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. `12 xx 10^(9) q NC^(-1)`
B. zero
C. `6 xx 10^(9) q NC^(-1)`
D. `4 xx 10^(9) q NC^(-1)`

Answer

To find the intensity of the electric field at 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. The electric field at point \( x = 0 \) will be the vector sum of the electric fields due to each individual charge.

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

### Step 3: Calculate the Electric Field Contributions
For the charges at positions \( x = 1, 2, 4, 8, \ldots \):
- The distance from the charge at \( x = 1 \) to the point \( x = 0 \) is \( 1 \) meter.
- The distance from the charge at \( x = 2 \) to the point \( x = 0 \) is \( 2 \) meters.
- The distance from the charge at \( x = 4 \) to the point \( x = 0 \) is \( 4 \) meters.
- The distance from the charge at \( x = 8 \) to the point \( x = 0 \) is \( 8 \) meters.
- And so on.

Thus, the electric field contributions from these charges at point \( x = 0 \) can be expressed as:
\[ E = kq \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \ldots \right) \]

### Step 4: 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 the first term \( a = 1 \) and common ratio \( r = \frac{1}{4} \).

### Step 5: Sum of the Geometric Series
The sum \( S \) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

### Step 6: Calculate the Total Electric Field
Now substituting the sum back into the expression for the electric field:
\[ E = kq \cdot \frac{4}{3} \]
Thus, we have:
\[ E = \frac{4kq}{3} \]

### Step 7: Substitute the Value of \( k \)
Substituting \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \):
\[ E = \frac{4 \cdot 9 \times 10^9 \cdot q}{3} = 12 \times 10^9 q \]

### Final Answer
The intensity of the electric field at point \( x = 0 \) due to the infinite series of charges is:
\[ E = 12 \times 10^9 q \]