An infinite number of electric charges each equal to `5` nano-coulomb (magnitude) are placed along `X`-axis at `x=1 cm, x=2 cm,x=8cm`……and so on. In the setup if the consecutive charges have opposite sign, then the electrical field in Newton/Coulomb at `x=0` is `(1/(4piepsilon_(0))= 9xx10^(9)N-m^(2)//C^(2))`

Options

A. `12xx10^(4)`
B. `24xx10^(4)`
C. `36xx10^(4)`
D. `48xx10^(4)`

Explanation

To find the electric field at \( x = 0 \) due to an infinite number of alternating charges along the x-axis, we can follow these steps:

### Step 1: Understand the charge configuration
We have an infinite series of charges of \( +5 \, \text{nC} \) and \( -5 \, \text{nC} \) placed at positions \( 1 \, \text{cm}, 2 \, \text{cm}, 4 \, \text{cm}, 8 \, \text{cm}, \ldots \) along the x-axis. The positions can be represented as \( 2^n \, \text{cm} \) for \( n = 0, 1, 2, \ldots \).

### Step 2: Set the electric field equation
The electric field \( E \) at a point due to a point charge is given by:
\[ E = \frac{k \cdot |Q|}{r^2} \]
where \( k = \frac{1}{4\pi \epsilon_0} \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point of interest.

### Step 3: Calculate the contributions from each charge
At \( x = 0 \):
1. The charge at \( x = 1 \, \text{cm} \) contributes:
\[ E_1 = \frac{k \cdot 5 \times 10^{-9}}{(0.01)^2} = \frac{k \cdot 5 \times 10^{-9}}{10^{-4}} = k \cdot 5 \times 10^{-5} \]
(directed to the right, positive)

2. The charge at \( x = 2 \, \text{cm} \) contributes:
\[ E_2 = -\frac{k \cdot 5 \times 10^{-9}}{(0.02)^2} = -\frac{k \cdot 5 \times 10^{-9}}{4 \times 10^{-4}} = -\frac{k \cdot 5 \times 10^{-5}}{4} \]
(directed to the left, negative)

3. The charge at \( x = 4 \, \text{cm} \) contributes:
\[ E_3 = \frac{k \cdot 5 \times 10^{-9}}{(0.04)^2} = \frac{k \cdot 5 \times 10^{-9}}{16 \times 10^{-4}} = \frac{k \cdot 5 \times 10^{-5}}{16} \]
(directed to the right, positive)

4. The charge at \( x = 8 \, \text{cm} \) contributes:
\[ E_4 = -\frac{k \cdot 5 \times 10^{-9}}{(0.08)^2} = -\frac{k \cdot 5 \times 10^{-9}}{64 \times 10^{-4}} = -\frac{k \cdot 5 \times 10^{-5}}{64} \]
(directed to the left, negative)

### Step 4: Sum the contributions
The total electric field \( E \) at \( x = 0 \) is the sum of all contributions:
\[ E = E_1 + E_2 + E_3 + E_4 + \ldots \]
This can be expressed as:
\[ E = k \cdot 5 \times 10^{-5} \left( 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \ldots \right) \]

### Step 5: Recognize the series
The series inside the parentheses is a geometric series with:
- First term \( a = 1 \)
- Common ratio \( r = -\frac{1}{4} \)

The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{1}{1 - (-\frac{1}{4})} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5} \]

### Step 6: Substitute back to find \( E \)
Now substituting back, we have:
\[ E = k \cdot 5 \times 10^{-5} \cdot \frac{4}{5} = k \cdot 4 \times 10^{-5} \]
Substituting \( k = 9 \times 10^9 \):
\[ E = 9 \times 10^9 \cdot 4 \times 10^{-5} = 36 \times 10^4 \, \text{N/C} \]

### Final Answer
Thus, the electric field at \( x = 0 \) is:
\[ E = 36 \times 10^4 \, \text{N/C} \]