Chapter 24 Problem 18

Problem

A large metal sheet of thickness $l$is placed between, and parallel to, the plates of the parallel-plate capacitor. It does not touch the plates, and extends beyond their edges.

(a) What is now the net capacitance in terms of $A$, $d$,and $l$?

(b) If $l=0.40d$ by what factor does the capacitance change when the sheet is inserted?

Solution

(a)

The negative charge in the inserted plate will be drawn towards the positive capacitor plate, leaving positive charge behind. Using Gauss's law, you can easily prove the same charge will be on each face of the plate as on the original capacitor plates. Similarly, the same electric field will be in the gaps as before the plate was inserted. Use that electric field to determine the potential difference between the two original plates, and the new capacitance. The parameters are indicated in the figure.

$E={\frac {\sigma }{\epsilon _{0}}}={\frac {Q}{A\epsilon _{0}}}$; $V_{1}=Ex={\frac {Qx}{A\epsilon _{0}}}$; $V_{2}=E(d-l-x)={\frac {Q(d-l-x)}{A\epsilon _{0}}}$

$\Delta V=V_{1}+V_{2}={\frac {Qx}{A\epsilon _{0}}}+{\frac {Q(d-l-x)}{A\epsilon _{0}}}$ $={\frac {Q(d-l)}{A\epsilon _{0}}}={\frac {Q}{C}}$

$\rightarrow C=\epsilon _{0}{\frac {A}{d-l}}$

(b)

$C_{\text{initial}}=\epsilon _{0}{\frac {A}{d}};$ $C_{\text{final}}=\epsilon _{0}{\frac {A}{d-l}};$ ${\frac {C_{\text{final}}}{C_{\text{initial}}}}$ $={\frac {\epsilon _{0}{\frac {A}{d-l}}}{\epsilon _{0}{\frac {A}{d}}}}$ $={\frac {d}{d-l}}$ $={\frac {d}{d-0.40d}}$ $={\frac {1}{0.60}}$ $=1.7$