Chapter 24 Problem 26

Problem

The variable capacitor

• Three conducting plates
• each of area A

(a) Is this a series or parallel arrangement?

(b) C as a function of $d_{1}$,$d_{2}$, and $A$. (Distance between the plates is much smaller than the area)

(c) The middle plate can be moved What are the minimum and maximum values of the total capacitance?

Summary

The figure shows a capacitor tapped in the middle by a conductor. The "left" plates and "right" plates of the capacitors are connected by a conductor.

(a)

The "high voltage plates" are at the same potential (the middle plate), the "low voltage plates" are at the same potential (the outer plates). Thus this is a parallel arrangement.

(b)

$C=C_{1}+C_{2}={\frac {\epsilon _{0}A}{d_{1}}}+{\frac {\epsilon _{0}A}{d_{2}}}=\epsilon _{0}A\left({\frac {1}{d_{1}}}+{\frac {1}{d_{2}}}\right)$

$C=\epsilon _{0}A\left({\frac {d_{1}+d_{2}}{d_{1}d_{2}}}\right)$

(c)

$l=d_{1}+d_{2}={\textrm {constant}}$

plugging in the result in (b)

$C=\epsilon _{0}A\left({\frac {l}{d_{1}(l-d_{1})}}\right)$

It is easy to see when $d_{1}\rightarrow 0$ capacitance goes $C\rightarrow \infty$. A real capacitor would break due to forces involved, but in our case

$C_{\textrm {max}}=\infty$

find the saddle point where ${\frac {dC}{d(d_{1})}}=0$

${\frac {dC}{d(d_{1})}}={\frac {d}{d(d_{1})}}\left[{\frac {\epsilon _{0}Al}{d_{1}l-d_{1}^{2}}}\right]=\epsilon _{0}Al{\frac {(l-2d_{1})}{(d_{1}l-d_{1}^{2})^{2}}}=0$ $\rightarrow d_{1}={\frac {1}{2}}l=d_{2}$

$C_{\text{min}}=\epsilon _{0}A\left({\frac {d_{1}+d_{2}}{d_{1}d_{2}}}\right)_{d_{1}={\frac {1}{2}}l}$ $=\epsilon _{0}A\left({\frac {l}{\left({\frac {1}{2}}l\right)\left({\frac {1}{2}}l\right)}}\right)$ $=\epsilon _{0}A{\frac {4}{l}}$

$C_{\text{min}}=\epsilon _{0}A{\frac {4}{d_{1}+d_{2}}}$