Electric field at a point of distance r from a uniformly charged wire of infinite length having linear charge density `lambda` is directly proportional to

Options

A. `r^(-1)`
B. `r`
C. `r^(2)`
D. `r^(-2)`

Answer

To solve the question regarding the electric field at a distance \( r \) from a uniformly charged wire of infinite length with linear charge density \( \lambda \), we will derive the formula step by step.

### Step-by-Step Solution

1. **Understand the Configuration**:
We have an infinite straight wire with a uniform linear charge density \( \lambda \). The goal is to find the electric field \( E \) at a distance \( r \) from the wire.

**Hint**: Visualize the wire and the point where you want to calculate the electric field.

2. **Use Gauss's Law**:
We will apply Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space \( \epsilon_0 \).

\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]

**Hint**: Remember that the total charge enclosed is related to the linear charge density and the length of the wire.

3. **Choose a Gaussian Surface**:
For an infinite wire, the most suitable Gaussian surface is a cylindrical surface coaxial with the wire. Let the length of the cylinder be \( L \) and its radius be \( r \).

**Hint**: The symmetry of the problem allows us to assume that the electric field \( E \) is constant over the curved surface of the cylinder.

4. **Calculate the Charge Enclosed**:
The charge enclosed by the Gaussian surface is given by:
\[ Q_{\text{enc}} = \lambda L \]

**Hint**: Linear charge density \( \lambda \) is charge per unit length.

5. **Calculate the Electric Flux**:
The electric flux through the curved surface of the cylinder is:
\[ \Phi_E = E \cdot A = E \cdot (2 \pi r L) \]
where \( A \) is the surface area of the cylindrical surface.

**Hint**: The electric field is perpendicular to the surface area vector on the curved surface.

6. **Apply Gauss's Law**:
Set the electric flux equal to the charge enclosed divided by \( \epsilon_0 \):
\[ E \cdot (2 \pi r L) = \frac{\lambda L}{\epsilon_0} \]

**Hint**: Notice that \( L \) cancels out from both sides.

7. **Solve for Electric Field \( E \)**:
Rearranging the equation gives:
\[ E = \frac{\lambda}{2 \pi \epsilon_0 r} \]

**Hint**: This shows how \( E \) depends on \( r \).

8. **Relate to Proportionality**:
The formula can be rewritten as:
\[ E \propto \frac{\lambda}{r} \]
This indicates that the electric field \( E \) is inversely proportional to the distance \( r \) from the wire.

**Hint**: In terms of proportionality, you can express this as \( E \propto r^{-1} \).

### Conclusion
Thus, the electric field at a point at a distance \( r \) from a uniformly charged wire of infinite length having linear charge density \( \lambda \) is directly proportional to \( \frac{\lambda}{r} \) or inversely proportional to \( r \).

### Final Answer
The electric field \( E \) is inversely proportional to the distance \( r \) from the wire.