A thin straight infinitely long conducti...

A thin straight infinitely long conducting wire having charge density `lambda` enclosed bya cylindrical surface of radius r and length `l_(r)`, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.

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To find the electric flux through the surface of a cylindrical surface surrounding an infinitely long conducting wire with charge density \( \lambda \), we can follow these steps: ### Step 1: Understand the Geometry We have a thin straight infinitely long conducting wire with a linear charge density \( \lambda \). The cylindrical surface has a radius \( r \) and a length \( l \) that coincides with the wire's length. **Hint:** Visualize the setup with the wire at the center of the cylinder. ### Step 2: Calculate the Charge Enclosed ...

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a point charge q is placed at the centre of a cylinder of length L and radius R the electric flux through the curved surface of the cylinder is

`(q)/(epsilon_(0))(L)/(sqrt(L^(2)+R^(2)))`

`(q)/(epsilon_(0))(L)/(sqrt(L^(2)+2R^(2)))`

`(q)/(epsilon_(0))(L)/(sqrt(L^(2)+4R^(2)))`

`(q)/(2epsilon_(0))`

A wire AB of length L has linear charge density lambda = Kx , where x is measured from the end A of the wire. This wire is enclosed by a Gaussian hollow surface. Find the expression for electric flux through the surface .

`(KL^2)/(2epsilon_0)`

`(KL)/(2epsilon_0)`

`(KL^2)/(epsilon_0)`

`(KL)/(epsilon_0)`

A cylinder of radius R and length l is placed in a uniform electric field E parallel to the axis of the cylinder. The total flux over the curved surface of the cylinder is

zero

`piR^2E`

`2piR^2E`

`E//piR^2`