The question asks about how the electric field varies with distance for an infinitely long thin charged wire, according to Gauss’s law. To determine this variation, we apply Gauss’s law, which is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the net charge enclosed within that surface.

Gauss's Law and Electric Field Derivation

Gauss's law is a powerful tool for calculating electric fields, especially for charge distributions with high symmetry. It states that the total electric flux (\(\Phi_E\)) through any closed surface (called a Gaussian surface) is proportional to the total electric charge (\(Q_{enclosed}\)) enclosed within that surface. Mathematically, Gauss's law is expressed as:

\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} \]

where:

• \( \vec{E} \) is the electric field vector.
• \( d\vec{A} \) is an infinitesimal area vector element on the Gaussian surface.
• \( Q_{enclosed} \) is the net charge enclosed by the Gaussian surface.
• \( \epsilon_0 \) is the permittivity of free space, a fundamental constant.

Electric Field of an Infinitely Long Charged Wire

Consider an infinitely long thin charged wire with a uniform linear charge density \( \lambda \) (charge per unit length). Due to the infinite length and uniform charge distribution, the electric field (\( \vec{E} \)) will have radial symmetry. This means that at any point, the electric field will be directed perpendicularly outward from the wire (if \( \lambda \) is positive) or inward (if \( \lambda \) is negative). The magnitude of the electric field will only depend on the perpendicular distance \( r \) from the wire, and not on the position along the wire or the angular position around the wire.

Gaussian Surface Selection

To apply Gauss's law effectively for this specific charge distribution, we must choose a Gaussian surface that takes advantage of the symmetry of the electric field. A cylindrical Gaussian surface is ideal for an infinitely long charged wire. We select a cylindrical surface of radius \( r \) and length \( L \) that is coaxial with the charged wire.

This cylindrical Gaussian surface has three parts:

• Two flat circular end caps (top and bottom).
• A curved cylindrical side surface.

Applying Gauss's Law for Flux Calculation

Let's calculate the electric flux through each part of the Gaussian cylinder:

• Flux through the end caps: For the end caps, the electric field (\( \vec{E} \)) is everywhere parallel to the surface (or perpendicular to the area vector \( d\vec{A} \)). Therefore, \( \vec{E} \cdot d\vec{A} = E dA \cos(90^\circ) = 0 \). The electric flux through both end caps is zero.
• Flux through the curved cylindrical surface: For the curved surface, the electric field (\( \vec{E} \)) is perpendicular to the surface at every point, meaning it is parallel to the area vector \( d\vec{A} \). Also, due to symmetry, the magnitude of \( \vec{E} \) is constant at all points on the curved surface at a distance \( r \) from the wire. So, \( \vec{E} \cdot d\vec{A} = E dA \cos(0^\circ) = E dA \).

The total electric flux through the curved surface is:

\[ \Phi_{curved} = \int E dA = E \int dA \]

The integral \( \int dA \) represents the total area of the curved surface of the cylinder, which is \( 2\pi r L \).

So, the total electric flux through the Gaussian cylinder is:

\[ \Phi_E = \Phi_{end\ caps} + \Phi_{curved} = 0 + E (2\pi r L) = E (2\pi r L) \]

Charge Enclosed by the Gaussian Cylinder

The charge enclosed (\( Q_{enclosed} \)) within the Gaussian cylinder is the charge present on the segment of the infinitely long thin charged wire of length \( L \). Since the linear charge density is \( \lambda \), the enclosed charge is:

\[ Q_{enclosed} = \lambda L \]

Deriving the Electric Field Relation

Now, we can substitute the expressions for \( \Phi_E \) and \( Q_{enclosed} \) back into Gauss's law:

\[ E (2\pi r L) = \frac{\lambda L}{\epsilon_0} \]

To find the electric field \( E \), we can rearrange the equation:

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

Notice that the length \( L \) cancels out, which is expected since the electric field of an infinitely long wire should not depend on the arbitrary length of our chosen Gaussian surface:

\[ E = \frac{\lambda}{2\pi \epsilon_0 r} \]

Conclusion on Electric Field Variation

From the derived expression, \( E = \frac{\lambda}{2\pi \epsilon_0 r} \), we can clearly see how the electric field \( E \) varies with the perpendicular distance \( r \) from the infinitely long thin charged wire.

Since \( \lambda \), \( 2\pi \), and \( \epsilon_0 \) are all constants, the electric field \( E \) is inversely proportional to the distance \( r \).

Therefore, the electric field due to an infinitely long thin charged wire varies as \( \frac{1}{r} \).