How do you derive this (E=v/d)?
Apr 26, 2009 #2 Feldoh 1,341 3 From the relation that electric potential is the negative of the line integral of the electric field... That is when the electric field and path are in the same direction. http://upload.wikimedia.org/math/6/a/1/6a1a8acf2fe85e80aa19a9885a205a69.png Sorry latex isn't working
From the relation that electric potential is the negative of the line integral of the electric field... That is when the electric field and path are in the same direction. http://upload.wikimedia.org/math/6/a/1/6a1a8acf2fe85e80aa19a9885a205a69.png Sorry latex isn't working
Apr 26, 2009 #3 dE_logics 741 0 aaaaa that sorta went over my head. Can you do it with this relation - V = k(q/d) i.e potential difference at a distance d from a charged particle q of a unit charge brought from infinity.
aaaaa that sorta went over my head. Can you do it with this relation - V = k(q/d) i.e potential difference at a distance d from a charged particle q of a unit charge brought from infinity.
Apr 26, 2009 #5 dE_logics 741 0 Yep got that -- In V = k(q/d) substitute q with (Ed2)/k (derived from E.F at a point from a source charge q) Solve and you get it.
Yep got that -- In V = k(q/d) substitute q with (Ed2)/k (derived from E.F at a point from a source charge q) Solve and you get it.