An infinite number of charges each equal to q, are placed along the X-axis at `x = 1, x = 2, x = 4, x = 8`,…….. and so on.
(i) find the electric field at a point `x = 0` due to this set up of charges.
(ii) What will be the elctric field if the above setup, the consecutive charges have opposite signs.

Explanation

To solve the problem step by step, we will analyze the electric field produced by an infinite series of charges placed along the x-axis at positions \( x = 1, 2, 4, 8, \ldots \) and then consider the case where the charges alternate in sign.

### Part (i): Electric Field at \( x = 0 \)

1. **Identify the Charge Distribution**:
The charges are placed at \( x = 1, 2, 4, 8, \ldots \). This is a geometric progression where the position of the \( n^{th} \) charge is given by \( x_n = 2^{n} \) for \( n = 0, 1, 2, \ldots \).

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:
\[ E = \frac{kq}{r^2} \]
where \( k \) is Coulomb's constant. At \( x = 0 \), the distances to the charges are:
- For the charge at \( x = 1 \): \( r_1 = 1 \)
- For the charge at \( x = 2 \): \( r_2 = 2 \)
- For the charge at \( x = 4 \): \( r_3 = 4 \)
- For the charge at \( x = 8 \): \( r_4 = 8 \)
- And so on.

Therefore, the electric fields from these charges at \( x = 0 \) are:
\[ E_1 = \frac{kq}{1^2}, \quad E_2 = \frac{kq}{2^2}, \quad E_3 = \frac{kq}{4^2}, \quad E_4 = \frac{kq}{8^2}, \ldots \]

3. **Sum the Electric Fields**:
The total electric field \( E_{\text{net}} \) at \( x = 0 \) is the sum of the electric fields from all charges:
\[ E_{\text{net}} = E_1 + E_2 + E_3 + E_4 + \ldots = kq \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \ldots \right) \]

4. **Recognize the Series**:
The series \( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \ldots \) can be expressed as:
\[ \sum_{n=0}^{\infty} \frac{1}{(2^n)^2} = \sum_{n=0}^{\infty} \frac{1}{4^n} \]
This is a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{1}{4} \). The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

5. **Final Expression for Electric Field**:
Substituting back into the expression for \( E_{\text{net}} \):
\[ E_{\text{net}} = kq \cdot \frac{4}{3} = \frac{4kq}{3} \]
Therefore, the electric field at \( x = 0 \) is:
\[ E_{\text{net}} = \frac{4kq}{3} \quad \text{(along the positive x-direction)} \]

### Part (ii): Electric Field with Alternating Charges

1. **Charge Configuration**:
Now, the charges are alternating in sign: \( +q, -q, +q, -q, \ldots \) at the same positions.

2. **Calculate the Electric Field**:
The electric fields will now alternate in direction:
\[ E_{\text{net}} = E_1 - E_2 + E_3 - E_4 + \ldots \]
This can be expressed as:
\[ E_{\text{net}} = kq \left( \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{4^2} - \frac{1}{8^2} + \ldots \right) \]

3. **Recognize the Series**:
The series \( 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \ldots \) is also a geometric series with first term \( a = 1 \) and common ratio \( r = -\frac{1}{4} \):
\[ S = \frac{1}{1 - (-\frac{1}{4})} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5} \]

4. **Final Expression for Electric Field**:
Substituting back into the expression for \( E_{\text{net}} \):
\[ E_{\text{net}} = kq \cdot \frac{4}{5} = \frac{4kq}{5} \]
Therefore, the electric field at \( x = 0 \) with alternating charges is:
\[ E_{\text{net}} = \frac{4kq}{5} \quad \text{(along the positive x-direction)} \]

### Summary of Results:
- For the first case (all charges positive): \( E = \frac{4kq}{3} \)
- For the second case (alternating charges): \( E = \frac{4kq}{5} \)