91Ó°ÊÓ

We have given that Two $2.0cm*2.0cm$cm metal electrodes are spaced $1.0mm$apart and connected by wires to the terminals of a $9.0V$battery.

step 1 : part a:

The potential diffrence is that supplied by the battery, $9V$.

the charge will be found as

$q=Câ–³V$

Find the capacitance. Since the capacitor is a parallel plate, and the plates are squares with sides of length L. the capacitance will be given by

$C={\mathrm{Î\mu }}_0\frac{L^2}{d}$.

The charge will be

$q=\frac{{\mathrm{Î\mu }}_0{L}^2â–³V}{d}$

$q=\frac{8.85·{10}^{-12}·0.{02}^2·9}{0.001}=32pC$

The battery was disconnected, the charge will remain the same $32pC$.

The potential difference, through the definition of the capacitance, can be found to be

$â–³V=\frac{q}{C}$

The capacitance will be half of the previous one, since the separation was doubled.

$C={\mathrm{Î\mu }}_0\frac{L^2}{2d}=\frac{C_1}{2}$

The potential difference after the increase in separation to be

$â–³V_2=\frac{q}{C_2}=\frac{C_1â–³V_1}{C_2}=2â–³V_1=12V$