A thin straight infinitely long conducting wire having charge density $\lambda$ is enclosed by a cylindrical surface of radius $r$ and length $\ell$, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.

$\phi = \dfrac{\lambda\ell}{\varepsilon_0}$.

CBSE Class 12 · 2011 · ElectrostaticsShort Answer

A thin straight infinitely long conducting wire having charge density

λ

is enclosed by a cylindrical surface of radius

r

and length

ℓ

, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.

Answer: $ϕ = \frac{λ ℓ}{ε _{0}}$ .

Solution

The charge enclosed by the cylinder is

q = λ ℓ

. By Gauss's law the flux through the closed cylindrical surface is

ϕ = \frac{q _{enc}}{ε _{0}} = \frac{λ ℓ}{ε _{0}}

. (The flat end faces contribute nothing, since the radial field is parallel to them; all the flux passes through the curved surface.)

CBSE Class 12 Board · 2011 · Subjective

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