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CBSE Class 12-science Answered

using gauss's law derive an expression for electric field intensity long, straight line of linear charge density ø per cm.
Asked by Balan | 26 Mar, 2018, 13:39: PM
answered-by-expert Expert Answer
Let us imagine a surface of cylinder symmetrically placed so that its axis coincides with the charged line as shown in figure. Let r be the radius of cylinder and l be its length. Very near to the charged line field lines are perpendicular to the charged line and no field lines are passing through the cross section of cylinder. There is a uniform electric field on the surface of cylinder.
As per Gauss law begin mathsize 12px style surface integral E times d s space equals space fraction numerator lambda space l over denominator epsilon subscript o end fraction end style...................(1)
Equation (1) can be written as E×2 π r×l = ( λ l ) / εo  
Hence begin mathsize 12px style E space equals space fraction numerator lambda over denominator 2 pi epsilon subscript o r end fraction end style


Answered by Thiyagarajan K | 26 Mar, 2018, 20:43: PM

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