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`

Explanation

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, each of charge \(1 \, \mu\text{C}\), located at positions \(x = 1, 2, 4, 8, \ldots\) on the x-axis.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- We have a charge \(Q = 1 \, \text{C}\) at the origin \((0, 0)\).
- There are infinite charges \(q = 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C}\) located at positions \(x = 1, 2, 4, 8, \ldots\).

2. **Calculate the Force from Each Charge**:
- The force \(F\) between two point charges is given by Coulomb's law:
\[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{|Q \cdot q|}{r^2} \]
- Here, \(r\) is the distance between the charges.

3. **Calculate the Distance for Each Charge**:
- For the charge at \(x = 1\):
\[ r_1 = 1 \]
\[ F_1 = \frac{1}{4\pi \epsilon_0} \cdot \frac{1 \cdot 1 \times 10^{-6}}{1^2} = \frac{1 \times 10^{-6}}{4\pi \epsilon_0} \]
- For the charge at \(x = 2\):
\[ r_2 = 2 \]
\[ F_2 = \frac{1}{4\pi \epsilon_0} \cdot \frac{1 \cdot 1 \times 10^{-6}}{2^2} = \frac{1 \times 10^{-6}}{4\pi \epsilon_0 \cdot 4} \]
- For the charge at \(x = 4\):
\[ r_3 = 4 \]
\[ F_3 = \frac{1}{4\pi \epsilon_0} \cdot \frac{1 \cdot 1 \times 10^{-6}}{4^2} = \frac{1 \times 10^{-6}}{4\pi \epsilon_0 \cdot 16} \]
- Continuing this pattern, we can generalize for the charge at position \(x = 2^n\):
\[ F_n = \frac{1 \times 10^{-6}}{4\pi \epsilon_0 \cdot (2^n)^2} = \frac{1 \times 10^{-6}}{4\pi \epsilon_0 \cdot 4^n} \]

4. **Sum the Forces**:
- The total force \(F_{\text{net}}\) is the sum of all these forces:
\[ F_{\text{net}} = \sum_{n=0}^{\infty} F_n = \sum_{n=0}^{\infty} \frac{1 \times 10^{-6}}{4\pi \epsilon_0 \cdot 4^n} \]
- This is a geometric series with first term \(a = \frac{1 \times 10^{-6}}{4\pi \epsilon_0}\) and common ratio \(r = \frac{1}{4}\).

5. **Calculate the Sum of the Geometric Series**:
- The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{\frac{1 \times 10^{-6}}{4\pi \epsilon_0}}{1 - \frac{1}{4}} = \frac{\frac{1 \times 10^{-6}}{4\pi \epsilon_0}}{\frac{3}{4}} = \frac{1 \times 10^{-6}}{3\pi \epsilon_0} \]

6. **Substituting the Value of \(\epsilon_0\)**:
- The value of \(\epsilon_0\) is approximately \(8.85 \times 10^{-12} \, \text{F/m}\).
- Therefore:
\[ F_{\text{net}} = \frac{1 \times 10^{-6}}{3\pi \times 8.85 \times 10^{-12}} \approx \frac{1 \times 10^{-6}}{8.3 \times 10^{-11}} \approx 12,000 \, \text{N} \]

### Final Result:
The net force acting on the \(1 \, \text{C}\) charge at the origin due to the infinite number of \(1 \, \mu\text{C}\) charges is approximately \(12,000 \, \text{N}\).