# Deriving value of capacitance using E = V/d

1. Apr 26, 2009

### dE_logics

This formula (E = V/d) has been used to derive the value of capacitance...but how is this applicable?...this is for point charges and here we're dealing with plates.

Any derivation to prove this is also applicable for plates?

2. Apr 26, 2009

### tiny-tim

Hi dE_logics!

V = ∫ E dx is the equation for the potential difference between any two points.

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

E = D/ε, and D = -Q/A.

3. Apr 27, 2009

### dE_logics

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 v1 - v2

4. Apr 27, 2009

### Born2bwire

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.

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. Apr 27, 2009

### tiny-tim

Hi dE_logics!
oh I see what you mean …

yes … V = -Q/r is the Coulomb potential at distance r from a point charge Q, relative to zero potential defined at infinity …

but that's because it's ∫ E dr along (any) path from ∞ to r (where of course E = Q/r2).

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 then you use V = ∫ E dr

6. Apr 27, 2009

### cabraham

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

Last edited: Apr 27, 2009
7. Apr 28, 2009

### dE_logics

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.

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.

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

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.

8. Apr 28, 2009

### Born2bwire

E = V/d is only valid for point charges if you have a constant electric field, it is not a general equation. You are getting too caught up on the definition of potential. Potential difference is the work it takes to move point charge from A to B in an electric field. That is the potential difference between A and B. You are asking for the capacitance of a parallel plate capacitor. This is derived specifically using the Laplacian equation as cabraham has worked out, but this assumes infinite plates. It is still a good estimate if you have large plates if you ignore fringing because the electric field is more or less constant between the plates.

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.

9. Apr 28, 2009

### tiny-tim

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

10. Apr 28, 2009

### dE_logics

Humm...then my source is wrong!

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

11. Apr 28, 2009

### Born2bwire

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. Apr 28, 2009

### tiny-tim

From the https://www.physicsforums.com/library.php?do=view_item&itemid=101"

Two ways of defining voltage:

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

Last edited by a moderator: Apr 24, 2017
13. Apr 28, 2009

### dE_logics

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

14. Apr 28, 2009

### Staff: Mentor

The potential difference V between two points a and b, for one-dimensional motion, is defined as

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

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

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

15. Apr 30, 2009

### dE_logics

That means the 2 are the same.

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

16. Apr 30, 2009

### dE_logics

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

17. Apr 30, 2009

### Born2bwire

If "a" is infinite, then the voltage would be infinite.

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

18. Apr 30, 2009

### Staff: Mentor

19. Apr 30, 2009

### robphy

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

20. Apr 30, 2009

### dE_logics

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

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.