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Any derivation to prove this is also applicable for plates?

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- Thread starter dE_logics
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Any derivation to prove this is also applicable for plates?

- #2

tiny-tim

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Any derivation to prove this is also applicable for plates?

Hi dE_logics!

V = ∫ E dx is the equation for the potential difference between

The plates (in a capacitor ) are only relevant for calculating E …

E = D/ε, and D = -Q/A.

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V = ∫ E dx is the equation for the potential difference betweenanytwo points.

V here is the potential of a coulomb charge brought from infinity to this point, d units from the charge source.

How can this be the potential difference?...it should be v

- #4

Born2bwire

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I'm not sure where you got E=V/d is for point charges or that V is only the potential brought from infinity. We are asking for the potential between the two plates, this inherently is asking for the potential difference. Referencing to bringing in from infinity is just another way to do potential difference by pinning infinity as your reference. Here, the reference is one of the plates.

- #5

tiny-tim

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V here is the potential of a coulomb charge brought from infinity to this point, d units from the charge source.

How can this be the potential difference?...it should be v_{1}- v_{2}

oh I see what you mean …

yes … V = -Q/r

but that's because it's ∫ E dr along (any) path from ∞ to r (where of course E = Q/r

But the E field there is not the same as the E field for a capacitor …

that comes from E = D/ε, and D = -Q/A …

and

- #6

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I worked this out, but I don't know where I stashed the sheet. I'll try to post it later tonight. The E = V/d is computed

by solving the Laplacian equation in 1 dimension, (del^2)V = 0 = Vxx + Vyy+ Vzz, where Vxx = (d^2)V/d(x^2), etc. In

1 dimension, x only, we have d2v/dx2 = 0. Integrating we obtain

dV/dx = c1. Integrating again we get V(x) = c1*x + c2. The boundary conditions are V(0) = 0 & V(d) = V1, so that V(0)

= 0 = c1*0 + c2 + c2, so that c2 = 0. Next V(d) = V1 = c1*d, which results in c1 = V1/d. Hence V(x) = (V1/d)*x.

Then for static charges we know that E = -grad V = -dV/dx. But dV/dx = -V1/d.

Hence |E| = V1/d. V1 = voltage on 1 plate with 0V on the other.

I'll attach a sheet later tonight.

Claude

by solving the Laplacian equation in 1 dimension, (del^2)V = 0 = Vxx + Vyy+ Vzz, where Vxx = (d^2)V/d(x^2), etc. In

1 dimension, x only, we have d2v/dx2 = 0. Integrating we obtain

dV/dx = c1. Integrating again we get V(x) = c1*x + c2. The boundary conditions are V(0) = 0 & V(d) = V1, so that V(0)

= 0 = c1*0 + c2 + c2, so that c2 = 0. Next V(d) = V1 = c1*d, which results in c1 = V1/d. Hence V(x) = (V1/d)*x.

Then for static charges we know that E = -grad V = -dV/dx. But dV/dx = -V1/d.

Hence |E| = V1/d. V1 = voltage on 1 plate with 0V on the other.

I'll attach a sheet later tonight.

Claude

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- #7

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Born2bwire said:E = V/d is only valid is we have a constant potential/electric field. With a parallel plates, ignoring fringing effects, the electric field is approximately the same, hence the integral Edx is the same as Ed.

That is the question, how do you know this?

The formula can be derived from v = kQ/r...where k is the dielectric stuff; this formula is applicable for a point charge, or a changing field, considering that E=V/d might not be applicable for a uniform field...I don't know...that's the question.

I'm not sure where you got E=V/d is for point charges or that V is only the potential brought from infinity.

Again considering the formula can be derived from v = kQ/r...here v is the potential of a unit charge brought from infinity to a distance r form a charged named/having an intensity Q.

We are asking for the potential between the two plates

Yeah...exactly how can you state that v will be the potential difference?

Referencing to bringing in from infinity is just another way to do potential difference by pinning infinity as your reference.

Taking infinity as a reference is something absolute, infact potential difference at 2 points in a field for a certain charge is defined as the difference in P.E stored between the 2 or the difference in the work done to bring each of the charges to different positions from infinity; i.e potential is more fundamental than P.D.

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Born2bwire

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But as for the intermediate step that E=V/d, V is defined as the potential difference between the plates. V is always potential difference and always requires a reference point. Sometimes we take infinite as the reference and use potential but that is generally chosen when we do not have a reference of consequence to use. When it comes to potential difference, it doesn't matter as the path for the potential from infinity can be made the same up until a desired point in space.

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tiny-tim

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E = V/d is only valid for point charges if you have a constant electric field, it is not a general equation.

Yup … if E isn't constant, you have to write V = ∫ E d(d)

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Any links to derivation of V = ∫ E d(d) ?...and derivation of capacitance? (I'm searching)

- #11

Born2bwire

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Any links to derivation of V = ∫ E d(d) ?...and derivation of capacitance? (I'm searching)

The simplest derivation would come from the Lorentz Force. In an electrostatic case, F = qE. So the work to move a charge 'q' would be the integral of the force over the distance. Hence, V = \int E \cdot d\ell . Where E and d\ell are vectors. The fact that for the parallel plate capacitor V = E/d is a happy coincidence. But if you wanted a more exact answer you would solve the 3D Laplacian (or is it Poisson's... screw it) for a finite parallel plate capacitor.

- #12

tiny-tim

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Any links to derivation of V = ∫ E d(d) ?

voltage = energy/charge = work/charge = force"dot"distance/charge = (from the Lorentz force) electric field"dot"distance, or dV = E.dr

but also voltage = energy/charge = (energy/time)/(charge/time) = power/current, or V = W/I

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Just confirm me one thing first, the expression ∫ E dr and Ed (= V) are different?

- #14

jtbell

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[tex]V = - \int_a^b {E(x) dx}[/tex]

If E does not depend on x, then you can pull it out of the integral:

[tex]V = - E \int_a^b {dx} = - E (b - a) = - E \Delta x[/tex]

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That means the 2 are the same.

If a is infinite we get V = Ed...right?

If a is infinite we get V = Ed...right?

- #16

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Any links to derive capacitance of a parallel plate air capacitor?...pls?

- #17

Born2bwire

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cabraham has the derivation, but I'm pretty sure it is derived in Griffith's text as well.

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jtbell

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That means the 2 are the same.

If a is infinite we get V = Ed...right?

d is the separation of the plates (from x=a to x=b)

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V = - E \int_a^b {dx} = - E (b - a) = - E \Delta x

[/tex]

Wait...you cant treat E(x) as a constant, it appears wrong to me.

Anyway...it does give out the same thing by any alternative technique.

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cabraham has the derivation, but I'm pretty sure it is derived in Griffith's text as well.

That is not so...check out the attached file.

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Thanks!...I'll see if I generate any problems.

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d is the separation of the plates (from x=a to x=b)

No, apparently it seems that formula set does not seem to work for a capacitor.

- #24

jtbell

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Wait...you cant treat E(x) as a constant, it appears wrong to me.

For the ideal parallel-plate capacitor, you can

http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/gaulaw.html#c1

applying it to a plane sheet of charge:

http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elesht.html#c1

and considering a parallel-plate capacitor as consisting of two oppositely-charged sheets.

http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elesht.html#c2

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For the ideal parallel-plate capacitor, you canderivethat E is uniform between the plates, by using Gauss's Law:

http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/gaulaw.html#c1

applying it to a plane sheet of charge:

http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elesht.html#c1

and considering a parallel-plate capacitor as consisting of two oppositely-charged sheets.

http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elesht.html#c2

That's ok, but my question is how can you take E(x) as a constant? It's a function of x, and you're differentiating WRT x...moreover this doesn't have to do with the derivation of capacitance...naively these set of equations are for point charges only.

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