An infinite number of charges, each of charge `1 mu C` are placed on the `x`-axis with co-ordinates `x=1,2,4,8….oo` If a charge of `1C` is kept at the origin, then what is the net force action on `1C` charge

Options

A. 9000 N
B. 12000 N
C. 24000 N
D. 36000 N

Answer

To solve the problem, we need to calculate the net force acting on a charge of \(1 \, \text{C}\) placed at the origin due to an infinite number of charges of \(1 \, \mu\text{C}\) located at positions \(x = 1, 2, 4, 8, \ldots\) on the x-axis.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- The charge at the origin is \(Q = 1 \, \text{C}\).
- The charges on the x-axis are \(q = 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C}\) located at \(x = 1, 2, 4, 8, \ldots\).

2. **Determine the Distance of Each Charge from the Origin**:
- The distances of the charges from the origin are \(r_1 = 1\), \(r_2 = 2\), \(r_3 = 4\), \(r_4 = 8\), and so on.

3. **Calculate the Force Due to Each Charge**:
- The force \(F\) between two point charges is given by Coulomb's law:
\[ F = k \frac{|Q \cdot q|}{r^2} \]
- Here, \(k\) is Coulomb's constant, approximately \(9 \times 10^9 \, \text{N m}^2/\text{C}^2\).

4. **Calculate the Net Force**:
- The net force acting on the charge at the origin due to all the other charges can be expressed as:
\[ F_{\text{net}} = k \cdot Q \cdot q \left( \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \ldots \right) \]
- Substituting the values:
\[ F_{\text{net}} = k \cdot (1 \, \text{C}) \cdot (1 \times 10^{-6} \, \text{C}) \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \ldots \right) \]

5. **Recognize the Series**:
- The series \( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \ldots \) can be rewritten as:
\[ S = 1 + \frac{1}{4} + \frac{1}{16} + \ldots \]
- This is a geometric series with the first term \(a = 1\) and common ratio \(r = \frac{1}{4}\).

6. **Sum the Infinite Series**:
- The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

7. **Substitute Back into the Force Equation**:
- Now substituting \(S\) back into the force equation:
\[ F_{\text{net}} = k \cdot (1) \cdot (1 \times 10^{-6}) \cdot \left(\frac{4}{3}\right) \]
- Therefore:
\[ F_{\text{net}} = 9 \times 10^9 \cdot (1 \times 10^{-6}) \cdot \left(\frac{4}{3}\right) \]

8. **Calculate the Final Result**:
- Simplifying gives:
\[ F_{\text{net}} = 9 \times 10^3 \cdot \frac{4}{3} = 12,000 \, \text{N} \]

### Final Answer:
The net force acting on the \(1 \, \text{C}\) charge at the origin is \(12,000 \, \text{N}\).