Question

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

Answer 1

Let's consider a line charge !

Now, field intensity due to the point charge can be calculated using GAUSS' LAW:

• We will consider a cylindrical Gaussian Surface around the line charge.

$\displaystyle \oint \vec{E}. \vec{ds} = \frac{ q_{e}}{ \epsilon_{0}}$

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

$\implies \displaystyle \oint E \times ds = \frac{ \lambda h}{ \epsilon_{0}}$

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

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

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

$\implies \displaystyle E \propto \frac{ 1}{r }$

So, electric field inversely proportional to distance.