Derivation for E = V/d? (capacitors)


by plazprestige
Tags: capacitors, derivation, difference, potential, separation
plazprestige
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#1
May3-12, 03:03 PM
P: 21
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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olivermsun
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May3-12, 03:12 PM
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Think of the definitions of electric field and electric potential, and then think of W = F d.
plazprestige
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#3
May3-12, 03:28 PM
P: 21
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.

jtbell
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#4
May3-12, 03:46 PM
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Derivation for E = V/d? (capacitors)


Quote Quote by plazprestige View Post
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...


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