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`

Answer

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 Setup
We have two charges:
- Charge \( q_1 = 2 \, \mu C \) located at \( x = 0 \)
- Charge \( q_2 = -1 \, \mu C \) located at \( x = 6 \)

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

### Step 3: Consider the Internal Point (Between the Charges)
Let’s denote the distance from charge \( q_2 \) (located at \( x = 6 \)) to point \( P \) as \( L \). Therefore, the distance from charge \( q_1 \) to point \( P \) will be \( 6 - L \).

The total potential at point \( P \) is:
\[ V_P = k \frac{2 \times 10^{-6}}{6 - L} + k \frac{-1 \times 10^{-6}}{L} \]
Setting \( V_P = 0 \):
\[ k \frac{2 \times 10^{-6}}{6 - L} - k \frac{1 \times 10^{-6}}{L} = 0 \]
We can cancel \( k \) (as it is a constant) from both sides:
\[ \frac{2 \times 10^{-6}}{6 - L} = \frac{1 \times 10^{-6}}{L} \]

### Step 4: Cross Multiply and Solve for \( L \)
Cross multiplying gives:
\[ 2L = 6 - L \]
Rearranging this:
\[ 3L = 6 \implies L = 2 \]

### Step 5: Calculate the Position of Point \( P \)
Since \( L = 2 \), the position of point \( P \) from the origin (where \( q_1 \) is located) is:
\[ x_P = 6 - L = 6 - 2 = 4 \]

### Step 6: Consider the External Point (Outside the Charges)
Now, let’s consider a point \( P \) located to the right of charge \( q_2 \). Let the distance from \( q_2 \) to point \( P \) be \( L \). The distance from \( q_1 \) to point \( P \) will then be \( 6 + L \).

The total potential at point \( P \) is:
\[ V_P = k \frac{2 \times 10^{-6}}{6 + L} + k \frac{-1 \times 10^{-6}}{L} \]
Setting \( V_P = 0 \):
\[ k \frac{2 \times 10^{-6}}{6 + L} - k \frac{1 \times 10^{-6}}{L} = 0 \]
Again, canceling \( k \):
\[ \frac{2 \times 10^{-6}}{6 + L} = \frac{1 \times 10^{-6}}{L} \]

### Step 7: Cross Multiply and Solve for \( L \)
Cross multiplying gives:
\[ 2L = 6 + L \]
Rearranging this:
\[ L = 6 \]

### Step 8: Calculate the Position of Point \( P \)
Since \( L = 6 \), the position of point \( P \) from the origin is:
\[ x_P = 6 + L = 6 + 6 = 12 \]

### Final Answer
The points where the electric potential is zero are:
1. \( x = 4 \)
2. \( x = 12 \)