Question

Find the intensity of the electric field due to the linear charge by the law of gauss and prove that the electric field is inversely proportional to the distance?​

Answer 1

Electric Field Intensity due to linear charge:

• Let's assume that an infinite linear charge with uniform linear charge density $\lambda$.

• Now, we will consider a CYLINDRICAL GAUSSIAN SURFACE as follows:

• ''q" refers to enclosed charge.

Applying Gauss' Law:

$\rm \displaystyle\oint \vec{E}. \vec{ds} = \dfrac{q}{ \epsilon_{0} }$

$\rm \implies \displaystyle\oint E \times ds \times \cos( {0}^{ \circ} ) = \dfrac{q}{ \epsilon_{0} }$

$\rm \implies \displaystyle\oint E \times ds = \dfrac{ \lambda \times l }{ \epsilon_{0} }$

$\rm \implies \displaystyle E \oint ds = \dfrac{ \lambda \times l }{ \epsilon_{0} }$

$\rm \implies \displaystyle E \times (2\pi rl) = \dfrac{ \lambda \times l }{ \epsilon_{0} }$

$\rm \implies \displaystyle E = \dfrac{ \lambda }{ 2\pi r \epsilon_{0} }$

$\rm \implies \displaystyle E \propto \: \dfrac{1}{r}$

[Hence Proved].