Question

prove that the electric field is inversely proportional to the distance​

Answer 1

CORRECT QUESTION:

• For a linear charge arrangement, prove that electric field intensity is inversely proportional to the distance.

Solution:

• Applying Gauss' Law considering a cylindrical Gaussian surface around the linear charge arrangement.

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

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

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

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

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

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

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

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

[Hence Proved].