91Ó°ÊÓ

We have given that A typical cell has a layer of negative charge on the inner surface of the cell wall and a layer of positive charge on the outside surface, thus making the cell wall a capacitor.

We need to find the the capacitance of a $50mm$-diameter cell with a $7.0nm$-thick cell wall whose dielectric constant is $9.0$.

We have a parallel plate capacitor, that has a surface of a sphere with a given diameter $D$, given the separation of $d$, and given dielectric constant $∈$. The capacitance of this capacitor, assuming it will be in parallel-plate mode,

localid="1648557227051" $C=∈{∈}_0\frac{A}{d}...\left(1\right)$,

where,

$A=4{\mathrm{Ï€¸é}}^2={\mathrm{Ï€\P Ù}}^2$

is the area of the plates. Using the value of$A$in equation$\left(1\right)$

$C=∈{∈}_o\frac{{\mathrm{Ï€\P Ù}}^2}{d}$

Using the given numerical values

$C=9·85·{10}^{-12}·\frac{\mathrm{π}·{\left(5·{10}^{-5}\right)}^2}{7·{10}^{-9}}=89pF$