An infinite number of electric charges each equal to 5 nano coulombs are placed along x-axis at x = 1 cm, x-2 cm, x -4 cm, x = 8 cm,.. and so on. In this setup, if the consecutive charges have opposite sign, then the electric field in newton/coulomb at x =0 is

Options

A. Infinity
B. zero
C. `(1)/(4 pi epsilon_(0)) ((2q)/(3))`
D. `(1)/(4 pi epsilon_(0)) (2q)`

Answer

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

### Step-by-Step Solution:

1. **Identify the Charge Configuration**:
The charges are placed at positions \( x = 1 \, \text{cm}, 2 \, \text{cm}, 4 \, \text{cm}, 8 \, \text{cm}, \ldots \) which are powers of 2. The charges alternate in sign: the charge at \( x = 1 \, \text{cm} \) is \( +5 \, \text{nC} \), at \( x = 2 \, \text{cm} \) is \( -5 \, \text{nC} \), at \( x = 4 \, \text{cm} \) is \( +5 \, \text{nC} \), and so on.

2. **Electric Field Due to a Point Charge**:
The electric field \( E \) due to a point charge \( Q \) at a distance \( r \) is given by:
\[ E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} \]
where \( \epsilon_0 \) is the permittivity of free space.

3. **Calculate the Electric Field Contributions**:
We will denote the electric field contributions from the charges at positions \( x = 1, 2, 4, 8, \ldots \) respectively as \( E_1, E_2, E_3, \ldots \).

- For \( x = 1 \, \text{cm} \):
\[ E_1 = \frac{1}{4 \pi \epsilon_0} \frac{5 \times 10^{-9}}{(0.01)^2} = \frac{5 \times 10^{-9}}{4 \pi \epsilon_0 \times 10^{-4}} \]

- For \( x = 2 \, \text{cm} \):
\[ E_2 = \frac{1}{4 \pi \epsilon_0} \frac{5 \times 10^{-9}}{(0.02)^2} = \frac{5 \times 10^{-9}}{4 \pi \epsilon_0 \times 4 \times 10^{-4}} = \frac{5 \times 10^{-9}}{16 \pi \epsilon_0 \times 10^{-4}} \]

- For \( x = 4 \, \text{cm} \):
\[ E_3 = \frac{1}{4 \pi \epsilon_0} \frac{5 \times 10^{-9}}{(0.04)^2} = \frac{5 \times 10^{-9}}{4 \pi \epsilon_0 \times 16 \times 10^{-4}} = \frac{5 \times 10^{-9}}{64 \pi \epsilon_0 \times 10^{-4}} \]

- For \( x = 8 \, \text{cm} \):
\[ E_4 = \frac{1}{4 \pi \epsilon_0} \frac{5 \times 10^{-9}}{(0.08)^2} = \frac{5 \times 10^{-9}}{4 \pi \epsilon_0 \times 64 \times 10^{-4}} = \frac{5 \times 10^{-9}}{256 \pi \epsilon_0 \times 10^{-4}} \]

4. **Sum the Electric Field Contributions**:
The total electric field at \( x = 0 \) is the sum of the contributions from all charges:
\[ E = E_1 - E_2 + E_3 - E_4 + \ldots \]

Factoring out \( \frac{5 \times 10^{-9}}{4 \pi \epsilon_0 \times 10^{-4}} \):
\[ E = \frac{5 \times 10^{-9}}{4 \pi \epsilon_0 \times 10^{-4}} \left( 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \ldots \right) \]

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

6. **Calculate the Total Electric Field**:
Substituting back into the equation for \( E \):
\[ E = \frac{5 \times 10^{-9}}{4 \pi \epsilon_0 \times 10^{-4}} \cdot \frac{4}{5} = \frac{1}{\pi \epsilon_0 \times 10^{-4}} \]

7. **Final Calculation**:
Using \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \):
\[ E = \frac{1}{\pi \times 8.85 \times 10^{-12} \times 10^{-4}} \approx 36 \times 10^{4} \, \text{N/C} \]

### Final Answer:
The electric field at \( x = 0 \) is approximately \( 36 \times 10^{4} \, \text{N/C} \).