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Derivation for E = V/d? (capacitors) 
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#1
May312, 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. 


#2
May312, 03:12 PM

P: 581

Think of the definitions of electric field and electric potential, and then think of W = F d.



#3
May312, 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. 


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