An infinite number of charges, each of magnitude q, are placed along x-axis at x = 1m, 2m, 4m, 8m, 16m and so on but the consecutive charges are of opposite sign starting with +q at x = 1m. A point charge `q_0` , kept at the origin, experiences a force of magnitude :

Options

A. `(qq_0)/(4 pi epsi_0)`
B. `(qq_0)/(5 pi epsi_(0))`
C. `(qq_(0))/(3pi epsi_0)`
D. `(qq_(0))/(2piepsi_(0))`

Answer

To solve the problem, we need to calculate the net force experienced by a point charge \( q_0 \) placed at the origin due to an infinite series of alternating charges placed along the x-axis. The charges are placed at positions \( x = 1m, 2m, 4m, 8m, 16m, \ldots \) with the first charge being \( +q \) at \( x = 1m \), the second charge being \( -q \) at \( x = 2m \), and so on.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- The charges are placed at positions \( x = 1, 2, 4, 8, 16, \ldots \) meters.
- The charge at \( x = 1m \) is \( +q \).
- The charge at \( x = 2m \) is \( -q \).
- The charge at \( x = 4m \) is \( +q \).
- The charge at \( x = 8m \) is \( -q \).
- This pattern continues indefinitely.

2. **Calculate the Force Due to Each Charge:**
- The force \( F \) between two charges is given by Coulomb's law:
\[ F = \frac{k \cdot |q \cdot q_0|}{r^2} \]
- Where \( k = \frac{1}{4 \pi \epsilon_0} \) is Coulomb's constant and \( r \) is the distance from the charge to the point charge \( q_0 \) at the origin.

3. **Calculate Individual Forces:**
- For the first charge at \( x = 1m \):
\[ F_1 = \frac{k \cdot q \cdot q_0}{1^2} = k \cdot q \cdot q_0 \]
- For the second charge at \( x = 2m \):
\[ F_2 = -\frac{k \cdot q \cdot q_0}{2^2} = -\frac{k \cdot q \cdot q_0}{4} \]
- For the third charge at \( x = 4m \):
\[ F_3 = \frac{k \cdot q \cdot q_0}{4^2} = \frac{k \cdot q \cdot q_0}{16} \]
- For the fourth charge at \( x = 8m \):
\[ F_4 = -\frac{k \cdot q \cdot q_0}{8^2} = -\frac{k \cdot q \cdot q_0}{64} \]
- This pattern continues for all charges.

4. **Sum the Forces:**
- The total force \( F_{\text{net}} \) is the sum of all these forces:
\[ F_{\text{net}} = F_1 + F_2 + F_3 + F_4 + \ldots \]
- Substituting the expressions we found:
\[ F_{\text{net}} = k \cdot q \cdot q_0 \left( 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \ldots \right) \]

5. **Recognize the Series:**
- The series inside the parentheses 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:
\[ S = \frac{a}{1 - r} = \frac{1}{1 - (-\frac{1}{4})} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5} \]

6. **Final Calculation:**
- Substitute back into the force equation:
\[ F_{\text{net}} = k \cdot q \cdot q_0 \cdot \frac{4}{5} \]
- Replacing \( k \) with \( \frac{1}{4 \pi \epsilon_0} \):
\[ F_{\text{net}} = \frac{q \cdot q_0}{4 \pi \epsilon_0} \cdot \frac{4}{5} = \frac{q \cdot q_0}{5 \pi \epsilon_0} \]

### Final Answer:
The magnitude of the force experienced by the charge \( q_0 \) at the origin is:
\[ F_{\text{net}} = \frac{q \cdot q_0}{5 \pi \epsilon_0} \]