, 0) \).

2. **Write the Expression for Electric Potential:**
- The electric potential \( V \) at a point due to a point charge \( Q \) at a distance \( r \) is given by:
\[ V = \frac{Q}{4\pi \epsilon_0 r} \]

3. **Calculate the Total Potential at a Point \( x \):**
- The total potential \( V \) at a point \( x \) on the x-axis due to both charges can be calculated by summing the potentials due to each charge:
\[ V(x) = V_q + V_{-2q} \]
- The distance from the charge \( q \) at \( (3a, 0) \) to the point \( x \) is \( |x - 3a| \).
- The distance from the charge \( -2q \) at \( (-3a, 0) \) to the point \( x \) is \( |x + 3a| \).
- Therefore, the potential at point \( x \) is:
\[ V(x) = \frac{q}{4\pi \epsilon_0 |x - 3a|} + \frac{-2q}{4\pi \epsilon_0 |x + 3a|} \]

4. **Set Up the Equation for Zero Potential:**
- To find the points where the potential is zero, we set:
\[ \frac{q}{|x - 3a|} = \frac{2q}{|x + 3a|} \]
- Simplifying this gives:
\[ |x - 3a| = 2|x + 3a| \]

5. **Solve for \( x \):**
- We will consider two cases based on the absolute values:
- **Case 1:** \( x - 3a = 2(x + 3a) \)
\[ x - 3a = 2x + 6a \implies -x = 9a \implies x = -9a \]
- **Case 2:** \( -(x - 3a) = 2(x + 3a) \)
\[ -x + 3a = 2x + 6a \implies 3a - 6a = 3x \implies -3a = 3x \implies x = -a \]

6. **Graphing the Potential:**
- The potential will be positive near the positive charge \( q \) and negative near the negative charge \( -2q \).
- The potential will approach infinity as we get close to the charges and will cross zero at the points found above.

### Final Graph Description:
- The graph of potential \( V \) versus position \( x \) will show:
- A steep rise towards positive infinity as \( x \) approaches \( 3a \) from the left.
- A drop to zero at \( x = -a \).
- A steep drop to negative infinity as \( x \) approaches \( -3a \) from the right.
- The potential will also cross zero at \( x = -9a \).