Infinite charges of magnitude `q` each are lying at `x= 1,2,4,8….meter` on `X`-axis. The value of intensity of electric field at point `x=0` due to these charges will be

Options

A. `12xx10^(9)qN//C`
B. Zero
C. `6xx10^(9)qN//C`
D. `4xx10^(9)qN//C`

Explanation

To find the electric field intensity at the point \( x = 0 \) due to an infinite series of charges located at \( x = 1, 2, 4, 8, \ldots \) meters on the x-axis, we can follow these steps:

### Step 1: Understand the Configuration
The charges are located at positions \( x = 1, 2, 4, 8, \ldots \) meters. Each charge has a magnitude \( q \). The electric field at point \( x = 0 \) will be the vector sum of the electric fields produced by each of these charges.

### Step 2: Calculate the Electric Field from Each Charge
The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is given by the formula:
\[ E = \frac{k \cdot q}{r^2} \]
where \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

### Step 3: Determine the Distance from Each Charge to the Point
For the charges located at \( x = 1, 2, 4, 8, \ldots \):
- The distance from the charge at \( x = 1 \) to \( x = 0 \) is \( 1 \) meter.
- The distance from the charge at \( x = 2 \) to \( x = 0 \) is \( 2 \) meters.
- The distance from the charge at \( x = 4 \) to \( x = 0 \) is \( 4 \) meters.
- The distance from the charge at \( x = 8 \) to \( x = 0 \) is \( 8 \) meters.
- And so on.

### Step 4: Write the Expression for the Total Electric Field
The total electric field at \( x = 0 \) is the sum of the electric fields due to each charge:
\[ E_{\text{total}} = E_1 + E_2 + E_3 + E_4 + \ldots \]
Substituting the distances:
\[ E_{\text{total}} = \frac{kq}{1^2} + \frac{kq}{2^2} + \frac{kq}{4^2} + \frac{kq}{8^2} + \ldots \]

### Step 5: Factor Out Common Terms
We can factor out \( kq \):
\[ E_{\text{total}} = kq \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \ldots \right) \]

### Step 6: Recognize the Series
The series inside the parentheses is:
\[ \sum_{n=0}^{\infty} \frac{1}{(2^n)^2} = \sum_{n=0}^{\infty} \frac{1}{4^n} \]
This is a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{1}{4} \).

### Step 7: Calculate the Sum of the Series
The sum \( S \) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

### Step 8: Substitute Back into the Electric Field Expression
Now substituting back into the expression for \( E_{\text{total}} \):
\[ E_{\text{total}} = kq \cdot \frac{4}{3} \]

### Step 9: Final Calculation
Substituting \( k = 9 \times 10^9 \):
\[ E_{\text{total}} = 9 \times 10^9 \cdot q \cdot \frac{4}{3} = 12 \times 10^9 q \, \text{N/C} \]

### Conclusion
Thus, the intensity of the electric field at point \( x = 0 \) due to the infinite charges is:
\[ E = 12 \times 10^9 q \, \text{N/C} \]