# Capacitance of two non parallel plates

What is the formula for capacitance of two non parallel plates at an angle with each other?If the plates were parallel then the value can be calculated as (PermittivityX area of one plate)/distance between them.But what happens in case the plates are tilted at an angle?The question came to mind while trying to understand electrostatic separator.What would be the derivation of the formula for capacitance of two non parallel plates placed at an angle?

I did get a method from http://www.davidpublishing.com/davidpublishing/upfile/12/15/2011/2011121573197833.pdf Equation 6 from the above link above helps but it is independent of the length of the plates which doesn't seem likely.

• Equation 6 in the paper you linked says $$C = \frac{\varepsilon_0 \varepsilon_1}{\alpha} \ln\left[1 + \frac{l}{d} \right]$$, which quite clearly is dependent on the length of the plates $l$ in the $\ln(1 + l/d)$ term. – Pranav Hosangadi Nov 23 '14 at 6:22
• Dear @Pranav.Pls go through the derivation."l"has been assumed to be the width and not the length.At the most this is for a unit length.I am looking for a practical formula.Would you be knowing whether equation 6 is the one universally accepted? – Chappy Nov 23 '14 at 7:00
• I think the equation gives the capacitance per unit width (into the paper). Dimensional analysis seems to validate my point. – Pranav Hosangadi Nov 23 '14 at 8:03
• Length and width would be interchangeaable depending on how one orients the plates so I guess it won't matter. – Chappy Nov 23 '14 at 9:29
• 13.02.2018: Link now dead. – Qmechanic Feb 13 '18 at 18:28

Let's do some calculus.

Suppose you have two plates, almost parallel (off by an angle $\alpha$). The plates lie in the XY plane, from $(0, 0)$ to $(x_1, y_1)$. At $x = 0$, the plates are separated by a distance $z_0$, and at $x = x_1$, the plates are separated by a distance $z_1$.

We'll now consider an infinitesimally small element of both plates. (Since parallel capacitances add, and all the infinitesimal pairs are in a parallel configuration, we can use integration)

\begin{align} \tan \alpha &= \frac{z_1 - z_0}{x_1} \\ \mathrm{d}C &= \varepsilon \frac{\mathrm{d}A}{\delta z} \\ \mathrm{d}A &= y_1 ~\mathrm{d}x \\ \delta z &= z_0 + x \tan \alpha \\ \therefore C &= \int\mathrm{d}C \\ &= \int\limits_A \varepsilon \frac{\mathrm{d}A}{\delta z} \\ &= \int\limits_0^{x_1} \varepsilon \frac{y_1 ~\mathrm{d}x}{z_0 + x \tan \alpha} \\ &= \varepsilon ~ y \left[ \cot \alpha \ln(z_0 \cos \alpha + x \sin \alpha) \right]_0^{x_1} \\ &= \varepsilon ~ y_1 \left(\frac{\ln(z_0 \cos \alpha + x_1 \sin \alpha)}{\tan \alpha} - \frac{\ln(z_0 \cos \alpha)}{\tan \alpha} \right) \\ &= \varepsilon ~ y_1 \left( \frac{\ln(1 + (x_1 / z_0) \tan \alpha)}{\tan \alpha} \right) \\ &= \frac{\varepsilon ~ y_1}{\tan \alpha} \ln \left( 1 + \frac{x_1}{z_0} \frac{z_1 - z_0}{x_1} \right) \\ &= \frac{\varepsilon ~ y_1}{\tan \alpha} \ln \left( \frac{z_1}{z_0}\right) \end{align}

If you assume $\alpha$ is small, then $\tan \alpha \approx \alpha$, which gives \begin{align} C &= \frac{\varepsilon ~ y_1}{\alpha} \ln \left(1 + \frac{x_1}{z_0} \right) \end{align}

This conclusion is the same as the Eq. 6 in the paper you linked.

• Why should $\delta C = \varepsilon \delta A/\delta z$? This formula only holds if $\delta z$ is small compared to the linear dimension of the plate $\delta A$, which is clearly not the case here. – Norbert Schuch Jan 10 '16 at 9:20

Assuming that the charge distribution is constant, using the knowledge that capacitance is added in parallel, you could treat your angled plate as being comprised of infinitely many parallel plates, approximating the angle of the plate you would like. You would then be able to integrate across this infinity of plates to find your answer.

As I said, this assumes charge distribution to be even across the plate, and so does not account for fringes. We can say that the charge will be even with a fair amount of certainty due to the knowledge that electric charges in conductors will position themselves to produce equilibrium.