Derivation for E = V/d? (capacitors)

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    Capacitors Derivation
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Discussion Overview

The discussion centers around the derivation of the formula E = V/d, which relates the electric field (E) between the plates of a capacitor to the potential difference (V) and the separation distance (d) between the plates. The scope includes theoretical reasoning and mathematical derivation related to capacitors.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant notes that the electric field between parallel plate capacitors is constant, except near the edges, and seeks to understand the derivation of E = V/d.
  • Another participant suggests considering the definitions of electric field and electric potential, referencing the work-energy relationship W = F d.
  • A participant outlines the definitions of electric field (E = F/q) and electric potential (V = W/q), and attempts to derive E = V/d through substitution and reasoning about work done.
  • There is a suggestion that W/q is equal to V, leading to a connection between work and electric potential, although the reasoning remains exploratory and not fully resolved.

Areas of Agreement / Disagreement

Participants are engaged in an exploratory discussion with no consensus reached on the derivation process. Multiple viewpoints and reasoning approaches are presented without resolution.

Contextual Notes

The discussion includes assumptions about the definitions of electric field and potential, and the relationship between work and energy, which may not be universally agreed upon or fully detailed.

plazprestige
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One of the formulas I came across while doing problems with simple parallel plate capacitors was E = V/d, where E is the magnitude of the electric field between the plates, V is the potential difference between the plates, and d is the separation of the plates. I'm wondering where this formula is derived from.

I know that the electric field between the two plates of a capacitor is constant (except near the edges), but am not sure how that would play into the explanation.
 
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Think of the definitions of electric field and electric potential, and then think of W = F d.
 
well the definition of an electric field is F/q, where q is in the field, and the definition of electric potential is electrical potential energy divided by charge.

E = F/q
V = E/q
W = Fd

So by substitution, W = Eqd

I want to get to E = V/d so I'll solve for E...

E = W/qd

So W/q is somehow equal to V? So W/q = E/q

And by the work energy theorem, W = delta E, and the voltage in E = V/d is in fact a potential difference.

Thanks! I literally reasoned that out while typing. Thanks for pointing me in the right direction.
 
plazprestige said:
well the definition of an electric field is F/q, where q is in the field, and the definition of electric potential is electrical potential energy divided by charge.

[...snippety snip...]

So W/q is somehow equal to V?

Like you said at the beginning of your post... :biggrin:
 

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