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

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Discussion Overview

The discussion centers around the derivation of the equation E = v/d, exploring the relationship between electric potential (V), electric field (E), and distance (d). Participants examine different approaches and mathematical relations relevant to this concept.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant suggests deriving E = v/d from the relation that electric potential is the negative of the line integral of the electric field, particularly when the electric field and path are aligned.
  • Another participant proposes using the relation V = k(q/d) to derive the equation, indicating that this represents the potential difference at a distance d from a charged particle q.
  • A later reply introduces a substitution of q with (Ed²)/k, derived from the electric field at a point from a source charge, to solve for the equation.
  • There are mentions of technical issues with LaTeX formatting, which may affect the clarity of the mathematical expressions presented.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on a single derivation method, as participants propose different approaches and express varying levels of understanding regarding the mathematical relationships involved.

Contextual Notes

Participants note limitations with LaTeX formatting, which may hinder the presentation of mathematical expressions. There is also a reliance on specific definitions and assumptions regarding electric potential and electric fields that are not fully explored.

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How do you derive this (E=v/d)?
 
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aaaaa :biggrin: 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.
 


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


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.
 

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