# Why can rolled up capacitors be modelled as flat parallel plates?

If you have two parallel plate of opposite charge with a dielectric in between, then

Capacitance, $C = \frac{A \epsilon_{0}}{d} K$, where $K$ is the dielectric constant.

This is used to calculate the capacitance of a capacitor. However, a real capacitor is actually two of those plates rolled up many times. Wouldn't the electric field $\vec{E}$ be different?

So why is the equation for the capacitance still valid?

You can still use this formula if the parallel plates are rolled enough times so that the curvature of the plates is minimal. Therefore, this formula still holds. Remember that this formula, even in the case of real parallel plates, requires that $A >> d$. So, it is not exact. There is also the boundary problem: close to the boundary of the capacitor $\vec{E}$ is not perpendicular to the plates.

All this to say that this equation is an approximation. How good of an approximation it is? That's an engineering question. In the lab, $C$ depends on the $\frac{A}{d}$ ratio, the magnitude of V, and yes, on the curvature of the plates, in case they are rolled up.

• If you roll it up enough times, wouldn't the center be very curved while the outer parts less curved? – user52874 Jul 15 '14 at 3:34
• That's true. I would assume that when manufacturing these, they don't roll it up all the way to the center. Maybe an electronic engineer could clarify this. – Physico Jul 16 '14 at 21:08

I think the capacitance of a rolled up capacitor is double to that of a plate capacitor of equal surface area - this is based on the fact that once rolled up, each layer has two surfaces facing an oppositely charged surface (one that is above and one below). Thus there would be + - - + + - - + + etc. layers.

Don't know if this helped but worth my five cents...

The answer to the question is that they don't use that formula for calculating the capacitance of the rolled up capacitors. The definition of the capacitance of a conductor is the amount of charge needed to bring a conductor to an electric potential, $V$, is proportional to the potential:$$Q = CV,$$ and the constant of proportionality is capacitance. For example, an isolated conducting sphere of radius $R$ has capacitance: $C = 4\pi R \epsilon_0$. You can do a similar calculation to show that the capacitance per unit length of an infinitely long cylinder with radius $a$ inside of a grounded cylinder of radius $b$ is: $$\frac{C}{L} = \frac{2\pi\epsilon_0}{ \ln\left(\frac{b}{a}\right)}.$$

So, a rolled up capacitor, with two commingled spirals, will have a somewhat complicated expression for the capacitance that can probably only be calculated numerically, if a prediction of a given form is desired. In practice, I would imagine that the firms responsible for making these devices have empirical measurements of capacitance for the rolled up geometries as a function of factors like the tightness of the winding, length of cylinder, etc.

Though the cylinder expression is straightforward to turn into an approximation that reduces to $C = \epsilon_0 A/d$ if you use $b=d + a$ and $d \ll a$. So, as long as the distance between the spirals is much smaller than all of:

• the width of the plates,
• the length of the spirals, and
• the radius of curvature of the spiral for the whole curve,

the parallel plate approximation should work. The extent to which these conditions do not hold lead to departures from the approximation, but not from the basic definition of capacitance, $Q = CV$.