# Why is the electric field strength ($V/d$) constant in a charging capacitor?

The electric field strength at a point in a charging capacitor $$=V/d$$, and is the force that a charge would experience at a point. This doesn't seem to make sense, as all the capacitor is is 2 plates, one positively and one negatively charged, and we have an equation to represent the electric field strength at a point between 2 charges.

This equation is $$kq1/r^2$$, and so for a point between 2 charges it would be $$kq1/r^2 + kq2/(d-r)^2$$. Using these equations the electric field at a point between 2 capacitors would look like this:

This graph clearly does not show a constant electric field, but instead one varying constantly.

I would assume that the source of the error is modelling the capacitor plates as 2 point charges, but I cant understand how whatever the difference is between the 2 scenarios would result in such a different shape field.

Edit: Considering I thought a source of error could be treating the capacitor as a single point, I again wrote an equation integrating all the forces over a plate. The plate has a radius of 100cm (area 10000$$\pi$$cm^2, and a distance between plates of 10cm.

$$f(x)=\int_0^{2\pi}\int_0^{100}{1\over r^2 +x^2}+{1\over r^2+(d-x)^2}drd\theta$$. The graph for this looks like this:

This is better than originally, but it still isn't nearly as flat as I would expect - moving from 5cm to 3cm has a difference in field strength of 5%, and from 5cm to 1cm has a difference of 20%. This feels like a very non-negligible difference, am I missing something further?

• For 2 plates you would have to add the field of all the point charges, so you have an integral over the area of the plates. Commented Mar 5 at 10:56
• Does that make a meaningful difference? I can't see how that would change it to become a constant value. Either way, I dont know how to integrate over a full area, but i've integrated over a line instead (approximating a plate as a line of charges instead of a plane of charges), and i get a very similar shape. Commented Mar 5 at 11:14
• Capacitors are not point like charges, look up Griffiths Intro to Electrodynamics or Reitz Commented Mar 5 at 11:23
• Sure, but they are just made up of many point like charges, and then when you integrate across them all (as i said in my previous comment), you still get the exact same shape graph. Commented Mar 5 at 11:28
• During charging V is changing and d is fixed. Commented Mar 5 at 11:36

## 4 Answers

Others have said this but the integral in the OP:

$$f_{\rm Wrong}(x)=\int_0^{2\pi}\int_0^{100}{1\over r^2 +x^2}+{1\over r^2+(d-x)^2}{\rm d}r{\rm d}\theta \ ,$$

is not quite correct. Firstly, the area element in 2D is $$r{\rm d}r{\rm d}\theta$$, secondly you are I suppose adding the x-components along a line joining the midpoints of two circular plates but the integrand has the magnitudes. Fixing these issues gives us:

$$E_x(x)=2\pi\int_0^R r\left[\frac{x}{(r^2 +x^2)^{3/2}}+\frac{d-x}{(r^2 +(d-x)^2)^{3/2}}\right]{\rm d}r \ .$$ note that I can define $$\xi=x/d$$ which goes from $$0$$ to $$1$$ and $$y=r/d$$ going from $$0$$ to $$\rho=R/d$$. With this notation change we have

$$E_x(\xi,\rho) = 2\pi\int_0^\rho y\left[\frac{\xi}{(y^2 +\xi^2)^{3/2}}+\frac{1-\xi}{(y^2 +(1-\xi)^2)^{3/2}}\right] { \rm d} y \ ,$$

which shows that the field has the nice property that it only depends on the aspect ratio $$\rho$$ of the plate radius to the separation, and the fraction of the separation one looks at. OPs original expression $$f_{\rm Wrong}(x)$$ lacks this property.

In fact, this function is pretty flat even for relatively small $$\rho$$!

• Thank you! This is exactly what I was looking for. I'm quite surprised that it changes so much when you treat it as many charges instead of just one average charge, but i can clearly see it does. Commented Mar 6 at 9:09

$$f(x)=\int_0^{2\pi}\int_0^{100}{1\over r^2 +x^2}+{1\over r^2+(d-x)^2}drd\theta$$

It's probably not the only problem with your analysis, but this formula is not correct. You're integrating the magnitude of the electric field from all the elements of charge on the plates. But electric fields don't add up as scalars, they add up as vectors. You should (for a position along the center-line of the structure) be integrating the x-components of the field, not the magnitude of the field.

The plate has a radius of 100cm (area 10000π cm^2, and a distance between plates of 10cm

This gives only a 10 to 1 ratio between the transverse and axial dimensions. A practical ceramic disc capacitor might have transverse dimensions of a couple millimeters, and separation between the plates measured in microns. So 100-to-1 or 1000-to-1 ratio between them.

If you want to find the result the "traditional" way, look in any electrostatics textbook in the chapter where Gauss's law is introduced. Solving this problem in the limit of infinite transverse dimensions is typically one of the examples given of the application of Gauss's law.

By "capacitor" we assume that:

1. It consists of two charged plates, one positive and one negative.
2. The distance between the two plates is small enough that fringing effects can be ignored.

By these assumptions, dealing capacitors as point charges does not hold. In most cases, assuming that capacitors have area $$A$$ and distance $$d$$ such that $$\sqrt{A} >> d$$ holds.

• I've just edited my post to show the field strength for an entire plate - it still doesnt feel like it's particularly constant. Commented Mar 5 at 15:21
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Commented Mar 5 at 16:45

Your method of integrating over a line along the radius and then over the complete disc to find the value of electric field at a point between the capacitor seems incorrect. An arbitrary point between the plates of the capacitors is not necessarily located symmetrically with respect to all the points of the capacitor plate. So considering its distance to be $$\sqrt{r^{2}+x^{2}}$$ from all the points of the capacitor is inappropriate. Moreover, electric field is a vector quantity, and its apparent that it is wrongly being considered as a scalar here.

Electric field between the plates of a capacitor is calculated by applying Gauss' Law while considering the plates as infinite sheets of charges.

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Commented Mar 5 at 16:44
• I think it's clear that OPs method can be fixed with a few minor modifications, and that then gives the desired result. OP is considering a point on a line joining the centres of the two plates, they just are missing a few factors, but otherwise they are doing the right thing for a finite area plate. Commented Mar 5 at 17:52