Answer
(a) $1.2\times 10^7N/C$; directed into the cell
(b) $97mV$; the outer wall
Work Step by Step
(a) We know that
$E=\frac{\sigma}{K\epsilon_{\circ}}$
We plug in the known values to obtain:
$E=\frac{0.58\times 10^{-3}C/m^2}{(5.5)(8.85\times 10^{-12}C^2/Nm^2)}$
$E=1.2\times 10^7N/C$ directed into the cell.
(b) We know that
$V=\frac{\sigma d}{K\epsilon_{\circ}}$
We plug in the known values to obtain:
$V=\frac{(0.58\times 10^{-3}C/m^2)(8.1\times 10^{-9}m)}{(5.5)(8.85\times 10^{-12}C^2/Nm^2)}$
$V=97\times 10^{-3}V$
$V=97mV$
Thus, the outer wall has the higher potential.
