A thin straight infinite long conducting …
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Posted by Hibu Nassung 6 years, 3 months ago
- 2 answers
Sia ? 6 years, 3 months ago
A thin straight conducting wire will be a uniform linear charge distribution. Let q charge be enclosed by the cylindrical surface.
Gauss' law states that electric flux through any closed surface is equal to the net charge enclosed inside the surface divided by permittivity of vacuum.
Linear charge density,
{tex}\lambda = \frac { q } { l }{/tex}
{tex}q = \lambda l{/tex} ....... (i)
By Gauss' theorem,
{tex}\therefore {/tex} Total electric flux through the surface of cylinder
{tex}\phi = \frac { q } { \varepsilon _ { 0 } }{/tex} [Gauss' theorem]
{tex}\therefore \quad \phi = \frac { \lambda l } { \varepsilon _ { 0 } }{/tex} [From Eq. (i)]
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Satyam Kumar 4 years, 9 months ago
1Thank You