An infinite number of charges, each equal to `Q=10 mu C` are placed along the x-axis at `x=1, 3, 9 …..m`. Calculate the magnitude of electric field at `x=0` if the consecutive charge have opposite signs.

Explanation

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

### Step 1: Identify the positions and signs of the charges
The charges are placed at positions \( x = 1, 3, 9, \ldots \) m, which can be expressed as \( x_n = 3^{n-1} \) for \( n = 1, 2, 3, \ldots \). The charges alternate in sign: \( +Q, -Q, +Q, -Q, \ldots \).

### Step 2: Determine the electric field contribution 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 (\( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)).

### Step 3: Calculate the distance from each charge to the point of interest
For the charge at position \( x_n = 3^{n-1} \):
- The distance to \( x = 0 \) is \( r_n = 3^{n-1} \).

### Step 4: Write the expression for the total electric field at \( x = 0 \)
The electric field contributions from the charges can be expressed as:
\[ E_{\text{total}} = \sum_{n=1}^{\infty} E_n \]
where \( E_n \) is the electric field due to the \( n \)-th charge. The contributions will alternate in sign:
\[ E_n = \frac{kQ}{(3^{n-1})^2} \cdot (-1)^{n+1} \]

### Step 5: Factor out common terms
We can factor out \( kQ \):
\[ E_{\text{total}} = kQ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(3^{n-1})^2} \]
This simplifies to:
\[ E_{\text{total}} = kQ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{3^{2n}} = kQ \sum_{n=0}^{\infty} \left(-\frac{1}{9}\right)^{n} \]

### Step 6: Recognize the series as a geometric series
The series \( \sum_{n=0}^{\infty} r^n \) converges to \( \frac{1}{1 - r} \) for \( |r| < 1 \). Here, \( r = -\frac{1}{9} \):
\[ \sum_{n=0}^{\infty} \left(-\frac{1}{9}\right)^{n} = \frac{1}{1 - \left(-\frac{1}{9}\right)} = \frac{1}{1 + \frac{1}{9}} = \frac{9}{10} \]

### Step 7: Substitute back to find \( E_{\text{total}} \)
Now substituting back into our expression for \( E_{\text{total}} \):
\[ E_{\text{total}} = kQ \cdot \frac{9}{10} \]

### Step 8: Calculate the final value
Given \( Q = 10 \, \mu C = 10 \times 10^{-6} \, C \):
\[ E_{\text{total}} = k \cdot 10 \times 10^{-6} \cdot \frac{9}{10} \]
\[ E_{\text{total}} = 8.99 \times 10^9 \cdot 10 \times 10^{-6} \cdot \frac{9}{10} \]
\[ E_{\text{total}} = 8.99 \times 9 \times 10^3 \, \text{N/C} = 80.91 \times 10^3 \, \text{N/C} \approx 8.1 \times 10^4 \, \text{N/C} \]

### Final Answer
The magnitude of the electric field at \( x = 0 \) is approximately \( 8.1 \times 10^4 \, \text{N/C} \).

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