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 problem of finding the relationship between the electric field \( E \) at a distance \( r \) from a uniformly charged wire of infinite length with linear charge density \( \lambda \), we can follow these steps:

### Step 1: Understand the Problem
We need to determine how the electric field \( E \) behaves with respect to the distance \( r \) from an infinitely long charged wire.

**Hint:** Recall that the electric field due to a charged object can be derived using Gauss's Law.

### Step 2: Set Up the Gaussian Surface
To apply Gauss's Law, we consider a cylindrical Gaussian surface that surrounds the charged wire. Let the cylinder have:
- Length \( L \)
- Radius \( r \)

**Hint:** The symmetry of the problem suggests that the electric field will be uniform and directed radially outward from the wire.

### Step 3: Apply Gauss's Law
Gauss's Law states:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
Where \( \Phi_E \) is the electric flux through the Gaussian surface and \( Q_{\text{enc}} \) is the charge enclosed by the surface.

**Hint:** The electric flux \( \Phi_E \) can be expressed as \( \int E \cdot dA \).

### Step 4: Calculate the Electric Flux
The electric flux through the curved surface of the cylinder is given by:
\[ \Phi_E = E \cdot A \]
Where \( A \) is the surface area of the cylindrical side, which is \( 2 \pi r L \). Therefore:
\[ \Phi_E = E \cdot (2 \pi r L) \]

**Hint:** Remember that the electric field \( E \) is constant over the curved surface of the cylinder.

### Step 5: Determine the Charge Enclosed
The total charge enclosed by the Gaussian surface is:
\[ Q_{\text{enc}} = \lambda \cdot L \]
Where \( \lambda \) is the linear charge density.

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

### Step 6: Substitute into Gauss's Law
Substituting the expressions for electric flux and charge into Gauss's Law gives:
\[ E \cdot (2 \pi r L) = \frac{\lambda L}{\epsilon_0} \]

### Step 7: Solve for Electric Field \( E \)
Now, we can solve for \( E \):
\[ E = \frac{\lambda}{2 \pi \epsilon_0 r} \]

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

### Step 8: Identify the Relationship
From the equation \( E = \frac{\lambda}{2 \pi \epsilon_0 r} \), we see that:
\[ E \propto \frac{1}{r} \]
This means that the electric field \( E \) is inversely proportional to the distance \( r \).

### Conclusion
Thus, the 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 \( r^{-1} \).

**Final Answer:** The electric field is proportional to \( r^{-1} \).