Question

Electric field due to infinite, straight uniformly charged wire varies with distance ‘r’ as

• A. r
• B. \(\frac{1}{r}\)
• C. \(\frac{1}{r^2}\)
• D. r²

Answer 1

The Correct Option is B

Step 1: Recall the formula for the electric field due to a uniformly charged infinite wire

For an infinitely long, straight uniformly charged wire, the electric field at a point at a distance $ r $ from the wire is given by Gauss's law.

Gauss's law states that:

$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}$

where $ Q_{\text{enclosed}} $ is the charge enclosed by the Gaussian surface, and $ \epsilon_0 $ is the permittivity of free space.

For an infinitely long wire, the electric field $ E $ at a distance $ r $ from the wire is radial and uniform in magnitude at all points on a circle around the wire.

Step 2: Use symmetry and apply Gauss's law

Consider a Gaussian surface in the form of a cylinder with its axis coincident with the wire. The radius of this cylindrical surface is $ r $, and the length of the cylinder is $ L $.

The electric flux through the curved surface of the cylinder is:

$\Phi_E = E \cdot 2 \pi r L$

Since the electric field is uniform and radial, the flux through the two ends of the cylinder is zero. The charge enclosed by the Gaussian surface is the charge on the length $ L $ of the wire, which is $ \lambda L $, where $ \lambda $ is the linear charge density.

By Gauss's law:

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

Step 3: Solve for the electric field

Solving for $ E $:

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

Thus, the electric field due to an infinitely long uniformly charged wire is inversely proportional to the distance $ r $ from the wire.

Final Answer: 
The electric field due to an infinite, straight uniformly charged wire varies with distance $ r $ as $ \frac{1}{r} $, which matches option (B).

Answer 2

The electric field due to an infinitely long, straight, uniformly charged wire is given by:

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

where:

• $\lambda$ is the linear charge density (charge per unit length),
• $\epsilon_0$ is the permittivity of free space, and
• $r$ is the perpendicular distance from the wire.

Since $\lambda$ and $\epsilon_0$ are constants, the electric field $E$ is inversely proportional to the distance $r$: $E \propto \frac{1}{r}$.

The correct answer is (B) $\frac{1}{r}$.