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 DUE TO LINEAR CHARGE:

• Let's consider a linear charge with an uniform linear charge density of $\lambda$.

• Now, consider a cylindrical Gaussian surface around the linear charge.

• ''q" refers to enclosed charge.

Applying Gauss' Law:

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

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

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

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

$\rm \implies\displaystyle E \times 2\pi rh= \dfrac{ \lambda h}{ \epsilon_{0}}$

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

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

[Hence Proved]