Question

7. Electric field due to infinite long straight wire of linear charge density , varies
with I distance of point from the wire as
O(a) rº
(b) r
(c) r2
(d) r3​

Answer 1

Given:

Infinite long straight wire of linear charge density $\lambda$.

To find:

Relationship of field intensity with distance?

Calculation:

• Consider a cylindrical Gaussian Surface around the linear charge system.

Applying Gauss' Law:

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

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

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

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

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

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

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

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

$\boxed{ \implies \displaystyle E \propto {r}^{ - 1} }$

So, field intensity is inversely proportional to r.