Derivation for E = V/d? (capacitors)

1. May 3, 2012

plazprestige

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. May 3, 2012

olivermsun

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

3. May 3, 2012

plazprestige

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.

4. May 3, 2012

Staff: Mentor

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