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Using Gauss's law, derive an expression ...

Using Gauss's law, derive an expression for the electric field intensity at any point near a uniformly charged thin wire of `charg e//l eng th = lambda C//m`.

Text Solution

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To calculate the electric field , imagine a cylindrical Gaussian surface , since , the field is ererywhere radial flux through two ends of the . Cylindrical Gaussian surface is zero
=flux through the curved cylindrical part of the surface
` Exx 2pi rl `
Appliying Gauss.s Law
Flux `phi =(q)/(in _0) `
Total charge enclosed `=lambda l`
` therefore phi =(lambda L)/( in _0) " "............(ii) `
Using Equations (i) & (ii)
` Exx 2pi rl =(lambda l )/( in _0) `
` rArr E= ( lambda )/( 2pi in n_0r) `
In vector notation
` oversetto E= (lambda )/( 2pi in _0r) hatn `
(Where n is unit vector normal to the charge )
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