There is an infinite straight chain of alternating charges `q` and `-q` . The distance between the neighbouring charges is equal to `a`. Find the interaction energy of each charge with all the others.
Instruction . Make use of the expansion of In `(1 + alpha)` in a power series in `alpha`.

Options

A. `- (2 q)/(4 pi epsilon_0 a)`
B. ` (2 q^2 log_e 2)/(4 pi epsilon_0 a)`
C. ` -(2 q^2 log_e 2)/(4 pi epsilon_0 a)`
D. none of these.

Explanation

To find the interaction energy of each charge in an infinite straight chain of alternating charges \( +q \) and \( -q \), we can follow these steps:

### Step 1: Understanding the Configuration
We have an infinite chain of charges alternating between \( +q \) and \( -q \), with a distance \( a \) between neighboring charges. The charges can be represented as follows:
\[ \ldots, +q, -q, +q, -q, +q, -q, \ldots \]

### Step 2: Calculating the Interaction Energy
The interaction energy \( U \) of a charge due to another charge is given by the formula:
\[ U = k \frac{q_1 q_2}{r} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

For a charge \( +q \) (let's consider the first charge in the chain), the interaction energy with the neighboring charges can be calculated as follows:

1. **First Neighbor** (\( -q \) at distance \( a \)):
\[ U_1 = k \frac{q \cdot (-q)}{a} = -\frac{k q^2}{a} \]

2. **Second Neighbor** (\( +q \) at distance \( 2a \)):
\[ U_2 = k \frac{q \cdot q}{2a} = \frac{k q^2}{2a} \]

3. **Third Neighbor** (\( -q \) at distance \( 3a \)):
\[ U_3 = k \frac{q \cdot (-q)}{3a} = -\frac{k q^2}{3a} \]

4. **Fourth Neighbor** (\( +q \) at distance \( 4a \)):
\[ U_4 = k \frac{q \cdot q}{4a} = \frac{k q^2}{4a} \]

Continuing this pattern, we can express the total interaction energy \( U \) for the charge \( +q \) as:
\[ U = U_1 + U_2 + U_3 + U_4 + \ldots \]
\[ U = -\frac{k q^2}{a} + \frac{k q^2}{2a} - \frac{k q^2}{3a} + \frac{k q^2}{4a} - \ldots \]

### Step 3: Factoring Out Common Terms
We can factor out \( \frac{k q^2}{a} \):
\[ U = \frac{k q^2}{a} \left(-1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots \right) \]

### Step 4: Recognizing the Series
The series inside the parentheses is an alternating series:
\[ S = -1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots \]
This series can be related to the logarithmic expansion:
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
For \( x = 1 \):
\[ \log(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \]
Thus, we can write:
\[ S = -\log(2) \]

### Step 5: Final Expression for Interaction Energy
Substituting \( S \) back into the expression for \( U \):
\[ U = \frac{k q^2}{a} (-\log(2)) \]
Since \( k = \frac{1}{4 \pi \epsilon_0} \):
\[ U = -\frac{q^2 \log(2)}{4 \pi \epsilon_0 a} \]

### Step 6: Considering Both Sides of the Charge
Since the interaction energy is symmetric, we need to consider both sides of the charge:
\[ U_{\text{total}} = 2U = -\frac{2q^2 \log(2)}{4 \pi \epsilon_0 a} \]

### Final Result
Thus, the total interaction energy of each charge with all the others is:
\[ U = -\frac{q^2 \log(2)}{2 \pi \epsilon_0 a} \]

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