dE_logics
- 742
- 0
How do you derive this (E=v/d)?
How do you derive this (E=v/d)?
- Context: High School
- Thread starter dE_logics
- Start date
-
- Tags
- Derive
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 10K views
Feldoh
- 1,336
- 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
dE_logics
- 742
- 0
aaaaa
that sort of 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
- 742
- 0
There's something wrong with latex...its showing my old codes.
dE_logics
- 742
- 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.