Answer the following:
(a) State Gauss' law. Using this law, obtain the expression for the electric field due to an infinitely long straight conductor of linear charge density `lamda`.
(b) A wire AB of length L has linear charge density `lamda = 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 the electric flux through this surface.

Explanation

let the field lines must be radically outward. Draw a cylinderical gaussian surface of radius r and length l. then, applying gauss' theorem

`intE.dS=(1)/(epsi_(0))lamdal`
or `E_(r)2pirl=(1)/(epsi_(0))lamdal impliesE_(r)=(lamda)/(2piepsi_(0)r)`
Hence, if `r_(0)` is the radius, `V(r)-V(r_(0))=-int_(r_(0))^(r)E.dl=(lamda)/(2piepsi_(0))ln(r_(0))/(r)`
Since, `int_(r_(0))^(r)(lamda)/(2piepsi_(0)r)dr=(lamda)/(2piepsi_(0))int_(r_(0))^(r)dr=(lamda)/(2piepsi_(0))ln(r)/(r_(0))`
For a given V,
`ln(r)/(r_(0))=-(2piepsi_(0))/(lamda)[V(r)-V(r_(0))]`
`impliesr=r_(0)e^(-2piepsi_(0)Vr_(0)//lamda)e+2piepsi_(0)V(r)//lamda`
`r=r_(0)e^(-2piepsi_(0)[V(r)-V_((r_(0)))]//lamda`.
The equipotential surfaces are cylinders. of radius.