Point charge `q_(1) = 2 mu C` and `q_(2) = - 1mu C` are kept at points `x = 0` and `x = 6` respectively. Electrical potential will be zero at points

Options

A. `x = 2 and x = 9`
B. `x = 1 and x = 5`
C. `x = 4 and x = 12`
D. `x = -2 and x = 2`

Explanation

To find the points where the electric potential is zero due to the two point charges \( q_1 = 2 \, \mu C \) and \( q_2 = -1 \, \mu C \), we can follow these steps:

### Step 1: Understand the Electric Potential Formula
The electric potential \( V \) at a point due to a point charge \( q \) is given by the formula:
\[ V = k \frac{q}{r} \]
where \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, N \cdot m^2/C^2 \)), \( q \) is the charge, and \( r \) is the distance from the charge to the point where the potential is being calculated.

### Step 2: Set Up the Problem
Given:
- Charge \( q_1 = 2 \, \mu C \) located at \( x = 0 \)
- Charge \( q_2 = -1 \, \mu C \) located at \( x = 6 \)

We need to find the points \( x \) where the total electric potential \( V_{total} = V_1 + V_2 = 0 \).

### Step 3: Write the Expression for Total Electric Potential
The total electric potential at a point \( x \) is:
\[ V_{total} = k \frac{q_1}{|x - 0|} + k \frac{q_2}{|x - 6|} \]
Substituting the values of the charges:
\[ V_{total} = k \frac{2 \times 10^{-6}}{|x|} + k \frac{-1 \times 10^{-6}}{|x - 6|} \]
We can factor out \( k \):
\[ V_{total} = k \left( \frac{2 \times 10^{-6}}{|x|} - \frac{1 \times 10^{-6}}{|x - 6|} \right) \]
Setting \( V_{total} = 0 \):
\[ \frac{2 \times 10^{-6}}{|x|} - \frac{1 \times 10^{-6}}{|x - 6|} = 0 \]

### Step 4: Solve the Equation
This simplifies to:
\[ \frac{2}{|x|} = \frac{1}{|x - 6|} \]
Cross-multiplying gives:
\[ 2 |x - 6| = |x| \]

### Step 5: Consider Cases for \( |x| \) and \( |x - 6| \)
#### Case 1: \( x < 0 \)
Here, \( |x| = -x \) and \( |x - 6| = 6 - x \):
\[ 2(6 - x) = -x \]
\[ 12 - 2x = -x \]
\[ 12 = x \]
This case does not yield a valid solution since \( x < 0 \).

#### Case 2: \( 0 \leq x < 6 \)
Here, \( |x| = x \) and \( |x - 6| = 6 - x \):
\[ 2(6 - x) = x \]
\[ 12 - 2x = x \]
\[ 12 = 3x \implies x = 4 \]

#### Case 3: \( x \geq 6 \)
Here, \( |x| = x \) and \( |x - 6| = x - 6 \):
\[ 2(x - 6) = x \]
\[ 2x - 12 = x \]
\[ x = 12 \]

### Step 6: Conclusion
The electric potential is zero at two points:
- \( x = 4 \, m \)
- \( x = 12 \, m \)