Question

The Electric field intensity due to an infinitely long wire having linear charge density LAMBDA is 1 mc/m at point p is

Answer 1

Given:

Linear charge density of the infinitely long wire is:

$\lambda=1cm$

To Find:

Electric field intensity due to this infinitely long wire at a point p is = ?

Solution:

let us consider a uniformly charged infinitely long thin wire of surface charge density= $\lambda$

Since, we have to find the electric field at any point P. So, let us take the point P, whose distance from the wire is = 'r'.

Let a cylinder of radius r and height l be the Gaussian surface.

According to Gauss law:

$\int \vec E. \vec ds=\frac{charge(enclosed)}{\epsilon_0}$

Here, $\epsilon_0$ is the permitivity of free space , $\vec E$ is the electric field and $\vec ds$ is the surface area element.

The top and the bottom faces of the cylinder i.e. the base and the top doesn't contribute to electric flux because the electric field vector and area vector are perpendicular to each other and hence their dot product is zero.

$E \times \int \vec ds = \frac{q}{\epsilon_0}\\Or, E \times 2\pi r l = \frac{q}{\epsilon_0}\\Or, E=\frac{q}{2\pi r l\epsilon_0}=(\frac{q}{\epsilon_0})\times \frac{1}{2 \pi \epsilon_0 r}=\frac{\lambda}{2 \pi \epsilon_0 r}$

On putting the value of $\lambda= 1cm$ , we get:

$E=\frac{1}{2\pi \epsilon_0 r}$

Hence, the electric field intensity due to the given wire at a point P is $E=\frac{1}{2\pi \epsilon_0 r}$ .