This question asks us to find the magnitude of the electric field ($E$) created by an infinitely long straight wire possessing a uniform linear charge density ($\lambda$) at a specific distance ($r$) measured perpendicularly from the wire.

Deriving Electric Field Using Gauss's Law

To solve this, we can effectively use Gauss's Law, a fundamental principle in electromagnetism. Gauss's Law relates the electric flux through a closed surface to the net charge enclosed within that surface.

1. Symmetry Considerations

• Due to the infinite length and uniform charge distribution, the electric field must point radially outwards from the wire (assuming $\lambda$ is positive).
• The strength of the electric field, $E$, will only depend on the radial distance $r$ from the wire.

2. Choosing a Gaussian Surface

We choose a cylindrical Gaussian surface. Imagine a cylinder of radius $r$ and length $L$, coaxial with the wire. This surface is chosen because its shape matches the symmetry of the problem.

3. Calculating Electric Flux ($\Phi_E$)

The electric flux is the measure of the electric field lines passing through a surface. For our chosen cylinder:

• The electric field ($E$) is perpendicular to the curved surface area ($A_{curved}$) and points radially outward.
• The electric field is parallel to the flat end caps of the cylinder. Thus, no electric flux passes through the end caps (the angle between $\vec{E}$ and the area vector $d\vec{A}$ is 90 degrees, making $\vec{E} \cdot d\vec{A} = 0$).
• The curved surface area of the cylinder is $A_{curved} = 2\pi r L$.
• The total electric flux through the Gaussian surface is $\Phi_E = E \times A_{curved} = E \times (2\pi r L)$.

4. Determining Enclosed Charge ($Q_{enc}$)

The charge enclosed within our Gaussian cylinder of length $L$ is the linear charge density ($\lambda$) multiplied by the length ($L$):

$Q_{enc} = \lambda L$

5. Applying Gauss's Law

Gauss's Law states that the electric flux ($\Phi_E$) through a closed surface is equal to the enclosed charge ($Q_{enc}$) divided by the permittivity of free space ($\epsilon_0$):

$\Phi_E = \frac{Q_{enc}}{\epsilon_0}$

Substituting the expressions for $\Phi_E$ and $Q_{enc}$:

$E \times (2\pi r L) = \frac{\lambda L}{\epsilon_0}$

6. Solving for Electric Field ($E$)

Now, we rearrange the equation to solve for $E$:

$E = \frac{\lambda L}{2\pi r L \epsilon_0}$

The length $L$ cancels out:

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Conclusion

The magnitude of the electric field at a distance $r$ perpendicular to an infinitely long straight wire with uniform linear charge density $\lambda$ is given by the formula $E = \frac{\lambda}{2\pi \epsilon_0 r}$. This matches the fourth option provided.