An infinite number of charges, each equal to `q`, are placed along the `x` -axis at `x=1, x=2, x=4, x=8`, etc. The electric field at the point `x=0` due to the set of charges is `(q)/(n pi varepsilon_(0))`. Find `n`.

Answer

To solve the problem, we need to calculate the electric field at the point \( x = 0 \) due to an infinite number of charges placed at positions \( x = 1, x = 2, x = 4, x = 8, \ldots \) along the x-axis. The positions of the charges can be described as \( x = 2^n \) for \( n = 0, 1, 2, 3, \ldots \).

### Step-by-Step Solution:

1. **Identify the Charge Positions**:
The charges are located at \( x = 1, 2, 4, 8, \ldots \) which can be expressed as \( x_n = 2^n \) where \( n = 0, 1, 2, \ldots \).

2. **Calculate 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 \( k = \frac{1}{4 \pi \varepsilon_0} \). At \( x = 0 \), the distance from the charge at \( x_n = 2^n \) to the point \( x = 0 \) is \( r_n = 2^n \).

3. **Write the Expression for the Electric Field at \( x = 0 \)**:
The total electric field \( E \) at \( x = 0 \) due to all the charges is the vector sum of the contributions from each charge. Since all charges are positive and located to the right of \( x = 0 \), their electric fields will point to the left (negative x-direction):
\[ E = -\sum_{n=0}^{\infty} \frac{kq}{(2^n)^2} \]
Simplifying this, we have:
\[ E = -kq \sum_{n=0}^{\infty} \frac{1}{4^n} \]

4. **Evaluate the Infinite Series**:
The series \( \sum_{n=0}^{\infty} \frac{1}{4^n} \) is a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{1}{4} \):
\[ \sum_{n=0}^{\infty} \frac{1}{4^n} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

5. **Substitute Back into the Electric Field Expression**:
Now substituting the sum back into the electric field expression:
\[ E = -kq \cdot \frac{4}{3} \]
Since \( k = \frac{1}{4 \pi \varepsilon_0} \):
\[ E = -\frac{q}{4 \pi \varepsilon_0} \cdot \frac{4}{3} = -\frac{q}{3 \pi \varepsilon_0} \]

6. **Relate to Given Electric Field Expression**:
We are given that the electric field at \( x = 0 \) is \( \frac{q}{n \pi \varepsilon_0} \). Setting the two expressions equal gives:
\[ -\frac{q}{3 \pi \varepsilon_0} = \frac{q}{n \pi \varepsilon_0} \]
This implies:
\[ -\frac{1}{3} = \frac{1}{n} \]
Therefore, solving for \( n \):
\[ n = 3 \]

### Final Answer:
Thus, the value of \( n \) is \( \boxed{3} \).