An infinitly long wire is charged uniformly with charge density λ and placed in air, the electric field at distance r from wire will be:
Understanding the Electric Field of an Infinite Wire
This question asks us to find the electric field produced by an infinitely long straight wire that is uniformly charged. The charge density along the wire is given as \(\lambda\), which means the charge per unit length. The wire is placed in air, so we use the permittivity of free space, \(\varepsilon_0\).
To find the electric field due to an infinitely long charged wire, we can use Gauss's Law. Gauss's Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the enclosed electric charge. The formula for Gauss's Law is:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\varepsilon_0} \]Where:
- \(\vec{E}\) is the electric field vector.
- \(d\vec{A}\) is a differential area vector element perpendicular to the surface.
- \(q_{enclosed}\) is the total charge enclosed within the closed surface.
- \(\varepsilon_0\) is the permittivity of free space.
Applying Gauss's Law for an Infinite Wire
Due to the cylindrical symmetry of an infinitely long straight wire, the electric field must be radial, pointing directly away from the wire if \(\lambda\) is positive, or towards the wire if \(\lambda\) is negative. The magnitude of the electric field will be the same at all points equidistant from the wire.
We choose a cylindrical Gaussian surface with its axis coinciding with the wire. Let this cylinder have a radius \(r\) (the distance at which we want to find the electric field) and a length \(L\).
Now, let's consider the electric flux through this Gaussian surface:
- Flux through the end caps: The electric field is radial, meaning it points perpendicularly away from the wire and parallel to the base of the cylinder. The area vectors for the end caps are perpendicular to the bases (and thus parallel to the wire). Since \(\vec{E}\) is perpendicular to \(d\vec{A}\) on the end caps, the dot product \(\vec{E} \cdot d\vec{A} = E dA \cos(90^\circ) = 0\). So, the flux through the end caps is zero.
- Flux through the curved surface: On the curved surface, the electric field \(\vec{E}\) is everywhere perpendicular to the surface and parallel to the area vector \(d\vec{A}\). Also, the magnitude of \(\vec{E}\) is constant at all points on the curved surface because they are all at the same distance \(r\) from the wire. Therefore, \(\vec{E} \cdot d\vec{A} = E dA \cos(0^\circ) = E dA\).
The total flux through the Gaussian surface is the sum of the flux through the end caps and the curved surface:
\[ \oint \vec{E} \cdot d\vec{A} = \int_{\text{end caps}} \vec{E} \cdot d\vec{A} + \int_{\text{curved surface}} \vec{E} \cdot d\vec{A} \] \[ \oint \vec{E} \cdot d\vec{A} = 0 + \int_{\text{curved surface}} E \, dA \]Since E is constant on the curved surface, we can take it out of the integral:
\[ \oint \vec{E} \cdot d\vec{A} = E \int_{\text{curved surface}} dA \]The integral of \(dA\) over the curved surface is the surface area of the curved part of the cylinder, which is \(2\pi r L\). So, the total flux is:
\[ \oint \vec{E} \cdot d\vec{A} = E (2\pi r L) \]Next, we need to find the total charge enclosed within the Gaussian cylinder. Since the wire has a uniform linear charge density \(\lambda\) and the cylinder has a length \(L\), the enclosed charge \(q_{enclosed}\) is simply:
\[ q_{enclosed} = \lambda \times L \]Now, substitute the total flux and the enclosed charge into Gauss's Law:
\[ E (2\pi r L) = \frac{\lambda L}{\varepsilon_0} \]We want to find the magnitude of the electric field \(E\). We can rearrange the equation to solve for \(E\):
\[ E = \frac{\lambda L}{2\pi \varepsilon_0 r L} \]The length \(L\) cancels out from the numerator and the denominator, which is expected because the electric field of an infinitely long wire should not depend on the arbitrary length \(L\) of our Gaussian surface. So, the electric field at a distance \(r\) from the infinitely long charged wire is:
\[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]Comparing this result with the given options, we find that it matches option 4.
Summary of Electric Field Calculation
Let's summarize the steps for finding the electric field using Gauss's Law for an infinitely long uniformly charged wire:
- Identify the symmetry of the charge distribution (cylindrical).
- Choose an appropriate Gaussian surface (a cylinder coaxial with the wire).
- Calculate the electric flux through the Gaussian surface (\(E \cdot 2\pi r L\)).
- Calculate the charge enclosed within the Gaussian surface (\(\lambda L\)).
- Apply Gauss's Law (\(E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0}\)).
- Solve for the electric field \(E\) (\(E = \frac{\lambda}{2\pi \varepsilon_0 r}\)).
The electric field strength decreases inversely with the distance \(r\) from the wire, unlike the \(1/r^2\) dependence for a point charge or a spherical charge distribution.
Revision Table: Electric Field Formulas
| Charge Distribution | Electric Field Formula (Magnitude) | Key Parameters |
|---|---|---|
| Point Charge (at distance r) | \( E = \frac{1}{4\pi \varepsilon_0} \frac{q}{r^2} \) | q = charge, r = distance |
| Infinitely Long Line Charge (at distance r) | \( E = \frac{\lambda}{2\pi \varepsilon_0 r} \) | \(\lambda\) = linear charge density, r = distance |
| Infinite Plane Sheet of Charge (near the sheet) | \( E = \frac{\sigma}{2 \varepsilon_0} \) | \(\sigma\) = surface charge density |
Additional Information on Electric Fields and Charge Density
Linear Charge Density (\(\lambda\)): This is defined as the amount of charge per unit length on a one-dimensional object like a wire. Its unit is Coulombs per meter (C/m). For a uniform charge distribution, \(\lambda = Q/L_{total}\), where Q is the total charge and \(L_{total}\) is the total length. For an infinite wire, we use the concept of charge per unit length as the total charge is infinite.
Electric Field: The electric field at a point is defined as the electric force experienced by a unit positive test charge placed at that point. It is a vector quantity, having both magnitude and direction. The direction of the electric field points in the direction of the force on a positive charge.
Gauss's Law: This law is particularly useful for calculating electric fields for charge distributions that have high degrees of symmetry, such as spherical symmetry (for point charges or charged spheres), cylindrical symmetry (for infinitely long wires or cylinders), and planar symmetry (for infinite sheets of charge).
In the case of the infinitely long charged wire, the electric field is radial and its magnitude depends only on the distance from the wire. This cylindrical symmetry allows us to use a cylindrical Gaussian surface to simplify the flux calculation significantly.
Important Questions from Applications of Gauss’s Law
- The electric field lines from an isolated positively charged conducting sphere are
A charge Q is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube.
- An infinitely long straight uniformly charged wire has a linear charge density of $\lambda$. Calculate the work done by the electric field when a point charge $q$ is moved from an initial distance $r_1$ to a final distance $r_2$ ($r_2 > r_1$) from the wire.
The variation of electric field with respect to distance from centre of a charged conducting spherical shell of radius R is given by :
According to Gauss’s law, the electric field due to an infinitely long thin charged wire varies as:
