Question

State and explain Gauss's law in

electrostatics and use it to determine the electric field
intensity due to a line charge.
​

Answer 1

To find:-

• To find the electric field due to a point charge using Gauss's Law .

Answer:-

Gauss's law in electrostatics states that the electric flux through a closed surface is equal to 1/ε0 times the charge enclosed by that surface, that is,

`\implies \displaystyle \phi_E = \oint \vec{E} .\vec{ds} \\`

Also ,

`\implies\displaystyle \oint \vec{E}.\vec{ds} =(q)/(\epsilon_0) \\`

Linear density of charge:-

It ise defined as charge per unit length. Mathematically,

`\implies\displaystyle \lambda =(q)/(\ell) \\`

where ,

• 
  `q` is the charge .
• 
  `\ell` is the length of the conductor.

For figure see attachment.

Now let us assume that the linear density of charge is
`\lambda` . There is a point P near the wire at a distance of " r " from it , where we are interested in calculating the electrical field. Imagine a coaxial Gaussian cylinder of length
`\ell` , and radius " r " such that the point P lines on its surface.

Calculating Electrical flux through the cylinder:-

`\implies\displaystyle \phi_E = 0 + 0 + \oint E.ds \\`

`\implies\displaystyle\phi_E = \oint E.ds \ \cos0^o \\`

`\implies\displaystyle \phi_E = \oint Eds \\`

`\implies\displaystyle \phi_E = E\oint ds \\`

`\implies\displaystyle \phi_E = E (2\pi r\ell) \ \ dots (1) \\`

From Gauss's law:-

`\implies\displaystyle \phi_E =(q)/(\epsilon_0) \\`

`\implies\displaystyle \phi_E = (\lambda \ell)/(\epsilon_0)\ \dots (2) \\`

Therefore , from equation (1) and (2) , we have;

`\implies\displaystyle E * 2\pi r \ell =(\lambda\ell)/(\epsilon_0)\\`

`\implies\displaystyle \underline{\underline{\green{ E =(1)/(4\pi \epsilon_0)(2\lambda)/(r)}}}\\`

And we are done!