How do you derive this (E=v/d)?

[dE_logics]

Main Question or Discussion Point

How do you derive this (E=v/d)?

 

[Feldoh]

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

 

[dE_logics]

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.

 

[dE_logics]

There's something wrong with latex...its showing my old codes.

 

[dE_logics]

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.