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Find the equation of the equipotential for an infinite cylinder of radius r_(0) carrying charge of linear density lambda. |
Answer» Solution :Take a Gaussian surface of radius R and length l  `int_(0)^(2pil)vecE.dvecS= (Q)/(in_(0))` `= (lambdal)/(in_(0))` From Gauss.s LAW `[ E_(r) S cos theta]_(0)^(2pirl) = (lambdal)/(in_(0))` `E_(r)xx2pirl=(lambdal)/(in_(0))[ theta=0 :. cos 0^(@)= 1]` `:. E_(r)= (lambda)/(2pi in_(0)r)` The radius of infinite cylinder is `r_(0)` `V(r)- V(r_(0))=-int_(r_(0))^(r)Edl` `= -(lambda)/(2piin_(0))"log"_(E)(r)/(r_(0))=(lambda)/(2piin_(0))="log"_(e)(r_(0))/(r)` because `int_(r_(0))^(r)(lambda)/(2piin_(0))dr= (lambda)/(2p in_(0))int_(r_(0))^(r)(1)/(r) dr` `V = (lambda)/(2pi in_(0))"log"_(e)(r)/(r_(0))` For given V , `"log"_(e)(r)/(r_(0))=-(2piin_(0))/(lambda)xx[V(r)-V(r_(0))]r=r_(0)e ^((2pi in_(0))/(lambda)[V(r)-V(r_(0))])` `:. r = r_(0)e^(-(2pi in_(0))/(lambda)[V(r)-V(r_(0))])` |
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