An infinite cylinder of radius `r_(o),` carrying linear charge density `lamda.` The equation of the equipotential surface for the cylinder is

Explanation

There is an infinite cylinder of radius `r_0` and having linear charge density `lamda` . Assume a Gaussian surface of radius r and length l. According of Gauss’s theorem,
`int E.ds=q/e_0`
`int E.ds =(lamda 1)/e_0 (because q=lamda 1)` (`because` Angle between E and ds is zero.)
`E int dS=E.2 pi rl= (lamda l)/e_0` [ where `(2pirl)` is area of the curved surface of the cylinder]
`E=lamda/(2pie_0r)`..........(i)
If the radius is `r_0`, to find the potential difference at distance r from the line, consider the electric field. According to the formula of potential gradient,
`V_(r)-V_(r_0)=int_(r_0)^r lamda/(2pie_0r).dr` (angle between E and dr to zero)
`=- lamda/(2pi e_0) int_(r_0)^r (dr)/r=-lamda/(2pie_0).[log_e r]_(r_0)^r=(-lamda)/(2pie_0)[log_e r-log_e r_0]=lamda/(2pi e_0).[ log_e r_0-log_e r]=lamda/(2pie_0).log r_0/r`
`log_e r_0/r=(2pir_0)/lamda[V_(r)-V_(r_0)], log_e r/r_0=-(2pie_0)/lamda[V_(r)-V_(r_0)]`
`r/r_=e^((pie_)/lamda) [V_(r)-V_(r_0)], r=r_0e^((2pie_0)/lamda) [V_(r)-V_(r_0)]`
This is the equation of required equipotential surfaces.