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)`.
At how many points on the x-axis, (at finite distance) electric potential will be zero?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To determine the points on the x-axis where the electric potential is zero due to the given point charges, we can follow these steps:

### Step 1: Understand the Setup
We have two point charges:
- Charge \( q \) located at \( (3a, 0) \)
- Charge \( -2q \) located at \( (-3a, 0) \)

### Step 2: Write the Expression for Electric Potential
The electric potential \( V \) at a point \( P(x, 0) \) due to a point charge \( Q \) is given by:
\[ V = \frac{Q}{4\pi \epsilon_0 r} \]
where \( r \) is the distance from the charge to the point where we want to calculate the potential.

### Step 3: Calculate the Distances
For a point \( P(x, 0) \):
- The distance from charge \( q \) at \( (3a, 0) \) is \( r_1 = |x - 3a| \)
- The distance from charge \( -2q \) at \( (-3a, 0) \) is \( r_2 = |x + 3a| \)

### Step 4: Write the Total Electric Potential at Point P
The total electric potential \( V \) at point \( P \) is the sum of the potentials due to both charges:
\[ V = V_q + V_{-2q} = \frac{q}{4\pi \epsilon_0 |x - 3a|} + \frac{-2q}{4\pi \epsilon_0 |x + 3a|} \]
Setting \( V = 0 \):
\[ \frac{q}{|x - 3a|} - \frac{2q}{|x + 3a|} = 0 \]

### Step 5: Simplify the Equation
We can cancel \( q \) (assuming \( q \neq 0 \)):
\[ \frac{1}{|x - 3a|} = \frac{2}{|x + 3a|} \]
Cross-multiplying gives:
\[ |x + 3a| = 2|x - 3a| \]

### Step 6: Consider Different Cases for \( x \)
#### Case 1: \( x > 3a \)
In this case, both distances are positive:
\[ x + 3a = 2(x - 3a) \]
Solving gives:
\[ x + 3a = 2x - 6a \implies x = 9a \]

#### Case 2: \( -3a < x < 3a \)
Here, \( |x - 3a| = 3a - x \) and \( |x + 3a| = x + 3a \):
\[ x + 3a = 2(3a - x) \]
Solving gives:
\[ x + 3a = 6a - 2x \implies 3x = 3a \implies x = a \]

#### Case 3: \( x < -3a \)
In this case, both distances are negative:
\[ -(x + 3a) = 2(-x + 3a) \]
This leads to:
\[ -x - 3a = -2x + 6a \implies x = -9a \]
However, this point is not at a finite distance since we are only considering points on the x-axis.

### Step 7: Conclusion
The electric potential is zero at two points on the x-axis:
1. \( x = 9a \)
2. \( x = a \)

Thus, the answer is that the electric potential will be zero at **two points** on the x-axis.

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